In prelims, define \(\int_a^b f(t)\mathrm{d}t\) when \(f:[a,b]→ℝ\) is Riemann-integrable

Theorem. Suppose \((f_n)_{n=1}^{∞}\) are Riemann-integrable on \([a,b]\) and \(f_n→f\) uniformly on \([a,b]\). Then \(f\) is Riemann-integrable and \(\int_a^b f(t)\mathrm{d}t=\lim_{n→∞}\int_a^b f_n(t)\mathrm{d}t\)

We want theorems of the form

Hypotheses will be needed.

\[f_n=\left\{\begin{array}{ll}n^2x&0⩽x⩽\frac{1}{n}\\n^2(1-x)&\frac{1}{n}⩽x⩽\frac{2}{n}\end{array}\right.\]

Comparing the following:

Theorem. (Bounded convergence theorem §5) Let \((f_n)_{n=1}^{∞}\) be Riemann/Lebesgue integrable on \([a,b]\). Let \(f_n→f\) pointwise on \([a,b]\). Let \(C > 0\) such that \({|f_n(t)|}≤ C\) for all \(n∈ℕ\) and \(t∈[a,b]\).

If \(f\) is Riemann-integrable then \(f\) is Lebesgue integrable and \(\int_a^b f(t)\mathrm{d}t=\lim_{n→∞}\int_a^b f_n(t)\mathrm{d}t\)

Aiming for 2 powerful convergence theorems

• Monotone convergence theorem (MCT) – §4

• Dominated convergence theorem (DCT) – §5

Plan

Part 1. Set up theory and reach MCT

Part 2. Application come quickly

Recommended books

M. Capinski and E. Kopp, Measure, Integration and Probability – easy

E. M. Stein & R. Shakarchi, Real Analysis – not easy

1Extended real numbers

Work with the extended reals \([-∞,∞]=\{-∞,∞\}∪ℝ\) and define

\begin{eqnarray*}x+∞ &=& ∞+x \eqcirc ∞\\ x-∞ &=&-∞+x=-∞\\ x. ∞ &=& ∞ .x=(-x) . (-∞)=\left\{\begin{array}{ll}∞ & (x > 0)\\-∞ & (x < 0)\\ 0 & (x=0) \end{array}\right.\end{eqnarray*}

Do not define \(∞-∞\)

Why define \(0. ∞=0\) here? Consider the area of \(x\) axis.

Warning. Multiplication is not continuous at \((∞,0)\).Be careful with continuity

\(x_n=n^2→∞\)

\(y_n=\frac{1}{n}→0\)

Any subset \(E⊆[-∞,∞]\) has a sup & inf.

\(\sup ∅=-∞\)

If \(E⊆ℝ\) is bounded then \(\sup E=∞\).

\((a_n)\) is an increasing sequence then \(\lim a_n=\sup\{a_n:n∈ℕ\}\)

Proposition 1.1. Let \((a_n)_{n=1}^{∞}\) be a sequence of non-negative terms, then

\(\displaystyle\sum_{n=1}^{∞}a_n=\sup\left\{\sum_{n∈J}a_n:J⊂ℕ\text{ is finite}\right\}\)

Definition. Let \((a_n)\) be a sequence in \([-∞,∞]\)

\begin{align*}\limsup_{n→∞}a_n &=\lim_{m→∞}\sup_{n⩾m}a_n\\ \liminf_{n→∞}a_n &=\lim_{m→∞}\inf_{n⩾m}a_n\end{align*}

These exist as \((\sup_{n⩾m}a_n)_{m=1}^{∞}\) is decreasing.

Example. \(\limsup(-1)^n=1,\liminf(-1)^n=-1\)

Example 1.2. \(\limsup\left(1+\frac{1}{n}\right)\sin n=1,\liminf\left(1+\frac{1}{n}\right)\sin n=-1\)

Example. \(a_n=\left\{\begin{array}{ll}1+2^{-n}&n\text{ prime}\\0&\text{otherwise}\end{array}\right.\)

\(\displaystyle\limsup a_n=1,\liminf a_n=0\)

Proposition 1.3.

limsup and liminf are useful for avoiding epsilontics. For example, consider the Sandwich Rule, i.e., suppose that \(a_n≤b_n≤c_n\) for all \(n\) and \(\lim a_n=\lim c_n\). Then

\begin{align*}\limsup b_n &≤\limsup c_n& \text{(Proposition 1.3(4))}\\ &=\lim c_n &(\text{Proposition}1.3 (3))\\ &=\lim a_n& \text{(assumption)}\\ &=\liminf a_n &(\text{Proposition 1.3(3))}\\ &≤ \liminf b_n& \text{(Proposition 1.3(4))}\\ &≤ \limsup b_n& \text{(Proposition 1.3(2)).}\end{align*}

Hence equality holds throughout, so \(\lim b_n=\lim a_n\), by Proposition 1.3(3).

2 Lebesgue measure

Two differences

Riemann: divide \(x\)-axis & use upper, lower sums

Lebesgue: divide \(y\)-axis & estimate for area \(\sum_{i=0}^k y_i \operatorname{length}(X_i)\)

split any function into its positive and negative parts, and integrate these separately.

Wishlist for a notion of length

should be a function \(m:\mathcal{P}(ℝ) \rightarrow [0, ∞]\) satisfying

1. \(m (∅)=m (\{x \})=0\)
2. \(m (I)=b - a\) if \(I\) is an interval with endpoints \(a, b\), where \(a<b\)
3. \(m (A + x)=m (A)\) for \(A⊆ℝ, x∈ℝ\)[\(m\) is invariant under translation]
4. \(m (\alpha A)=| \alpha | m (A)\) for \(\alpha∈ℝ, A⊆ℝ\)
5. \(m (A) ≤ m (B)\) if \(A⊆B\) [\(m\) is monotone]
6. \(m \left( ⋃_{n=1}^∞A_n \right)=\sum_{n=1}^∞m (A_n)\) where \(A_1, A_2, …\) are pairwise disjoint [\(m\) is countably additive]
   too optimistic, no such function exists. Weaker: finite additivity
   (Banach-Tarski) no finite additive volume on \(ℝ^3\)

Some redundancy: additivity + nonnegativity ⇒ monotonicity

Outer measurable 

Suppose temporarily there is such a function and \(A⊆⋃_{n=1}^∞I_n\) for intervals \(I_n\)

Write \(I_n'=I_n ∖ (I_1∪…∪I_{n - 1})\) for \(n ≥ 2\) , then \(I_n'\) are pairwise disjoint

We would have \(m (A) ≤ m \left( ⋃_{n=1}^∞I_n \right)=\)\(m \left( ⋃_{n=1}^∞I_n' \right)=\sum_{n=1}^∞m(I_n') ≤ \sum_{n=1}^∞m (I_n)\)

Definition. For a bounded interval \(I\) with endpoints with \(b > a\), write \(| I |=b - a\). For an unbounded interval \(I\), \(| I |=∞\).

For \(A⊆ℝ\) define Lebesgue outer measure of \(A\),

\[m^*(A)=\inf \left\{\sum_{n=1}^∞| I_n |:I_n \text{ intervals and }A⊆⋃_{n=1}^∞I_n \right\}\]

It matters that we use countably many intervals – finitely many intervals give rise to the upper Jordan content – See [S&S] Chapter 1 Exercise 14

Proposition 1.

• \(m^*\) satisfies conditions (i)-(v)
• \(m^*\) is countably subadditive \(m \left( ⋃_{n=1}^∞A_n \right) ≤ \sum_{n=1}^∞m (A_n)\) for sets \(A_1, A_2, …∈ℝ\)

Proof. (ii) Given that the result is false for ℚ, the proof must use completeness of \(ℝ\).

For \(I=[a, b]\).

Certainly \(m^*(I) ≤ b - a\) as \(I⊆I∪∅∪∅∪…\)

Suppose \([a, b]\) is covered by bounded intervals \(I_n\) with endpoints \(a_n<b_n\)

Fix \(ε > 0\) and set \(J_n=(a_n - ε 2^{- n}, b_n + ε 2^{- n})\)

By Heine-Borel Theorem, \([a, b]\) is compact, \(J_n\) is open, so \(∃N\) such that \([a, b]∈⋃_{n=1}^N J_n\)

Order endpoints \(\{c_n, d_n:n∈ℕ\}\) as

\[x_1<x_2<⋯<x_k\]

So \(x_1<a<b<x_k\). So \(b - a<\sum_{i=1}^{n - 1}(x_{i + 1}- x_i)\)

Each \((x_i, x_{i + 1})\) is contained in some \(J_n=(c_n, d_n)\) with \(c_n=x_{l_n}, d_n=x_{m_n}\)

\[b - a<\sum_{i=1}^{n - 1}(x_{i + 1}- x_i) ≤ \sum_{n=1}^N\sum_{i=l_n}^{m_n - 1}(x_{i + 1}- x_i)=\sum_{n=1}^N | J_n |\]

Finally\[\sum_{n=1}^∞| I_n | ≥ \sum_{n=1}^N | I_n |=\sum_{n=1}^N (| J_n | - 2 ε 2^{- n}) > b - a - 2 ε\]

As \(ε\) is arbitrary, \(b - a ≤ m^*(I)\). ◻

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Definition 2.1. \(E⊆ℝ\) is a null set if \(m^*(E)=0\)

Corollary 2.2.

1. subsets of null sets are null

2. countable union of null sets are null

3. countable sets are null

Proof. (2)Let \((E_n)_{n=1}^∞\) be null, fix \(ε > 0\).

For each \(n\) find \((I_{n,m})_{m=1}^∞\) such that \(E_n ⊆\bigcup_{m=1}^∞ I_{n,m}\) and \(\sum_{m=1}^∞|I_{n,m}|<ε .2^{-n}\).

Now \(\{I_{m,n}:m,n=1,2,…\}\) is a countable family of intervals covering \(\bigcup E_n\), and \(\sum_{n=1}^∞\sum_{m=1}^∞|I_{n,m}|<\sum_{n=1}^∞ ε .2^{-n}=ε\). Hence \(m^*\left(\bigcup E_n\right)=0\). ◻

Example 2.3. (Cantor's middle third set)null sets can be uncountable.

\[C_0=[0,1],C_1=\left[0,\frac{1}{3}\right]∪\left[\frac{2}{3},1\right],C_2=\left[0,\frac{1}{9}\right]∪\left[\frac{2}{9},\frac{1}{3}\right]∪\left[\frac{2}{3},\frac{7}{9}\right]∪\left[\frac{8}{9},1\right],…\]

Let \(C=\bigcap_{n=1}^∞ C_n\). Then \(C\) is nonempty by finite intersection property of compact sets.

\(C\) is uncountable by Cantor's diagonal argument.

\(C=\left\{x∈[0,1]:x\text{ has an expansion }x=\sum_{n=1}^∞ a_n 3^{-n},a_n∈\{0,2\}\right\}\)

As each \(C_n\) has \(2^n\) intervals of length \(3^{-n}\), \(m^*(C)≤\left(\frac{2}{3}\right)^n→0\), so \(m^*(C)=0\).

Say a property \(Q\) of the real numbers holds almost everywhere (a.e.) if \(\{x:Q\text{ does not hold at }x\}\) is null. E.g. \(χ_C=0\) (a.e.) as \(C\) is null.

Example 2.4. (Vitali set)Consider the partition of \(ℝ\) into its cosets for \(ℚ\), \((x+ℚ)_{x∈ℝ}\)

Let \(A ⊆[0,1]\) consists of exactly one element from each coset.

1. \(x,y∈A,x≠y⇒x-y∉ℚ\)

2. \(∀x∈[0,1],∃q∈ℚ,\) s.t. \(x+q∈A\)

Now \([0,1]⊆\bigcup_{q∈ℚ∩[0,1]}(A-q)⊆[-1,2]\)

By (i), \((A-q)_{q∈ℚ∩[0,1]}\) are pairwise disjoint.

If \(m^*\) was countably additive,

\[1=m^*[0,1]≤\sum_{q∈ℚ∩[-1,1]}m^*(A-q)=\sum_{q∈ℚ∩[-1,1]}m^*(A)≤3\]

contradiction as \(1≤∞⋅m^*(A)≤3\). What is \(m^*(A)\)?

Definition. A set \(E ⊆ℝ\) is Lebesgue measurable if

\[m^*(A)=m^*(A∩E)+m^*(A ∖E)\]

for all subsets \(A\) of \(ℝ\). Write \(ℳ_\text{Leb}\) for the collection of Lebesgue measurable sets.

