1. Let $m$ denote the Lebesgue measure on $ℝ$.
   1. 1. Define the Lebesgue outer measure $m^*$ and what it means for $E ⊆ ℝ$ to be Lebesgue measurable.
      2. Show that for any subsets $E_1, E_2, …$ of $ℝ$, $$ m^*\left(⋃_{n=1}^∞ E_n\right) ⩽ \sum_{n=1}^∞ m^*\left(E_n\right) $$
      3. Let $F_1 ⊇ F_2 ⊇ ⋯$ be Lebesgue measurable sets with $m\left(F_1\right)<∞$. Use countable additivity to show that $$ m\left(⋂_{n=1}^∞ F_n\right)=\lim _{n →∞} m\left(F_n\right) . $$
      4. Let $\left(E_n\right)_{n=1}^∞$ be a sequence of Lebesgue measurable sets with $\sum_{n=1}^∞ m\left(E_n\right)<∞$ and set $E=⋂_{n=1}^∞ ⋃_{r ⩾ n} E_r$. Show that $m(E)=0$.
   2. Let $\left(f_n\right)_{n=1}^∞$ be a sequence of measurable functions from $[0,1]$ to $ℝ$.
      1. Show that for each $n ∈ ℕ$, there exists $a_n ∈ ℝ$ with $$ m\left(\left\{x ∈[0,1]:\left|f_n(x)\right|>a_n / n\right\}\right)<2^{-n} $$
      2. Deduce that for almost all $x ∈[0,1]$, $$ \frac{f_n(x)}{a_n} → 0   \text { as } n →∞ $$
2. 1. State Fubini’s and Tonelli’s theorems for a measurable function $f:ℝ^2→ℝ$
   2. Let $a>1$.
      1. Show that the function $f(x, y)=e^{-x y}$ is integrable on $[0,∞)×[1, a]$.
      2. Show that $$ ∫_0^∞ \frac{e^{-x}-e^{-a x}}{x} \mathrm{~d} x=\log a $$
      3. Is the function $g(x, y)=e^{-x y}-a e^{-a x y}$ integrable over $[0,1]×[1,∞)$ ? Justify your answer.
   3. Let $f:[0,1] → ℝ$ be measurable. Suppose that the function $$ g(x, y)=f(x)-f(y) $$ is integrable on $[0,1]×[0,1]$. Must $f$ be integrable on $[0,1]$ ? Justify your answer.
3. 1. 1. Define a simple function $ϕ:ℝ→[0,∞]$, and state an expression for the Lebesgue integral of $ϕ$.
      2. Let $f:ℝ→[0,∞]$ be measurable. Explain how to define its Lebesgue integral $∫_ℝf$.
      3. Let $f:ℝ→[-∞,∞]$ be measurable. What does it mean for $f$ to be Lebesgue integrable, and in that case define $∫_ℝf$.
      4. Suppose that $f,g:ℝ→[-∞,∞]$ are Lebesgue integrable with $f(x)⩽g(x)$ for all $x$. Show that $∫_ℝf⩽∫_ℝg$.
   2. Determine whether the following measurable functions are Lebesgue integrable over the specified set. Justify your answers.
      1