Note \(m^*(A)≤m^*(A∩E)+m^*(A ∖E)\) always, so to prove \(E\) is Lebesgue measurable, it suffices to show the reverse inequality

\[m^*(A∩E)+m^*(A∖E)≤m^*(A) \text{for all }A\]

Proposition 2.5. See Exercise 9, [Capinski&Kopp], [Stein&Shakarachi Chapter 6.1]

1. if \(E\) is null then \(E∈ℳ_\text{Leb}\)

2. if \(I\) is an interval, then \(I∈ℳ_\text{Leb}\)

3. if \(E∈ℳ_\text{Leb}\), then \(ℝ∖E∈ℳ_\text{Leb}\)

4. if \((E_n)_{n=1}^∞\) are Lebesgue measurable, \(\bigcup_{n=1}^∞ E_n∈ℳ_\text{Leb}\).

5. if \((E_n)_{n=1}^∞\) are pairwise disjoint, then \(m^*\left(\bigcup_{n=1}^∞ E_n\right)=\sum_{n=1}^∞ m^*(E_n)\)

Write \(m(E)=m^*(E)\) when \(E∈ℳ_\text{Leb}\) and \(m:ℳ_\text{Leb}→[0,∞]\) is the Lebesgue measure on \(ℝ\)–it is countably additive.

Note \(ℳ_\text{Leb}\) is closed under countable intersections because \(\bigcap_{n=1}^∞ E_n=ℝ ∖\left(\bigcup_{n=1}^∞(ℝ∖E_n)\right)\)

Corollary 2.6. As every open subsets of \(ℝ\) is a countable union of open intervals, open sets are Lebesgue measurable.

Corollary 2.7. If \(E∈ℳ_\text{Leb}\), then for \(ε > 0\), ∃open sets \(U ⊆ℝ\), such that \(E ⊆U\) and \(m(U∖E)<ε\).

If \(E∈ℳ_\text{Leb}\), then ∃\(G\) a countable intersection of open sets such that \(G⊇E\) and \(G∖E\) is null.

Proof. If \(m(E)<∞\), \(ε > 0\), let \(U\) be a countable union of disjoint open intevals such that \(E⊆U,m(U)<m(E)+ε\), therefore \(m(E)+ε > m(U)=m(U∩E)+m(U ∖E)=m(E)+m(U∖E)\), therefore \(m(U ∖E)<ε\).

If \(m(E)=∞\), let \(E_n=E∩[-n,n]\), find \(U_n\) open, \(U_n⊇E_n\) such that \(m(U_n∖E_n)< ε .2^{-n}\), therefore \(E ⊆\bigcup_{n=1}^∞ U_n=U\), \(m(U∖E)≤m\left(\bigcup_{n=1}^∞(U_n∖E_n)\right)≤\sum_{n=1}^∞ m(U_n ∖E_n)=ε\). ◻

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Corollary 2.7 \(E\) measurable\(⇔∀ε > 0, ∃U\) open, \(U⊇E\), \(m (U∖E)<ε\)

Proposition. \(E\) is measurable\(⇔∃G\) a \(G_δ\)-set(a countable intersection of open sets)such that \(G⊇E\) and \(m (G∖E) = 0\)

Proof. (⇒) In Corollary 2.7 take $ϵ=\frac1n$ and take intersection of $U$ over $n=1$ to ∞.
(⇐) Since \(G∖E\) is null, \(G∖E\) is measurable. Also \(G\) is measurable. So \(E = G∖(G∖E)\) is measurable.\(\Box\)

What about \(ℝ^n, n≥2\)?

Say a set of the form \(R = I_1×I_2×⋯×I_n\) is a rectangle if each \(I_i\) is an interval.

Define area of \(R\) to be \(| R | = \prod_{i = 1}^n | I_i |\)

For \(A⊆ℝ^n\), \(m^*(A) = \inf \left\{ \sum_{r = 1}^∞| R_r | : R_r \text{ is a rectangle in } ℝ^n \text{ and } A⊆\bigcup_{r = 1}^∞R_r \right\}\)

Define Lebesgue measurable sets in the same way: \(E⊆ℝ^n\) is measurable if and only if \(m^*(A) = m^*(A \cap E) + m^*(A∖E)\) and everything works essentially identically to the case of \(ℝ\).

3Measure spaces & Measurable sets

Definition. Let Ω be a set. A σ-algebra on Ω is \(ℱ⊆𝒫 (Ω) = \{ A : A⊆Ω \}\) such that

1. \(∅∈ℱ\)

2. \(E∈ℱ \Rightarrow Ω∖E∈ℱ\)

3. If \(E_1, E_2, …∈ℱ\), then \(\bigcup_{n = 1}^∞E_n∈ℱ\)

\((Ω, ℱ)\) is a measurable space, and sets in \(ℱ\) are \(ℱ\)-measurable.

Note. σ-algebras are closed under countable intersections by De Morgan's law.

Definition. A measure \(\mu\) on a measurable space \((Ω, ℱ)\) is \(μ: ℱ→[0, ∞]\) such that

• \(μ(∅) = 0\)

• If \((E_n)_{n = 1}^{∞}\) lie in \(ℱ\) and are pairwise disjoint, then \(μ\left( \bigcup_{n = 1}^∞E_n \right) = \sum_{n = 1}^∞μ(E_n)\)

\((X, ℱ, \mu)\) is a measure space. Often we just refer to \(X\).

Say \(\mu\) is finite, if \(μ(Ω)\) is finite. \(\mu\) is a probability measure if \(μ(Ω) = 1\).

Say \(\mu\) is σ-finite, if \(∃E_n∈ℱ\) such that \(μ(E_n)<∞\) and \(\bigcup_{n = 1}^∞E_n = Ω\).

Example 3.1.

• \((ℝ, ℳ_\text{Leb}, m)\) is σ-finite measure space.

• Ω finite, \(ℱ=𝒫 (Ω),μ(E) = {|E|}\) is counting measure on Ω.

• Lebesgue-Stieltjes measure on \(ℝ\), see notes.

Proposition 3.2. Let \((Ω, ℱ)\) be a measurable space.

1. If \(A, B∈ℱ\) and \(A⊆B\), then \(μ(A)≤μ(B)\).

2. \(A_1⊆A_2⊆A_3⊆…\) in \(ℱ\), then \(μ\left( \bigcup_{n = 1}^∞A_n \right) = \lim_{n→∞}μ(A_n)\). [say \(μ\) is continuous]

3. \(A_1⊇A_2⊇A_3⊇…\) in \(ℱ\) and \(m (A_1)<∞\), then \(μ\left( \bigcap_{n = 1}^∞A_n \right) = \lim_{n→∞}μ(A_n)\).

Remark. In 3, the condition \(m (A_1)<∞\) avoids the case \(A_n = [n, ∞), \bigcap_{n = 1}^∞A_n = ∅\).

Proof. (2)Let \(A_1' = A_1, A_n = A_n∖A_{n - 1}\), so \((A_n')\) are pairwise disjoint.

\[μ\left( \bigcup A_n \right) = \sum_{r = 1}^∞μ(A_r') = \lim_{n→∞} \sum_{r = 1}^nμ(A_r') = \lim_{n→∞}μ(A_n)\] \(\Box\)

Suppose \(ℬ⊆𝒫 (Ω)\), \(ℬ\) generates a σ-algebra \(ℱ_ℬ\) on Ω (the smallest σ-algebra on Ω containing \(ℬ\))

1. \(ℬ⊆ℱ_ℬ\)

2. If \(ℱ\) is a σ-algebra on Ω such that \(ℱ⊇ℬ\) then \(ℱ⊇ℱ_ℬ\)

Take \(ℱ_ℬ = \bigcap_{\substack{\text{ℱ is a σ-algebra on Ω}\\ ℱ⊇ℬ }} ℱ\)

Definition. Define \(ℳ_\text{Bor}\) to be the σ-algebra on \(ℝ\) generated by intervals. This is the Borel σ-algebra on \(ℝ\).

Proposition 3.3.

1. Each of the following classes generates \(ℳ_\text{Bor}\)

   • intervals \((a, ∞)\)

   • intervals \([a, b]\)

   • open sets

2. \(ℳ_\text{Bor}⫋ℳ_\text{Leb}\)see appendix

3. \(E∈ℳ_\text{Leb} ⇔∃A, B∈ℳ_{\text{Bol}}\) such that \(A⊆E⊆B\) and \(B∖A\) is null. \(ℳ_\text{Leb}\) is a completion of \(ℳ_\text{Bor}\) by adding in null sets.

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Definition. Let \((Ω, ℱ)\) be a measurable space. \(f : Ω→ℝ\) is \(ℱ\)-measurable if \(f^{-1}(I)∈ℱ\) for any interval \(I⊆ℝ\).

Proposition 3.4. Let \(ℬ\) be a class of sets in \(ℝ\) generating \(ℳ_\text{Bor}\) [any class from 3.3(1)] Then \(f : Ω→ℝ\) is measurable \(⇔f^{-1}(G)∈ℱ\) for all \(G∈ℬ\).

Proof. \(f_*(ℱ) := \{ G⊆ℝ: f^{-1}(G)∈ℱ \}\) is a σ-algebra over \(ℝ\).

\(ℬ⊆f_*(ℱ)⇔ℳ_\text{Bor}⊆f_*(ℱ)⇔I⊆f_*(ℱ)\) for all intervals \(I⊆ℝ\).

\(∴f : Ω→ℝ\) is measurable\(⇔f^{-1}(G)\) measurable for all open \(G⇔f^{-1}(G)\) measurable for Borel \(G⊆ℝ\).◻

Another important measurable space is obtained by restricting Lebesgue measure to measurable subsets \(E⊆ℝ\), typically intervals.

\(ℳ_\text{Leb}|_E = \{ F⊆E : F∈ℳ_\text{Leb}\} = \{ F∩E : F∈ℳ_\text{Leb}\}\)

\(\left(E, ℳ_\text{Leb}|_E, m |_{ℳ_\text{Leb}|_E}\right)\) is a measure space.

in this case give \(f : E→ℝ\) define \(\tilde{f} : ℝ→ℝ\) by \(\tilde{f} (x) = \begin{cases}f (x) & x∈E\\ 0 & x∉E\end{cases}\)

Note. \(f\) is \(ℳ_\text{Leb}|_E\)-measurable\(⇔\tilde{f}\) is \(ℳ_\text{Leb}\)-measurable.

Example 3.5.

1. Constant functions are measurable.

2. Characteristic function \( χ_A\) of a subset \(A\) of \(ℝ\) is measurable\(⇔A\) is measurable.

3. Continuous functions \(f : ℝ→ℝ\) measurable.

4. Monotone functions \(f : ℝ→ℝ\) measurable.

5. \(f : ℝ→ℝ\) continuous a.e.\(⇒f\) is measurable.

6. \(f, g : ℝ→ℝ\), \(f\) is measurable and \(f = g\) a.e.\(⇒g\) is measurable.

   This strictly contains 5: \( χ_ℚ = 1\) a.e. but \( χ_ℚ\) is nowhere continuous.

Proposition 3.6. Let \((Ω, ℱ)\) be a measurable space and \(f, g : Ω→ℝ\) measurable. Then

• \(f + g, fg\) and \(\max (f, g)\) are measurable.

• \(h∘f\) is measurable whenever \(h : ℝ→ℝ\) is continuous.

Proof. \(f + g\) measurable: Fix \(a∈ℝ\)

\[(f + g)^{-1}((a, ∞)) = \bigcup_{q∈ℚ} [f^{-1}((q, ∞))∩g^{-1}((a - q, ∞))]\]

As \(f (x) + g (x) > a⇔f (x) > a - g (x)⇔∃q∈ℚ: f (x) > q > a - g (x)\)

\(∴f + g\) is measurable.

\(fg\) and \(\max (f, g)\) are measurable: Exercise

\(h∘f\) is measurable: As for \(G⊆ℝ\) open, \((h∘f)^{-1}(G) = f^{-1}(h^{-1}(G))\) measurable.

As \(h^{-1}(G)\) open by continuity.◻

Note. \(h∘f\) is measurable when \(h\) is Borel measurable[$G∈ℳ_\text{Bor}$ implies $h^{-1}(G)∈ℳ_\text{Bor}$] and \(f\) is measurable[$G∈ℳ_\text{Bor}$ implies $f^{-1}(G)∈ℳ_\text{Leb}$].

For \(f : Ω→[- ∞, ∞]\) say \(f\) is measurable\(\overset{\text{def}}⇔f^{-1}((a, ∞])\) is measurable for all \(a∈ℝ\).

\(⇔f^{-1}(G)\) is measurable \(∀G⊆ℝ\) Borel and \(f^{-1}(\{ ∞ \}), f^{-1}(\{ - ∞ \})\) are measurable.

Proposition 3.7. Let \((f_n)_{n = 1}^∞\) be a sequence of measurable functions \(ℝ→ℝ\). Then \(\sup_n f_n, \inf_n f_n, \limsup_n f_n, \liminf_n f_n\) are measurable. Hence, if \(f (x) = \lim_{n→ ∞} f_n (x)\) a.e. then \(f\) is measurable.

Proof. Fix \(a∈ℝ\), \(\sup_n f_n (x) > a⇔∃n∈ℕ: f_n (x) > a\). So

\[(\sup_n f_n)^{-1}(a, ∞] = \bigcup_n f_n^{-1}((a, ∞]) \text{ is measurable}\]

So \(\sup_n f_n\) is measurable. So \(\inf_n f_n = - \sup_n (- f_n)\) is measurable. Finally,

\[\limsup_n f_n = \inf_m\sup_{n≥m} f_n\]

is measurable by previous results.

If \(\lim_n f_n (x) = f (x)\) a.e. then \(f (x) = \limsup_n f_n (x)\) a.e. so is measurable by Prop. 3.3(6).◻

Definition. \(ϕ : ℝ→ℝ\) is simple if it is measurable and takes finitely many values.

Example 3.8. \( χ_E\) is simple when \(E∈ℳ_\text{Leb}\)

\(ϕ,ψ\) simple\(⇒ϕψ, ϕ +ψ, \max (ϕ,ψ), h∘ϕ\) are simple for any \(h\)

\(\sum_{j = 1}^n β_j χ_{E_j}\) are simple for \(E_j∈ℳ_\text{Leb}\) and \(β_j∈ℝ\).

Given any simple function \(ϕ\), let \(α_1,…, α_n\) be the non-zero values of \(ϕ\) and \(β_i = ϕ^{-1}(\{ x_i \})\). Then

\[\tag{*}ϕ = \sum_{i = 1}^n α_i χ_{B_i}\]

Here the \(B_1,…, B_n\) are pairwise disjoint and \(α_i\) distinct.

We call (*) the standard form of \(ϕ\).

\(ϕ_{(0, 2)} + ϕ_{[1, 3]} = χ_{(0, 1)∪[2, 3]} + 2 χ_{[1, 2)}\) is the standard form.

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Simple functions: measurable functions with finitely many values

Step functions: finite linear combinations of \(χ_I\) for \(I\) a bounded interval

Proposition 3.8. Step functions are simple.

\(χ_{ℚ∩[0, 1]}\) is simple but not step function.

Proposition 3.9. Let \(f : ℝ→[0,∞]\) be measurable, then ∃ simple functions \(0⩽ϕ_1⩽ϕ_2⩽⋯\) such that \(\lim_{n→∞} ϕ_n (x) = f (x)\;\color{#f00}∀x\)

Theorem 3.10. Let \(f : ℝ→ℝ\) be measurable, then ∃ step functions \((ϕ_n)_{n = 1}^{∞}\) such that \(f (x) = \lim_{n→∞} ϕ_n (x)\) a.e. (x)

Proof. (3.9)For \(n∈ℕ, k = 0, 1, 2, …, 4^n - 1\), let \(B_{k, n} = \left\{ x∈ℝ : \frac{k}{2^n}⩽f (x)<\frac{k + 1}{2^n} \right\}\)

If we define \(ϕ_n = \sum_{k = 0}^{4^{n - 1}} \frac{k}{2^n} χ_{B_{k, n}} + 2^n χ_{\{ x : f (x)⩾2^n \}}\)

\[ϕ_n (x) = \begin{cases} k 2^{- n} & \text{if } x∈B_{k, n}\\ 2^n & \text{if } x∉\bigcup_{k = 0}^{4^n - 1} B_{k, n} \end{cases}\]

We have \(ϕ_n⩽ϕ_{n + 1}⩽⋯\)

If \(∃n∈ℕ\) such that \(f (x)⩽2^n\), then for all \(m⩾n\), \(| ϕ_n (x) - f (x) |⩽2^{- n}→0\)

i.e. If \(f (x)<∞\), then \(ϕ_m (x)→f (x)\) as \(m→∞\)

If \(f (x) =∞\), then \(ϕ_m (x) = 2^m→∞\) as \(m→∞\)◻

4Lebesgue integral: non-negative functions

Throughout §4 and §5 we work with \((ℝ,ℳ_\text{Leb}, m)\).

But mostly works in full generality i.e. for \((Ω,ℱ, μ)\).

	Approximate by	How to approximate
Riemann	Step functions	Simultaneously from above & below
Lebesgue	Simple functions	positive area from above negative area from below

Definition. If \(ϕ : ℝ→[0,∞]\) is a non-negative simple function, with standard form \(ϕ (x) = \sum_{i = 1}^n α_i χ_{B_i}, α_i > 0, B_i\) disjoint measurable. Define\[\int ϕ = \int_{ℝ} ϕ = \int_{-∞}^{∞} ϕ (x) \mathrm{d} x = \int_{-∞}^{∞} ϕ (x) \mathrm{d} m (x) := \sum_{i = 1}^n α_i m (B_i)∈[0,∞]\]

Proposition 4.1. Let \(ϕ, ψ : ℝ→[0,∞]\) be simple.

1. If \(ϕ (x) = \sum_{j = 1}^m β_jχ_{E_j} (x)\) with \(β_j⩾0\), then \(\int ϕ = \sum_{j = 1}^m β_j m (E_j)\)

2. \(\int (ϕ + ψ) = \int ϕ + \int ψ\)\(\int α ϕ = α \int ϕ\)

3. If \(ϕ⩽ψ\), then \(\int ϕ⩽\int ψ\)

Definition. Let \(f : ℝ→[0,∞]\) be measurable, define

\[\int f = \int_{ℝ} f = \int_{-∞}^{∞} f = \int_{-∞}^{∞} f (x) \mathrm{d} m (x) := \sup \left\{ \int ϕ : 0⩽ϕ⩽f, ϕ \text{ is simple} \right\}\]

If \(E∈ℳ_\text{Leb}\), \(f : E→[0,∞]\) is measurable, define \(\hat{f} (x) = \begin{cases} f (x) & x∈E\\ 0 & x∉E \end{cases}\) and \(\int_E f = \int_{ℝ} \hat{f}\)
If \(g : ℝ→[0,∞]\) is measurable. Define \(\int_E g = \int_{ℝ} g χ_E\)

Definition. Let \(f : ℝ→[0,∞]\) be measurable, \(E∈ℳ_\text{Leb}\), we say \(f\) is integrable on \(E\) if \(\int_E f<∞\).

Given \(f, g : ℝ→[0,∞]\) be measurable.

1. If \(α⩾0\), then \(\int α f = α \int f\)

2. If \(f⩽g\), then \(\int f⩽\int g\)

3. \(\int (f + g) = \int f + \int g\)

   ⩾ follows from definition

   ⩽ is fiddly

Theorem 4.2(Monotone Convergence Theorem). Let \((f_n)_{n = 1}^{∞}\) be an increasing sequence of non-negative measurable functions, \(f (x) = \lim_{n→∞} f_n (x)\). Then \(\lim_{n→∞} ∫f_n = ∫f\)

Proof. \(f_n⩽f⇒∫f_n⩽∫f⇒\lim_{n→∞} ∫f_n⩽∫f\). For the reverse inequality,

\(0⩽ϕ⩽f\), \(ϕ\) simple, we aim to show \(∫ϕ⩽\lim_{n→∞}∫f_n\)

For \(0<α<1\), consider \(B_n = \{ x : f_n (x)⩾α ϕ (x) \}∈ℳ_\text{Leb}\) as \(f_n - α ϕ\) is measurable.

Note \(B_1⊆B_2⊆…\) and \(\bigcup_{n = 1}^{∞} B_n =ℝ\) as if \(ϕ (x) \neq 0\), \(α ϕ (x)<f\) so \(∃n_0\) such that \(f_n (x) > α ϕ (x)\) for \(n⩾n_0\). If \(ϕ (x) = 0, x∈B_1\).

Note that \(α ϕ χ_{B_n}⩽f_n χ_{B_n}⇒α \int_{B_n} ϕ⩽\int_{B_n} f_n⩽\int_{ℝ} f_n\)

Claim: \(\int_{B_n} ϕ↗∫ϕ\) as \(n→∞\)

For \(E∈ℳ_\text{Leb}\), \(\int_{B_n} χ_E = m (E∩B_n)↗m (E)\) by Prop 3.2(2)

Write \(ϕ = \sum_{i = 1}^k α_i χ_{E_i}\) and \(\int_{B_n} ϕ = \sum_{i = 1}^k α_i \int_{B_n} χ_{E_i}→∫ϕ\)

This proves the claim \(α ∫ϕ⩽\lim_{n→∞}∫f_n\)

Let \(α↗1\), then \(∫ϕ⩽\lim_{n→∞}∫f_n\).◻

Dini's Theorem

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Corollary 4.3. (Baby MCT) Let \(f\) be a non-negative measurable function, \((E_n)\) be an increasing sequence of measurable sets, and \(E=\bigcup_{i=1}^∞ E_n\). Then \(\int_E f=\lim_{n→∞} \int_{E_n} f=\sup_n \int_{E_n} f\).

Proof. Apply MCT to \((f χ_{E_n})\). ◻

Corollary 4.4. (Additivity) For \(0⩽f, g\) measurable, \(\int (f+g)=\int f+\int g\).

Proof. By Prop 3.9 there exists simple functions \(0⩽ϕ_1⩽ϕ_2⩽⋯↗f\) and \(0⩽ψ_1⩽ψ_2⩽⋯↗g\)

i.e. \(ϕ_n (x)↗f (x)\) and \(ψ_n (x)↗g (x)\) for all \(x\).

\[\int (f+g)=\lim \int (ϕ_n+ψ_n) \xlongequal[\text{4.1}]{} \lim \left( \int ϕ_n+\int ψ_n \right)=\int f+\int g\]◻

Corollary 4.5. (MCT for series) Let \((f_n)^∞_{n=1}\) be non-negative measurable functions, \(f=\sum_{n=1}^∞ f_n\). Then \(\int f=\sum_{n=1}^∞ f_n\). [\(f\) is integrable\(⇔\sum_{n=1}^∞ f_n\) converges]

Corollary 4.6. Let \(f : [a, b]→[0,∞)\) be continuous. Then \(\int_{[a, b]} f=\int_a^b f\) is equal to the R-integral of \(f\) on \([a, b]\).

Proof. Sucessively bisecting \([a, b]\). Let \(ϕ_n\) be the step function corresponding to the lower Riemann sum for the \(n\)-th partition. Then \(0⩽ϕ_1⩽ϕ_2⩽⋯↗f\). The R-integral is equal to \(\lim_{n→∞} \int_a^b ϕ_n\) which is \(\int_a^b f\) by MCT. ◻

Example 4.7. \(f (x)=(1-x)^{-1/2}\) on \([0, 1)\). \(f\) is continuous on \([0, 1)\) so measurable. By Baby MCT \(\int_0^1 (1-x)^{-1/2} \mathrm{d} x=\lim_{n→∞} \int_0^{1-\frac{1}{n}} (1-x)^{-1/2} \mathrm{d} x=\lim_{n→∞} 2 (1-n^{-1/2})=2\) by FTC.

On the other hand, \((1-x)^{-1/2}=\sum_{n=0}^∞ \frac{(2 n) !}{4^n (n!)^2} x^n\) for \(0⩽x<1\)

By MCT for series, \(2=\int_0^1 (1-x)^{-1/2} \mathrm{d} x=\sum_{n=0}^∞ \frac{(2 n) !}{4^n (n!)^2} \int_0^1 x^n \mathrm{d} x=\sum_{n=0}^∞ \frac{(2 n) !}{4^n n! (n+1) !}\)

The fact that the series above converges to 2 can be obtained directly from the Binomial Expansion of \((1-x)^{1/2}\), via Abel's theorem.

Example 4.8. \(\lim_{n→∞} \int_0^{nπ} \cos \left( \frac{x}{2 n} \right) x^2 \mathrm{e}^{-x^3} \mathrm{d} x\)

Let \(f_n (x)=\cos \left( \frac{x}{2 n} \right) x^2 \mathrm{e}^{-x^3} χ_{[0, nπ]} (x)\)

These are measurable as \(\cos \left( \frac{x}{2 n} \right) x^2 \mathrm{e}^{-x^3}\) is continuous.

\(0⩽f_1⩽f_2⩽⋯\) as \(\cos \left( \frac{x}{2 n} \right)⩽\cos \left( \frac{x}{2 (n+1)} \right)\) for \(0⩽x⩽nπ\)

By MCT \(\lim_{n→∞} \int_0^{nπ} \cos \left( \frac{x}{2 n} \right) x^2 \mathrm{e}^{-x^3} \mathrm{d} x=\int_0^∞ x^2 \mathrm{e}^{-x^3} \mathrm{d} x\) as \(f_n (x)→x^2 \mathrm{e}^{-x^3}\) pointwise.

Now \(\int_0^∞ x^2 \mathrm{e}^{-x^3} \mathrm{d} x=\lim_{n→∞}\int_0^n x^2 \mathrm{e}^{-x^3} \mathrm{d} x\) by Baby MCT

\(=\lim_{n→∞}\frac{1-\mathrm e^{- n^3}}{3}=\frac{1}{3}\) by FTC.

5 Lebesgue Integral: General functions

Definition. For \(f : ℝ→[-∞,∞]\) measurable, set \(f^+=\max (f, 0), f^-=\max (- f, 0)\) both are non-negative and measurable. Note \({|f|}=f^++f^-\) and \(f=f^+-f^-\). Say \(f\) is measurable over \(ℝ\), written as \(f =ℒ^1(ℝ)\). If \(f\) is measurable and \(\int f^+, \int f^-\) are finite. In this case define \(\int f=\int f^+-\int f^-\).

Say \(E∈ℳ_\text{Leb}\), \(f\) is integrable over \(E\), write \(f∈ℒ^1(E)\), when \(f χ_E∈ℒ^1(ℝ)\). In this case \(\int_E f=\int_ℝf χ_E\).

Proposition 5.1.

1. If \(f\) is integrable, then \(|f|\) is integrable. [drawback of Lebesgue theory]
2. \({|f|}∈ℒ^1(ℝ)\) and \(f\) is measurable\(⇒f∈ℒ^1(ℝ)\).
3. (Comparison Test) Let \(f : ℝ→[-∞,∞]\) be measurable,
   • If \({|f|}⩽g\), for some \(g∈ℒ^1(ℝ)\), then \(f∈ℒ^1(ℝ)\).
   • Suppose \({|f|}⩾g⩾0\) for some measurable \(g\) which is not integrable, then \(f\) is not integrable.
4. If \(f, g∈ℒ^1(ℝ)\) and \(f+g\) is defined. Then \(f+g∈ℒ^1(ℝ)\) and \(αf∈ℒ^1(ℝ)\) for \(α∈ℝ\). Moreover, \(∫(f+g)=∫f+∫g,∫αf=α∫f\).
5. If \(f∈ℒ^1(ℝ)\) and \(g=f\) a.e. then \(g∈ℒ^1(ℝ)\) and \(∫f=∫g\).
6. If \(f\) is integrable\(⇒f (x)∈ℝ\) a.e.
7. If \(f\) is integrable \(∫{|f|}=0⇒f=0\) a.e.
8. If \(f\) is integrable over \(E∈ℳ_\text{Leb}\) and \(E_1⊆E_2⊆E_3⊆…\) are measurable with \(\bigcup_{n=1}^∞ E_n=E\). Then \(∫_E f=\lim_{n→∞}∫_{E_n} f\).

Proof. (1) and (2) follow from $∫f^±≤∫{|f|}=∫f^++∫f^-$. (3) follows from ${|f|}≤g⇒∫{|f|}≤∫g$. (4) follows from $(f+g)^±≤f^±+g^±$ and $(f+g)^++f^-+g^-=(f+g)^-+f^++g^+$. (5): Since ${|g-f|}=0$ a.e., any simple function $φ$ with $0≤φ≤{|g-f|}$ is a.e. 0, so its integral is 0. Hence $∫{|g-f|}=0$. (6), (7): Exercise, Sheet 2 Q9. (8): Apply Baby MCT to $f^+$ and $f^-$.

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\(f:ℝ→[-∞,∞]\) integrable\(⇔f\) is measurable and \(\int f^+, \int f^- <∞\) Then \(\int f=\int f^+-\int f^-\)

This works generally on (Ω, ℱ, μ)

• \((ℕ, 𝒫 (ℕ), μ)\), \(μ\) is counting measure

  Any \(f:ℕ→ℝ\) is measurable

  and \(f\) is integrable\(⇔\sum_{n∈ℕ}{|f(n)|}<∞⇔f\) is absolutely integrable

  \(\int_{ℕ} f \mathrm{d} μ=\sum_{n=1}^∞f (n)\) when \(f\) is intergrable

• \((Ω, 𝒫, ℙ)\) is probability space

  random variable \(X:Ω→ℝ\) are measurable functions and \(\int X\) is expectation

  \(X\) is integrable\(⇔𝔼[X]<∞\)

Corollary 5.2. By comparison test

1. \(g:ℝ→[-∞,∞]\) is integrable and \(h:ℝ→ℝ\) is measurable and bounded, then \(gh∈ℒ^1 (ℝ)\)

2. \(g∈ℒ^1 (ℝ)\) and \(E∈ℳ_\text{Leb}⇒ g |_E∈ℒ^1 (E)\)

3. Bounded measurable functions are integrable over measurable sets of finite measure

   Let \(f:[a, b]→ℝ\) be bounded. \(f\) is R-integrable\(⇔f\) is bounded and \(f\) is continuous a.e.

   Stein & Sharkarchi Exercise 1.7.4

   Can deduce measurability from this (overkill) or directly–see notes.

   wlog \(f:[a, b]→[0,∞)\)\[\int_{[a, b]}^R f=\sup\Set{\int_a^b ϕ:ϕ⩽f, ϕ\text{ step function}}⩽\sup\Set{\int_a^b ϕ:ϕ⩽f, ϕ\text{ simple function}}=\int_a^b f⩽\inf\Set{\int_a^bψ: f⩽ψ,ψ\text{ step function}}=\int_{[a, b]}^R f\]So equality holds throughout. The Riemann- & Lebesgue- integral agree.

Example 5.3.

1. \(f(x)=x^α\) on \([0, 1]\), \(α∈ℝ\)

   Note \(f\) is continuous so measurable.

   If \(α>0\), \(f\) is bounded so integrable on \((0, 1)\).

   If \(α<0\), We use Baby MCT and FTC.

   By FTC \(\int_{\frac{1}{n}}^1 x^α=\begin{cases}\log n & α=- 1\\ \frac{1}{α+1} (1-n^{- (α+1)}) & α \neq-1 \end{cases}\)

   By Baby MCT \(f∈ℒ^1 ((0, 1))\) and \(α >-1\) in this case.

   \(\int_0^1 f=\lim_{n→∞} \int_{\frac{1}{n}}^1 x^α=\frac{1}{α+1}\)

2. \(f(x)=x^α\) on \([1,∞)\) which is continuous so measurable

   \(f∈ℒ^1 ([1,∞))⇔α<- 1\)

   By Baby MCT, to work with intervals \([1, n]\) and FTC.

   Suppose we are given \(f:I→[-∞,∞]\). How to decide if \(f\) is integrable? Justify measurability.

   Justify measurability §3.5

   \(⇒\)Replace by \(|f|⇒\bigg\{\)

   (\(f\) bounded and \(I\) bounded) \(f\) is integrable

   (\(f\) or \(I\) unbounded) Decompose \(I=\bigcup I_n\), \(I_1⊆I_2⊆…\) Apply baby MCT

   Comparison Test, look for \(g\), \(0⩽{|f|}⩽g\) or \(0⩽g⩽{|f|}\)

3. \(f(x)=\frac{x^α}{1+x^{β}}\) over \((0,∞)\), \(α∈ℝ, β>0\)

   Split \((0,∞)\) as \((0, 1) \cup [1,∞)\)

   On \((0, 1)\): \(\frac{1}{2} x^α⩽f_n⩽x^α\), so \(f\) is integrable on \((0, 1) ⇔α >-1\).

   On \((1,∞):\frac{x^{α-β}}{2}⩽f(x)⩽x^{α-β}\), \(f\) is integrable on \((1,∞) ⇔α-β<- 1\)

   \(∴f∈ℒ^1 ((0,∞))⇔β-1>α >-1\).

4. \(\frac{\sin x}{x}\) is integrable on \((0, 1)\) as

   it is continuous so measurable and bounded on \((0, 1)\) as \(\frac{\sin x}{x}→1\) as \(x→0\)

5. \(\frac{\sin x}{x}\) not integrable on \((0,∞)\)

   For \(n∈ℕ, \int_{rπ}^{(r+1)π} \left| \frac{\sin x}{x} \right|⩾\frac{1}{(r+1)π} \int_{rπ}^{(r+1)π}\left|\sin x\right|=\frac{2}{(r+1)π}\)

   Hence \(\lim_{n→∞} \int_π^{(n+1)π} \left| \frac{\sin x}{x} \right|=∞\) using harmonic series.

Theorem 5.4. (Fundamental Theorem of Calculus)Let \(g:[a, b]→ℝ\) has continuous derivative on \([a, b]\) Then \(g'\) is integrable and \(\int_a^b g'=g (b)-g (a)\)

Example. \(f(x)=\begin{cases} \frac1x \sin \frac1x & x≠0\\ 0 & x=0 \end{cases}\) is continuous on \([0, 1]\), differentiable on \((0, 1]\) and \(f'\) is continuous on \((0, 1]\), but \(f'(x)=\sin\frac1x-{\color{red}{\frac{\cos x}x}}∉ℒ^1 ((0, 1))\) as \(\frac{\cos x}x≥\frac{\cos1}x\).

Example. Cantor-Lebesgue function \(ϕ:[0, 1]→[0, 1]\)

\(x∈C\), write \(x=\sum a_n 3^{- n}, a_n∈\{ 0, 2 \}\) and set \(ϕ(x)=\sum \frac{a_n}{2} 2^{- n}\)

For \(x∈\left[ \frac{1}{3}, \frac{2}{3} \right]\), \(ϕ(x)=\frac{1}{2}\)

For \(x∈\left[ \frac{1}{9}, \frac{2}{9} \right]\), \(ϕ(x)=\frac{1}{4}\)

On \([0, 1]∖C\), \(ϕ\) is monotone, continuous, differentiable with \(ϕ'(x)=0\)

But \(\int_0^1 ϕ≠ϕ(1)-ϕ(0)=1\)

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Theorem 5.6. Let \(f, g\) continuously differentiable on \([a, b]\) then

\[\int_a^b f(x) g'(x) \mathrm{d} x=f(b) g(b)-f(a) g(a)-\int_a^b f'(x) g(x) \mathrm{d} x\]

Example 5.7. (cannot infer the existence of one integral from the existence of the other)Integrating by parts,

\[\int_1^n \frac{\sin x}{x}=\cos 1-\frac{\cos n}{n}-\int_1^n \frac{\cos x}{x^2}\]

As \({|\cos x|}⩽1\), \(\left|\cos x\over x^2\right|⩽\frac{1}{x^2}\) which, is integrable on \([1, ∞)\), \(\frac{\cos x}{x^2}\) is integrable on \([1, ∞)\) by comparison.

By Prop 5.1(8) \(\int_1^∞\frac{\cos x}{x^2}\) exists and is equal to \(\lim_{n→∞}\int_0^n \frac{\cos x}{x^2}\)

\[∴\lim_{n→∞}\int_1^n \frac{\sin x}{x}=\cos 1-\int_1^∞\frac{\cos x}{x^2}\]

but \(\int_1^∞\frac{\sin x}{x}\) does not exist; \(\frac{\sin x}{x}\) is not integrable on \([1, ∞)\).

like the harmonic series is not integrable on the natural number with counting measure (not absolutely convergent)

Theorem 5.8. (Substitution)Let \(g:I→ℝ\) be monotone with continuous derivative on an interval \(I\), and let \(J\) be the interval \(g(I)\). For a measurable function \(f:J→ℝ\), \(f∈ℒ^1(J)⇔(f∘g){|g'|}∈ℒ^1(I)\). In this case \(\int_J f(y) \mathrm{d} y=\int_I f(g(x)){|g'(x)|}\mathrm{d} x\)

[Write \(\int_a^b\) for \(\int_{[a, b]}\) when \(a<b\); for \(-\int_{[b, a]}\) when \(a > b\)]

Road for proving theorems like this about all integrable functions \(f\)

Step 1. Show result for \(f=χ_E\) for \(E∈ℳ_\text{Leb}\)

Use Cor 2.7 about general form of Lebesgue measurable sets (in Bonus problem)

Step 2. Use linearity to show for simple non-negative \(f\)

Step 3. Use MCT to show for integrable non-negative \(f\)

Step 4. Use linearity to show for general integrable \(f\)

Example 5.9. \(I=(0, 1)\), \(g:I→(1, ∞)\) be \(g(y)=\frac{1}{y}\)

Consider \(f(x)=x^{α}\), \(f∈ℒ^1(J)⇔α<-1\)

\(f(g(y)){|g'(y)|}=y^{-α-2}∈ℒ^1(I)⇔-α-2 >-1⇔α<-1\)

6 Convergence theorems

If \(f_n(x)→f(x)\) a.e. when does \(\int f_n(x)→\int f(x)\)?

• MCT
• Fatou's lemma—mainly for technical use
• Dominated Convergence Theorem (DCT)

Theorem 6.1. (MCT)Let \((f_n)_{n=1}^∞\) be integrable such that

1. \(∀n, f_n⩽f_{n+1}\) a.e.
2. \(\sup_n \int f_n<∞\).

Then \(∃f\) such that \(f_n→f\) a.e.
\(f\) is integrable and \(\int f=\lim\int f_n\)

Proof. By Prop 5.1(6) \(f_n∈ℝ\) a.e.

For each \(n\), let \(N_n\) be null such that \(∀x \not\in N_n, f_n(x)⩽f_{n+1}(x)\) and \(f_n(x)∈ℝ\)

Redefine \(f_n(x)=0\) for any \(n\), and \(x∈\bigcup_{m=1}^∞N_m\). This doesn't change integrability or the values of the integrals. [to avoid \(∞-∞\) in \(f_n-f_1\)]

Then \((f_n-f_1)_{n=1}^∞\) is monotonic & non-negative, so by MCT \(\int(f-f_1)=\lim\int f_n-\int f_1<∞\)

Add \(\int f_1\) to get result. ◻

Theorem 6.2. (Fatou's lemma)Let \((f_n)_{n=1}^∞\) be non-negative measurable functions. Then

\[\int \liminf_{n→∞}f_n⩽\liminf_{n→∞}\int f_n\]

[in hat example, each \(\int f_n\) is 1, and \(f_n\) will converge to zero function, shows inequality can be strict.]

Proof. Let \(g_r=\inf_{n⩾r} f_n\) which is measurable, \(g_r⩽g_{r+1}\) and \(g_r→\liminf_{n→∞}f_n\)

Note \(g_r⩽f_r\ ∀r\) and hence \(\int g_r⩽\int f_r\)

\[\int \liminf f_r=\int \lim g_r \mathop{=}\limits^\text{MCT}\lim_{r→∞}\int g_r=\limsup_{r→∞}\int g_r⩽\liminf_{r→∞}\int f_r\]◻

One can also have $\int\limsup_{n→∞} f_n>\limsup _{n→∞} \int f_n$ – for example, $f_n(x)=\sin^2(x+n)$ on (0, π), each \(\int f_n\) is π/2, and \(\limsup_{n→∞} f_n(x)=1\)

Theorem 6.3. (Dominated Convergence Theorem)Let \((f_n)_{n=1}^∞\) be measurable functions such that \(f_n(x)→f(x)\) a.e. & \(∃g∈ℒ^1\) such that \({|f_n(x)|}⩽g(x)\) a.e. \(∀n\)

Then \(f∈ℒ^1\) and \(\int f=\lim\int f_n\)

Proof. \(f\) is measurable by Prop 3.7

\({|f(x)|}⩽g(x)\) a.e. so \(f∈ℒ^1\) by comparison.

Apply Fatou to \(g±f_n⩾0\) a.e.

\(∴\int g+f⩽\int g+\liminf\int f_n\) & \(\int g-f⩽\int g-\limsup \int f_n\)

Subtract \(\int g\), \(\int f⩽\liminf\int f_n⩽\limsup \int f_n⩽\int f\)

\(∴\limsup \int f_n=\liminf\int f_n=\int f, ∴\lim\int f_n=\int f\). ◻

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$\def\d{\mathrm{d}}$

Example 6.4. \(\int_0^1 \frac{n^{3/2}xe^x}{1+n^2 x^2}\d x=\int_0^1 \frac{(nx)^{3/2}}{1+n^2 x^2}\frac{e^x}{x^{1/2}}\d x\)

Consider \(g(y)=\frac{y^{3/2}}{1+y^2}→0\) as \(y→∞\). As \(g\) is continuous, \(g\) is bounded.

Let \(C=\sup_{[0,1]}g\) then \(\frac{n^{3/2}xe^x}{1+n^2 x^2}≤\frac{Ce^x}{x^{1/2}}≤\frac{Ce}{x^{1/2}}\) and \(x^{-1/2}\) is integrable on \((0, 1)\)

By DCT \(\int_0^1 \frac{n^{3/2}xe^x}{1+n^2 x^2}\d x→0\) as \(n→∞\).

Corollary 6.5. (Bounded Convergence Theorem)Let \(I\) be a bounded interval, \((f_n)\) be a sequence in \(ℒ^1(I)\) converging a.e. to \(f\), and suppose that there is a constant \(C\) such that \(| f_n(x) |≤C\) a.e. \(∀n\)

Then \(f∈ℒ^1\) and \(\int_I f=\lim_{n→∞}\int_I f_n\)

\(ℂ\)-valued functions: If \(f:Ω→ℂ\) with \((Ω, ℱ, μ)\) a measure space, we say \(f\) is integrable iff \(\operatorname{Re}f, \operatorname{Im}f\) are integrable. In that case \(\int f=\int \operatorname{Re}f+i \int \operatorname{Im}f\)

Example 6.6. Let \(γ_r\) be the semi-circular contour \(\{re^{iθ}:0≤θ≤π\}\), and consider

\[\int_{γ_r}\frac{e^{iz}}{z}\d z=i \int_0^{π}e^{ir \cosθ}e^{-r \sinθ}\dθ\]

Note \(e^{ir \cosθ}e^{-r \sinθ}→\left\{\begin{array}{ll}0 & r→∞\\ 1 & r→0 \end{array}\right.\) and uniformly bounded: \(| e^{ir \cosθ}e^{-r \sinθ}|≤1\)

\[\int_{γ_{R_n}}\frac{e^{iz}}{z}\d z→0 \text{ as }R_n→∞ \text{ by BCT}\]\[\int_{γ_{ε_n}}\frac{e^{iz}}{z}\d z→i π \text{ as }ε_n→0 \text{ by BCT}\]

By Cauchy's theorem

\[0=\int_{γ_{R_n}}\frac{e^{iz}}{z}\d z-\int_{γ_{ε_n}}\frac{e^{iz}}{z}\d z+\int_{ε_n}^{R_n}\frac{e^{ix}-e^{-ix}}{x}\d x\]

Let \(n→∞, R_n→∞ {,}ε_n→0\),

\[\lim \int_{ε_n}^{R_n}\frac{\sin x}{x}=\frac{π}{2}\]

So

\[\lim_{n→∞}\int_0^n \frac{\sin x}{x}=\frac{π}{2}\]

but the function is not integrable on \([0, ∞)\).

Theorem 6.7. MCT for Series (Corollary 4.5 above).

Theorem 6.8. (Lebesgue Series/Beppo Levi Theorem)Let \((g_n)\) be a sequence of integrable functions such that \(\sum_{n=1}^∞\int | g_n |<∞\)

Then \(\sum_{n=1}^∞g_n\) converges to an integrable function and \(\int \sum_{n=1}^∞g_n=\sum_{n=1}^∞\int g_n\)

Proof. By MCT for series, \(\sum_{n=1}^∞| g_n |\) converges a.e. to an integrable function.

By DCT with \(\sum_{n=1}^∞| g_n |\) as dominating function, the result follows.◻

Theorem 6.9. If \((g_n)_{n=1}^{∞}\) are integrable and \(\sum | g_n |\) is integrable, then \(g=\sum g_n\) converges a.e. and \(\int g=\sum \int g_n\)

Example 6.10. Let \(α>0\), and consider \(\int_0^1 x^{α-1}e^{-x}\d x\).

Let \(g_n(x)=(-1)^n \frac{x^{α+n-1}}{n!}\) so \(\sum g_n(x)=x^{α-1}e^{-x}\)

\(\int_0^1 | g_n(x) |=\frac{1}{n!(α+n)}\) since \(\sum \int | g_n |<∞\)

\(\therefore \int_0^1 x^{α-1}e^{-x}=\sum_{n=0}^∞\int_0^1(-1)^n \frac{x^{α+n-1}}{n!}\d x=\sum_{n=0}^∞\frac{(-1)^n}{n!(α+n)}\)

Example 6.11. For \(s∈ℝ\), \(\int_{-∞}^∞e^{-isx}e^{-x^2}\d x\)

Note the integrand is continuous so measurable. \(| e^{-isx}e^{-x^2}|=e^{-x^2}∈ℒ^1(ℝ)\) [See this by noting e\(^{-x^2}<e^{-| x |}\) for \(| x |>1\)]

Let \(g_n(x)=\frac{(-isx)^n}{n!}e^{-x^2}\)

\(\sum_{n=0}^∞g_n(x)=e^{-isx}e^{-x^2}\) [convergence by power series expansion]

\(\sum_{n=0}^∞| g_n(x) |=e^{| sx |-x^2}≤e^{s^2/2}e^{-x^2/2}\) [using \(| sx |≤\frac{x^2+s^2}{2}\)]

\(\therefore \sum | g_n |∈ℒ^1(ℝ)\)

By Theorem 6.8[term by term integration]

\[\int_{-∞}^∞e^{-isx}e^{-x^2}\d x=\sum_{n=0}^∞\int_0^∞\frac{(-isx)^n}{n!}e^{-x^2}\d x\]

As \(\int_{-∞}^∞e^{-x^2}\d x=\sqrt{π}\) [See later]

and \(\int_{-∞}^∞\frac{(-isx)^n}{n!}e^{-x^2}=0\) for \(n\) odd, can carefully use parts & induction

\[\int_{-∞}^∞\frac{(-isx)^n}{n!}e^{-x^2}=\left\{\begin{array}{ll}\frac{(2 m) ! \sqrt{π}}{4^m m!}& n=2 m, \text{even}\\ 0 & n \text{ odd}\end{array}\right.\]Thus\[\int_{-∞}^∞e^{-isx}e^{-x^2}\d x=\sum_{m=0}^∞\frac{(-is)^{2 m}\sqrt{π}}{4^m m!}=\sqrt{π}e^{-s^2/4}\]

7Differentiation under ∫ sign

Consider \(F(y)=\int f(x, y) \d x\)

For \(F:ℝ^2→ℝ\) such that \(∀y∈ℝ: x↦f(x, y)\) is integrable in \(x\)

When \(F\) is continuous? When \(F\) is differentiable and \(\frac{\d F}{\d y}=\int \frac∂{∂y}f(x, y) \d y\)?

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Example 7.1. \(f (x, y)=ye^{-x^2 y^2}\) is continuous on \(ℝ×ℝ\)

\(F (y)\) exists for all \(y\), \(F (0)=0\)

For \(y≠0\), substitute \(t=yx\): \(F (y)=\int ye^{-x^2 y^2} \mathrm{d}x=\int e^{-t^2} \mathrm{d}t=\sqrtπ\)

\(\therefore F\) is not continuous at 0.

Theorem 7.2. (Continuous-parameter DCT) Let \(I, J\) be intervals in \(ℝ\). \(f:I×J→ℝ\) such that

1) \(∀y∈J, x↦f (x, y)\) is integrable over $I$

2) \(∀y∈J\), \(\lim_{y'→y} f (x, y')=f (x, y)\) a.e. \(x∈I\)

3) \(∃g∈ℒ^1 (I)\), \(∀y∈J\), \({|f (x, y)|}≤g (x)\) a.e. \(x∈I\)

Then \(F (y)=\int_I f (x, y) \mathrm{d}x\) is continuous on \(J\).

Proof. Let \(y_n→y\) in \(J\) and \(f_n (x)=f (x, y_n)\)

so \({|f_n (x)|}≤g (x)\) a.e. $x$ for all \(n\)

By DCT, \(F (y_n)=\int_I f_n (x)→\int_I f (x, y) \mathrm{d}x=F (y)\)

\(∴ F\) is continuous at \(y∈J\) and so \(F\) is continuous on $J$.\(\Box\)

Example 7.3. Γ function

\[F (y)=\int_0^∞\mathrm{e}^{-x} x^{y-1} \mathrm{d}x \text{ for } y∈(0,∞)\]

Set \(f (x, y)=\mathrm{e}^{-x} x^{y-1}, I=(0,∞), J=(0,∞)\)

1) Exercise use standard integrals & exp beat powers

2) is immediate

3) Consider \(g (x)=\sup_{y>0} f (x, y)=\begin{cases}x^{-1} \mathrm{e}^{-x} & 0<x≤1\\∞ & x>1\end{cases}\)which is not integrable. But continuity is a local property, so fix \(b∈(0,∞)\), set \(J_b=\left( \frac{b}{2}, 2 b \right) \subseteq J\), set \(g_b (x)=\sup_{y∈J_b} f (x, y)=\begin{cases}x^{\frac{b}{2}-1} \mathrm{e}^{-x} & 0<x≤1\\ x^{2 b-1} \mathrm{e}^{-x} & x>1\end{cases}\)

Now \(g_b\) is integrable on \((0,∞)\) using standard estimate integrals. By Continuous-parameter DCT, \(F\) is continuous on \(J_b\), so continuous at \(b\). As \(b∈J\) is arbitrary, \(F\) is continuous on \((0,∞)\).

Corollary 7.4. Let \(I, J\) be intervals and \(f:I×J→ℝ\) satisfy 1) and 2) in Thm 7.4 and

3') \(∀b∈J, ∃\)open neighborhood \(b∈J_b \subseteq J\) and \(g_b∈ℒ^1 (I)\) such that \(∀y∈J_b, | f (x, y) |≤g_b (x)\) a.e. $x∈I$

Then \(F (y)=\int_I f (x, y) \mathrm{d}x\) is continuous on \(J\).

Theorem 7.5. Let \(I, J\) be intervals in \(ℝ\) and \(f:I×J→ℝ\) satisfy

1) \(∀y∈J, x↦f (x, y)\) is integrable over \(I\)

2) \(∀y∈J\), \(\frac{∂f}{∂y} (x, y)\) exists a.e. $x∈I$

3) \(∃g∈ℒ^1 (I),∀y∈J\), \(\left| \frac{∂f}{∂x} (x, y) \right|≤g (x)\) a.e. $x∈I$

Then \(F (y)=\int_I f (x, y) \mathrm{d}x\) is differentiable on \(J\) and \(F' (y)=\int_I \frac{∂f}{∂y} (x, y) \mathrm{d}x\).

Corollary 7.7 Suppose \(f\) is as above satisfying 1) & 2) &

3') \(∀b∈J \exists\)open neighborhood, \(b∈J_b⊂J\) and \(g_b∈ℒ^1 (I)\)

such that \(∀y∈J_b\), \(\left| \frac{∂f}{∂y} (x, y) \right|≤g_b (x)\) a.e. \(x∈I\)

Then \(F\) is differentiable on \(J\) and \(F' (y)=\int_I \frac{∂f}{∂y} (x, y) \mathrm{d}x\)

Proof of 7.5

Fix \(y∈J\), let \(y_n→y, y_n≠y, y_n∈J\)

Set \(g_n (x)=\frac{f (x, y_n)-f (x, y)}{y_n-y}\) so \(g_n∈ℒ^1 (I)\)

\(g_n (x)→\frac{∂f}{∂y} (x, y)\) for almost all \(x\), \(∀y∈J\)

By the mean value theorem, \(∃ξ_{x, n}\) between \(y_n\) and \(y\) such that \(\left| g_n (x)\right|=\left| \frac{∂f}{∂y} (x, ξ_{x, n}) \right|≤g (x)\) by 3.

DCT gives \(\frac{F (y_n)-F (y)}{y_n-y}=\int g_n→\int_I \frac{∂f}{∂y} (x, y) \mathrm{d}x\)

By sequential characterisation of limits, \(F\) is differentiable at \(y\) with the stated derivative.\(\Box\)

Example 7.6 Fourier transform of Gaussian, again

Let \(f (x, s)=\mathrm{e}^{-\mathrm{i} sx} \mathrm{e}^{-x^2}\) and \(F (s)=\int_{-∞}^∞f (x, s) \mathrm{d}x\)

We know \(F (s)\) exists and \(\frac{∂f}{∂s} (x, s)=-\mathrm{i} x \mathrm{e}^{-\mathrm{i} sx} \mathrm{e}^{-x^2}\)

\(\therefore \left| \frac{∂f}{∂s} (x, s) \right|={|x|} \mathrm{e}^{-x^2}\) is integrable on ℝ.

Thm 7.5 applies, \(F\) is differentiable on ℝ and integrating by parts\begin{align*}F' (s)&=-\mathrm{i} \int_{-∞}^∞x \mathrm{e}^{-\mathrm{i} sx} \mathrm{e}^{-x^2} \mathrm{d}x\\&=\frac{\rm i}2\mathrm{e}^{-\mathrm{i} sx} \mathrm{e}^{-x^2}|_{-∞}^∞-\frac12\int_{-∞}^∞s\mathrm{e}^{-\mathrm{i} sx} \mathrm{e}^{-x^2} \mathrm{d}x\\&=-\frac{s}{2} F (s)\end{align*}

Since \(F (0)=\sqrtπ, F (s)=\sqrtπ \mathrm{e}^{-s^2 / 4}, s∈ℝ\).

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[Stein] Proposition 3.6. \(E_1, E_2⊂ℝ\) is measurable\(⇒E_1×E_2\) is measurable

8 Double integrals

Take \(f∈ℒ^1 (ℝ)\) write \(\int_{ℝ^2} f\) or \(\int f\) or \(\int_{ℝ^2} f (x, y) \mathrm{d} (x, y)\) for the integration

Theorem 8.1. (Tonelli, 1909)Let \(f:ℝ^2→[0, ∞]\) be measurable. Then

1) \(x↦f (x, y)\) measurable for almost all \(y∈ℝ\)

2) \(y↦\int_ℝf (x, y) \mathrm{d} x\) is measurable

3) \(\int_ℝf \mathrm{d} (x, y)=\int_ℝ\left(\int_ℝf (x, y) \mathrm{d} x\right) \mathrm{d} y\) [including \(∞=∞\)]

[can also change roles of \(x\) and \(y\) leading to \(\int_ℝf \mathrm{d} (x, y)=\int_ℝ\left(\int_ℝf (x, y) \mathrm{d} y\right) \mathrm{d} x\)]

In (1) we cannot guarantee that \(x↦f (x, y)\) is measurable for all \(y\)

Converse Theorem 1 is false: Let \(A∈[0, 1]\) be non-measurable and \(B⊂ℝ\) be null, then \(A×B\) is null in \(ℝ^2\) so \(A×B\) is measurable. \(f=χ_{A×B}\) is non-negative and measurable. \(x↦f (x, y)=\begin{cases} χ_A (x) & y∈B\\ 0 & y∉B \end{cases}\)

• Let \(E⊂ℝ^2\) define \(E^y=\{x∈ℝ:(x, y)∈E\}, E_x=\{y∈ℝ:(x, y)∈E\}\)

  8.1 gives that \(E\)-measurable\(⇒E^y\)&\(E_x\) measurable AE(\(x\)), AE(\(y\)).

• Converse false. Assuming suitable set theoretic axioms

  \(\exists E⊂[0, 1]×[0, 1]\) such that all slices \(E_x, E^y\) are measurable and \(m (E_x)=1∀x, m (E^y)=0∀y\)

  \(E\) is not measurable

  By 8.1 As \(\int_ℝ\left(\int_ℝχ_E (x, y) \mathrm{d} x\right) \mathrm{d} y≠\int_ℝ\left(\int_ℝχ (x, y) \mathrm{d} y\right) \mathrm{d} x\)

  [S&S Ex 2.6.5]

Proof. Ideas

• Every non-negative measurable function is an increasing limit of simple non-negative functions (3.4) via MCT this reduces 8.1 to show the case \(f=χ_E\) for \(E∈ℳ_{\text{Leb}} (ℝ^2)\).

  Then check 8.1 for

  • \(χ_E\) with \(E\) open, \(m (E)<∞\)

  • \(χ_{\bigcap_{n=1}^∞U_n}\) with \(U_n\) open, \(m (U_n)<∞\) via MCT

  • \(χ_N\) for \(N\) null

• Use Cor 2.7 (\(ℝ^2\) version)

  \(E\) measurable\(⇔E⊂\bigcap_{n=1}^∞U_n\), \(U_n\) open, such that \(\bigcap_{n=1}^∞U_n∖E\) is null

 \(\Box\)

Theorem 8.2. (Fubini's theorem, 1907)Let \(f:ℝ^2→ℝ\) be integrable. Then for almost all \(y∈ℝ\), \(x↦f (x, y)\) is integrable in \(x\) and defining \(F (y)=\int_ℝf (x, y) \mathrm{d} x\), \(F\) is integrable and\[\int_{ℝ^2} f (x, y) \mathrm{d} (x, y)=\int_ℝF (y) \mathrm{d} y=\int_ℝ\left(\int_ℝf (x, y) \mathrm{d} x\right) \mathrm{d} y\]

Similarly \(\int_{ℝ^2} f (x, y) \mathrm{d} (x, y)=\int_ℝ\left(\int_ℝf (x, y) \mathrm{d} y\right) \mathrm{d} x\) with similar comments about the definition of the RHS.

Proof. Apply 8.1 to \(f^+\) and \(f^-\). \(\Box\)

Theorem 8.3. (Tonelli's theorem, 1909)Let \(f:ℝ^2→ℝ\) be measurable. Suppose \(\int_ℝ\left(\int_ℝ|f (x, y)|\mathrm{d} x\right) \mathrm{d} y<∞\) or \(\int_ℝ\left(\int_ℝ|f (x, y)|\mathrm{d} y\right) \mathrm{d} x<∞\) then \(f\) is integrable. Hence, Fubini's theorem is applicable to both \(f\) and \(|f|\).

Proof. By Tonelli's theorem, \(\int_{ℝ^2}|f (x, y)|\mathrm{d} (x, y)<∞\), so \(f∈ℒ^1 (ℝ^2)\). \(\Box\)

Example 8.4. \(f (x, y)=\frac{x-y}{(x+y)^3}\) on \((0, 1)×(0, 1)\) is continuous so measurable and \(\int_0^1 \left(\int_0^1 f (x, y) \mathrm{d} x\right) \mathrm{d} y≠\int_0^1 \left(\int_0^1 f (x, y) \mathrm{d} y\right) \mathrm{d} x\), so \(f\) is not integrable on \((0, 1)×(0, 1)\).

Warning. Even if \(\int_0^1 \left(\int_0^1 f (x, y) \mathrm{d} x\right) \mathrm{d} y=\int_0^1 \left(\int_0^1 f (x, y) \mathrm{d} y\right) \mathrm{d} x\) the function \(f\) might not be integrable. We'll see that in next lecture.

Example 8.5. \(\int_0^1 \left(\int_0^x \left(\frac{1-y}{x-y}\right)^{\frac{1}{2}} \mathrm{d} y\right) \mathrm{d} x\)

working on \(T=\{(x, y)∈ℝ^2:0<y<x, 0<x<1\}\)

Note \(f (x, y)=\left(\frac{1-y}{x-y}\right)^{1/2}\) is continuous on \(ℝ^2\) except for the null set \(\{(x, y):x=y\}\) and so measurable and hence \(f χ_T\) is measurable.

Note \(f (x, y) \geqslant 0\) on \(T\) so Thm 8.1 can be used to interchange order of integration.

Now the reversed repeated integral is:

\begin{align*} \int_0^1 \left(\int_y^1 \left(\frac{1-y}{x-y}\right)^{1/2}\mathrm dx\right)\mathrm dy &=\int_0^1 [2 (1-y)^{1/2} (x-y)^{1/2}]_{x=y}^{x=1}\mathrm dy\\ &=\int_0^1 2 (1-y)\mathrm dy=1 \end{align*}

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\(E⊆ℝ^2\) measurable\(⇒E^y\) measurable a.e. \(x\), \(E_x\) measurable a.e. \(y\)

Example 8.6. \(\int_0^∞e^{-x^2}=\frac{\sqrt{π}}{2}\)

Let \(f(x,y)=ye^{-y^2(1+x^2)}\) continuous so measurable

Consider \(\int_{(0,∞)×(0,∞)}f(x,y)\mathrm{d}(x,y)\). Note \(f⩾0\) on \((0,∞)×(0,∞)\).

Fix \(x∈(0,∞)\), let \(t=y(1+x^2)^{1/2}\)

\[\int_0^∞ye^{-y^2(1+x^2)}\mathrm{d}y=\int_0^∞\frac{te^{-t^2}}{1+x^2}\mathrm{d}t=\lim_{k→∞}\left[-\frac{e^{-t^2}}{2(1+x^2)}\right]_{t=0}^{t=k}=\frac{1}{2(1+x^2)}\]by FTC & Baby MCT.

Now \(\int_0^∞\int_0^∞f(x,y)\mathrm{d}y\mathrm{d}x=\frac{1}{2}\int_0^∞\frac{1}{1+x^2}\mathrm{d}x=\frac{π}{4}\) by FTC & Baby MCT.

By Tonelli(8.1) \(\int_0^∞\left(\int_0^∞f(x,y)\mathrm{d}x\right)\mathrm{d}y=\frac{π}{4}\)

For \(y > 0\), set \(u=xy\), \(\frac{π}{4}=\int_0^∞\left(\int_0^∞e^{-y^2-u^2}\mathrm{d}u\right)\mathrm{d}y=\left(\int_0^∞e^{-y^2}\mathrm{d}y\right)\left(\int_0^∞e^{-u^2}\mathrm{d}u\right)=\left(\int_0^∞e^{-u^2}\mathrm{d}u\right)^2\)

Therefore \(\int_0^∞e^{-u^2}\mathrm{d}u=\frac{\sqrt{π}}{2}\). More natural derivation later in the lecture.

Example 8.7. \(f(x,y)=\frac{xy}{x^4+y^4}\)

For fixed \(y\), \(\int_{ℝ}f(x,y)\mathrm{d}x=0\) because \(f\) is odd function

Note \(f\) is integrable, by comparison to \(\frac{1}{x^4}\) on \((0,∞)\)

For fixed \(x\), \(\int_{ℝ}f(x,y)\mathrm{d}y=0\).

So \(\int\left(\int f(x,y)\mathrm{d}x\right)\mathrm{d}y=\int\left(\int f(x,y)\mathrm{d}y\right)\mathrm{d}x=0\) but \(f∉ℒ^1(ℝ^2)\).

Note on (0,∞)×(0,∞), \(f(x,y)⩾0\)

Fix \(x > 0\), \(y=xt\), \(\int_0^∞f(x,y)\mathrm{d}y=\int_0^∞\frac{x^3 t}{x^4(1+t^4)}\mathrm{d}t=C\frac{1}{x}\) For \(C=\int_0^∞\frac{t}{1+t^4}\mathrm{d}t > 0\)

Since \(\frac{1}{x}∉ℒ^1(ℝ)\), \(f\) is not integrable on \(ℝ^2\).

Example 8.8. See Example 8.8 for an application of 8.3 using comparison.

Theorem 8.9. Let \(E⊂ℝ^2\) be measurable and set \(E'=\{(r,θ):0⩽r,0⩽θ<2 π,(r\cos θ,r\sin θ)∈E\}\)

Given a function \(f:E→ℝ\) be measurable, we define \(g:E'→ℝ,g(r,θ)=f(r\cos θ,r\sin θ)\).

Then \(f\) is integrable over \(E\) iff \(g\) is integrable over \(E'\) and \(\int_E f(x,y)\mathrm{d}(x,y)=\int_{E'}g(r,θ)r\mathrm{d}(r,θ)\).

Proof. (ad hoc) Check the theorem holds for characteristic functions on rectangles then prove the result for open sets, apply MCT.

Best way to prove is to apply a general theorem of product of measure spaces. ◻

Note \(\|(x,y)\|=(x^2+y^2)^{1/2}\),

\[\int_{B_{ℝ^2}(0,1)}\|(x,y)\|^{α}\mathrm{d}(x,y)<∞⇔\int_0^{2 π}\int_0^1 r^{α}r\mathrm{d}r\mathrm{d}θ<∞⇔r^{α+1}∈ℒ^1((0,1))⇔α+1 >-1⇔α >-2\]

Similarly

\[\int_{ℝ^2 ∖ B_{ℝ^2}(0,1)}\|(x,y)\|^{α}\mathrm{d}(x,y)<∞⇔α<-2\]

Example 8.10. (revisiting 8.6)\(\int_0^∞e^{-x^2}\mathrm{d}x=\frac{\sqrt{π}}{2}\)

Set \(E=(0,∞)×(0,∞)\), \(f(x,y)=e^{-x^2-y^2}⩾0\)

\[\int_0^∞\left(\int_0^∞e^{-x^2-y^2}\mathrm{d}y\right)\mathrm{d}x=\left(\int_0^∞e^{-x^2}\mathrm{d}x\right)^2\]

By Thm 8.9, \(\left(\int e^{-x^2}\mathrm{d}x\right)^2=\int_0^{\frac{π}{2}}\int_0^∞re^{-r^2}\mathrm{d}r\mathrm{d}θ=\frac{π}{4}\) by Baby MCT.

Therefore \(\int_0^∞e^{-x^2}=\frac{\sqrt{π}}{2}\)

Example 8.11. Example 8.7 in polar form the corresponding \(g\) is

\(g(r,θ)=\frac1r⋅\frac{\sinθ \cosθ}{\sin^4θ+\cos^4θ}\) not integrable over [0, ∞) × [0, 2π), because r⁻¹ is not integrable over [0, ∞).

Example 8.12. omitted

Given a change of variables \(T:(u,v)↦(x,y)\)

\(x,y\) is differentiable wrt \(u,v\).

form Jacobian \(J_T=\pmatrix{\frac{∂x}{∂u}&\frac{∂x}{∂v}\\\frac{∂y}{∂u}&\frac{∂y}{∂v}}=\frac{∂(x,y)}{∂(u,v)}\)

Theorem 8.13. \(E'⊂ℝ^2\) open set, \(T:E'→ℝ^2\) differentiable and injective with \(E=T(E')\). Let \(f:E→ℝ\) be measurable. Then \(f∈ℒ^1(E)⇔f∘T∈ℒ^1(E')\) and in that case \(\int_E f=\int_{E'}f∘T |\det J_T |\).

More generally, given \(σ\)-finite measure spaces \((Ω_i,ℱ_i,μ_i)_{i=1,2}\) can define \((Ω_1×Ω_2,ℱ_1⊗ℱ_2,μ_1×μ_2)\). \(ℱ_1⊗ℱ_2\) should contain \(E_1×E_2\) for \(E_i∈ℱ_i\). Define an outer measure \(μ^*\) on \(Ω_1×Ω_2\) by

\[μ^*(A)=\inf\left\{\sum_{n=1}^∞μ_1(E_{1,n})μ_2(E_{2,n}):A⊂\bigcup_{n=1}^∞E_{1,n}×E_{2,n},E_{i,n}∈ℱ_i\right\}\] See §6.1, 6.3 in Stein & Shakarchi.

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\(C[-1,1]\) complete with sup norm

not complete in \({‖f‖}_1=\int_{-1}^1{|f|}\)

\(f_n=\left\{\begin{array}{ll}1&x >\frac{1}{n}\\nx&|x|⩽\frac{1}{n}\\-1&x<-\frac{1}{n}\end{array}\right.\)

does not converge in \(C([-1,1],{‖⋅‖}_1)\)

9 \(L^p\)-spaces

For \(f\) integrable on a measurable set \(E\).

Define \({‖f‖}_1=\int{|f|}\)

\(‖α f‖=|α|‖f‖\)

\(‖f+g‖_1⩽‖f‖_1+‖g‖_1\)

This is almost a norm-space:

• integrable functions (value can take ∞) aren't a vector space (because $∞-∞$ undefined), but the integrable functions \(E→ℝ\) are a vector space.

• \(‖f‖_1=0⇔f=0\) a.e.

so \({‖⋅‖}_1\) is only a semi-norm: \(‖f‖⩾0∀f\). Fix by identifying functions which agree a.e.

Write \(f∼ g\) if \(f=g\) a.e. and \([f]\) for the equivalence class of \(f\)

Let \(L^1(E)=\{[f]:f∈ℒ^1(E)\}=ℒ^1(E)/∼\). This is a vector space.

\[L^1(E)=\{f:E→ℝ|f\text{ is integrable}\}/[0]\text{ as a vector space}\]

We drop the notation \([f]\) immediately and typically write \(f∈L^1(E)\).

\(L^1(E)\) is a normed space with \(‖f‖_1=\int_E|f|\)

\(d_1(f,g)=‖f-g‖_1\) and \(f_n→f\) in \(L^1⇔‖f_n-f‖_1→0\).

Goal: 9.6: \(L^1(E)\) is complete

But first \(L^p(E),p≠1\)

Fix \(p > 0\), set \(ℒ^p(E)\) to be the set of measurable functions \(f:E→ℝ\) such that \(|f|^p\) is integrable.

If \(f,g∈ℒ^p(E)\), \(|f+g|^p⩽2^p\max(|f|,|g|)^p⩽2^p(|f|^p+|g|^p)\)

By comparison, \(f+g∈ℒ^p(E)\)

\(ℒ^p(E)\) is a vector space.

Define \(‖f‖_p=\left(\int|f|^p\right)^{1/p}\) so \(‖α f‖=|α|‖f‖_p\)

\(‖f‖_p=0\) for \(f=0\) a.e.

\(L^p(E)=\{[f]:f∈ℒ^p(E)\}\) where \(f∼ g⇔f=g\) a.e. in \(L^p(E)\)

\(=ℒ^p(E)/[0]\) and \(‖[f]‖_p=‖f‖_p\) is well-defined

For \(p=2\), consider \(L^2_ℂ(E)\)

\(f,g∈ℒ^2_ℂ(E)⇒|f\bar{g}|⩽\frac{1}{2}(|f|^2+|g|^2)\)

\(∴f\bar{g}\) is integrable and can define \(⟨f,g⟩=f\bar{g}\)

\(∴L^p(E)\) inherits well-defined inner product \(⟨f,g⟩=\int_E f\bar{g}\)

\(⟨f,f⟩=\int|f|^2=‖f‖_2^2⩾0\) with inequality iff \(f=0\) in \(L^2(E)\)

For \(p=2\), \(‖f+g‖_2⩽‖f‖_2+‖g‖_2\) is a consequence of Cauchy-Schwarz

For \(0<p<1\), \(‖⋅‖_p\) does not satisfy the triangle rule.

Proposition 9.1. (Minkowski's inequality)Let \(E\) be a measurable set, \(1⩽p<∞\), \(f,g∈L^p(E)\), then \(‖f+g‖_p⩽‖f‖_p+‖g‖_p\). Therefore \((L^p(E),‖⋅‖_p)\) is a normed space for \(1⩽p<∞\).

Proof. Assume \(f,g≠0\). Set \(α=‖f‖_p\), \(β=‖g‖_p > 0\).

\(p≥1⇒f(r)=r^p\) is convex, ie. \((λ s+(1-λ)t)^p⩽λ s^p+(1-λ)t^p\) for all \(0<λ<1,s,t∈ℝ\).

Fix \(x∈E\). Let \(λ=\frac{α}{α+β}\), \(s=\frac{|f(x)|}{α}\), \(t=\frac{|g(x)|}{β}\),

\[\left(\frac{|f(x)|+|g(x)|}{α+β}\right)^p⩽\frac{1}{α+β}\left(\frac{α}{α^p}|f(x)|^p+\frac{β}{β^p}|g(x)|^p\right)\]

\(∴|f+g|^p⩽(|f|+|g|)^p⩽(α+β)^{p-1}\left(\frac{|f|^p}{α^{p-1}}+\frac{|g|^p}{β^{p-1}}\right)\)

Integrating\[‖f+g‖_p^p⩽(α+β)^p\]taking \(p\)th roots gives the result. ◻

can also consider \(L^{∞}\)

Let \(ℒ^∞(E)\) be essentially bounded functions[the set of measurable functions \(f:E→ℝ\) such that \(∃C > 0:|f|⩽C\) a.e.]

\[‖f‖_{∞}=\operatorname{esssup}|f|=\inf\{C:|f|⩽C\text{ a.e.}\}\]

this is a semi-norm on \(ℒ^∞(E)\) which induces the normed \(L^∞(E)\).

Proposition 9.2. (Hölder's inequality)Let \(p,q∈(1,∞)\) with \(\frac{1}{p}+\frac{1}{q}=1\). [The pair \((p,q)\) are called Hölder conjugates]

Let \(E\) be measurable, if \(f∈L^p(E),g∈L^q(E)\), then \(fg∈L^1(E)\) and \(‖fg‖_1⩽‖f‖_p‖g‖_q\).

Proof. in the notes—its concavity trick ◻

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Corollary 9.3. Let \(a<b\) and \(E=(a,b)\) and \(1⩽p_1<p_2<∞\)

If \(f∈L^{p_2}(a,b)\) then \(f∈L^{p_1}(a,b)\) and \({‖f‖}_{p_1}⩽(b-a)^{\frac{1}{p_1}-\frac{1}{p_2}}{‖f‖}_{p_2}\)

Proof. Apply Hölder to \({|f|}^{p_1}\) and \(χ_{(a,b)}\) with \(p=\frac{p_2}{p_1},q=\left(1-\frac{1}{p}\right)^{-1}\)

\({‖f‖}_{p_1}^{p_1}=\int|f|^{p_1}⩽‖|f|^{p_1}‖_p‖χ_{(a,b)}‖_q=\left(\int|f|^{p_1 p}\right)^{\frac{1}{p}}(b-a)^{1-\frac{p_1}{p_2}}\)

\(∴‖f‖_{p_1}⩽‖f‖_{p_2}(b-a)^{\frac{1}{p_1}-\frac{1}{p_2}}\) ◻

This holds whenever \(μ(E)<∞\)

\(∴\) in this case \(L^{p_2}(E)⊂L^{p_1}(E)\) and the inclusion is a continuous linear map. If \(f_n→f\) in \(L^{p_2}(E)\) ie. \(‖f_n-f‖_{p_2}→0\) then \(‖f_n-f‖_{p_1}→0\) so \(f_n→f\) in \(L^{p_1}(E)\)

Warning. This fails if \(μ(E)=∞\), and \(\frac{1}{x}∈L^2(1,∞)\) is not contained in \(L^1(1,∞)\).

Goal:\(L^p(E)\) is complete.

Example 9.4. 1. Convergence a.e. does not imply convergence in \(L^p\)-norm: If \(f_n(x)=n^2 x^n(1-x) (0⩽x⩽1)\), then \(f_n(x)→0\) a.e., but \(‖f_n‖_1→1\).

2. \(∃(f_n)^{∞}_{n=1}\) in \(L^1[0,1]\) s.t. \({‖f_n‖}_1→0\) but \(f_n↛0\) a.e.

\(f_1=χ_{[0,1]},f_2=χ_{\left[0,\frac{1}{2}\right]},f_3=χ_{\left[\frac{1}{2},1\right]},f_4=χ_{\left[0,\frac{1}{4}\right]}\)

By construction \({‖f_n‖}_1→0\), but \((f_n(x))_{n=1}^{∞}\) does not converge for any \(x\).

Note: there exists subsequence \((f_{n_k})\) such that \((f_{n_k})\) such that \(f_{n_k}→0\) a.e.

Theorem 9.5. (Riesz–Fischer)Let \(p∈[1,∞)\) and \(E\) be a measurable set. Suppose \((f_n)_{n=1}^{∞}\) is Cauchy in \(L^p(E)\) [ie. \(∀ε>0,∃N\) such that \(n,m⩾N\), \(‖f_n-f_m‖_p<ε\)]. Then \(∃f∈L^p(E)\) such that

Proof. \(∃n_1<n_2<⋯\) such that \(‖f_{n_{r+1}}-f_{n_r}‖_p<2^{-r}\). Consider \(g_k(x)=|f_{n_1}(x)|+\sum_{r=1}^k|f_{n_{r+1}}(x)-f_{n_r}(x)|\) and \(g(x)=|f_{n_1}(x)|+\sum_{r=1}^{∞}|f_{n_{r+1}}(x)-f_{n_r}(x)|\)

\(g_r(x)\nearrow g(x)\) and \(‖g_k‖_p⩽‖f_{n_1}‖_p+\sum_{r=1}^k 2^{-r}⩽‖f_{n_1}‖_p+1\) for all \(k\).

By MCT, \(\int|g|^p⩽(‖f_{n_1}‖_p+1)^p<∞\) ∴\(g∈L^p\)

∴ \(|g|^p\) and hence \(|g|\) is finite almost everywhere.

Let \(f(x)=f_{n_1}(x)+\sum_{r=1}^∞f_{n_{r+1}}(x)-f_{n_r}(x)\) which is absolutely convergent a.e. so converges a.e. ie. \(f_{n_r}(x)→f(x)\) a.e. and \(f∈L^p(E)\) by comparison. Hence 1 holds.

Certainly \(|f_{n_r}(x)-f(x)|→0\) a.e. As \(|f(x)-f_{n_k}(x)|^p=\left|\sum_{r=k}^∞f_{n_{r+1}}(x)-f_{n_r}(x)\right|^p⩽g(x)\).

As \(g⩾0\) and \(\int g^p<∞\). By DCT \(\int|f-f_{n_k}|^p→0\) ie. \(‖f_{n_k}-f‖^p_p→0\)

So \(f_{n_k}→f\) in \(L^p(E)\).

By a standard metric spaces fact, every Cauchy sequence with a convergent subsequence converges, so \(f_n→f\) in \(L^p(E)\). ◻

A normed space is complete iff absolutely convergent sequences are convergent.

Corollary 9.6. 1) For \(1⩽p<∞\), suppose \(f_n→f\) in \(L^p(E)\) then \(∃\) subsequence \((f_{n_k})\) of \((f_n)\) such that \(f_{n_k}→f\) a.e.

2) If \(f_n→f\) in \(L^p(E)\), \(f_n→g\) a.e. then \(g=f\) a.e.

Theorem 9.7. See Probability & martingales

Theorem 9.8. For \(1⩽p<∞\) the following set are dense in \(L^p(E)\) where \(E⊂ℝ\) is measurable.

1) step functions ie. for all \(f∈L^p(E)\) for all \(ε>0\) \(∃φ\) a step function such that \(‖f-φ‖_p<ε\)

See Stein & Shakarchi §2.4

2) Let \(E\) be an interval. \(∀f∈L^p(E),∀ε>0,∃g:E→ℝ\) continuous and \(K⊂E\) compact such that \(g(x)=0\) if \(x∉K\) and \(‖f-g‖_p< ε\)

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Theorem 9.8.For \(1⩽p<∞\) and the following are dense in \(L^p(I)\), \(I\) an interval

• step functions
• continuous functions with compact support
• test functions

Translation is continuous in $L^p$: For \(f:ℝ→ℝ\) measurable, define \(f_s(x)=f(x-s)\). If \(f∈L^p(ℝ)\) for \(1⩽p<∞\), then \({‖f_s-f‖}_p→0\) as \(s→0\).

Hint: cannot always find dominating function for $f$ suggests that it is naive and wrong to use dominated convergence theorem to $f_s→f$.

Proof. Fix \(ε>0\), by Thm 9.8 we can find \(g:ℝ→ℝ\) continuous and with compact support (ie, \(∃K⊂ℝ\) compact such that \(g(x)=0∀x∉K\)) such that \({‖f-g‖}_p⩽\frac{ε}{3}\). Since translation doesn't change $p$-norm, \({‖f_s-g_s‖}_p⩽\frac{ε}{3}\).

As \(g\) is continuous on \(K\), it is uniformly continuous.

\[\int_ℝ{|g-g_s|}^p→0\text{ as }s→0\text{ by the estimation lemma}\]

Fix \(δ>0\) such that \({‖g-g_s‖}_p⩽\frac{ε}{3}\) if \({|s|}<δ\), so \({‖f-f_s‖}_p⩽{‖f-g‖}_p+{‖g-g_s‖}_p+{‖g_s-f_s‖}_p⩽\frac{3 ε}{3}=ε\) ◻

Definition. For \(f∈L^1(ℝ)\), Fourier transform \(\hat{f}(s)=\int_{ℝ}f(x)e^{-isx}dx\)

Theorem. Let \(f∈L^1(ℝ)\)

1. \({|\hat{f}(s)|}⩽{|f|}_1\) by comparison with \(|f|\)
2. \(\hat{f}\) is continuous. By continuous parameter DCT (using \(|f|\) as the dominating function)

3. [Riemann-Lebesgue lemma] \(\hat{f}(s)→0\) as \({|s|}→∞\)

   \(\hat{}:L^1(ℝ)→C_0(ℝ)=\{g:ℝ→ℝ\text{ continuous and }g(s)→0\text{ as }{|s|}→∞\}\) is linear and continuous from \((L^1(ℝ),‖⋅‖_1)→(C_0(ℝ),‖⋅‖_{∞})\).

   As if \(f_n→f\) in \(L^1\) then \(\sup_{s∈ℝ}{|\widehat{f_n}(s)-\hat{f}(s)|}⩽{‖f_n-f‖}_1→0\) ie. \({‖\widehat{f_n}-\hat{f}‖}_\text{sup}→0\).

   Proof of 3. \(\hat{f}(s)=\int_ℝf(x)e^{-isx}dx=-\int_ℝf\left(x-\fracπs\right)e^{-isx}dx=\frac12\int_ℝ\left[f(x)-f\left(x-\fracπs\right)\right]e^{-isx}dx\)

   ∴\({|\hat{f}(s)|}⩽\frac12\left\|f-f_{\fracπs}\right\|_1→0\) as \({|s|}→∞\) by continuity of translations in \(L^1(ℝ)\). ◻

   Alternatively, when $f=χ_{(a,b)},\hat{f}(s)=\frac{i\left(e^{-i s b}-e^{-i s a}\right)}{s}→0$ as ${|s|}→∞$. This extends to step functions, by linearity. For general $f∈ ℒ^1(ℝ)$ and $ε>0$, there is a step function $φ$ such that ${‖f-φ‖}_1<ε$ by Theorem 9.8 , and there exists $K$ such that ${|\hatφ(s)|}<ε$ whenever ${|s|}>K$. Then$${|\hat{f}(s)|}≤{|\hat{f}(s)-\hatφ(s)|}+{|\hatφ(s)|}≤{‖f-φ‖}_1+{|\hatφ(s)|}<2 ε$$

4. Let \(g(x)=xf(x)\). Suppose \(g∈L^1(ℝ)\).

   Then \(\hat{f}\) is differentiable and \(\hat{f}'(s)=-i\hat{g}(s)\)

   Proof. \(g\) is the dominating function for the differentiation under integral theorem. ◻

5. If \(\hat{f}\) has continuous derivative \(f'∈L^1(ℝ)\) then \(\widehat{f'}(s)=is\hat{f}(s)\)

   Proof. Using integration by parts over intervals $[a_n,b_n]$ where $a_n→-∞,f(a_n)→0,b_n→∞$ and $f(b_n)→0$.

Topological group \(G\) which is locally compact and Hausdorff, \(∃\)a Borel measure \(μ\) on \(G\) (Harr measure) such that \(μ(E)=μ(g⋅E),g⋅E=\{gh:h∈E\}\) for all measurable set \(E\), if \(G\) is compact \(μ(G)=1\).

10 Absolutely continuous functions  

[§3 of Stein]

Question. If \(f∈L^1(ℝ)\) os \(F(x)=\int_0^x f(y)\mathrm{d}y\) differentiable almost everywhere? Is \(F'(x)=f(x)\) almost everywhere?

Theorem. (Lebesgue differentiation theorem)Yes.

Need \(\frac{1}{h}\int_x^{x+h}f(y)\mathrm{d}y→f(x)\) almost everywhere.

Question. Let \(F:[a,b]→ℝ\). Can we describe 1) when \(F\) is differentiable almost everywhere?

When 1 holds 2) \(F(b)-F(a)=\int_a^b F'(x)\mathrm{d}x\)?

Yes.

Definition. Say \(F:[a,b]→ℝ\) is absolutely continuous if for all \(ε>0\) there is \(δ>0\) such that

\[\sum_{r=1}^n|F(b_r)-F(a_r)|<ε\text{ whenever }(a_1,b_1),…,(a_r,b_r)\text{ are disjoint intervals in }[a,b]\text{ such that }\sum_{r=1}^n|b_r-a_r|<δ\]

Theorem 10.2. \(f∈L^1(ℝ)\), and set \(F(x)=\int_0^x f(y)\mathrm{d}y\), then \(F\) is absolutely continuous.

Theorem 10.3. Let \(F\) be absolutely continuous on \([a,b]\), then \(F\) is differentiable almost everywhere on \([a,b]\) and \(\int_a^b F'(x)\mathrm{d}x=F(b)-F(a)\)

Proof. See Stein. ◻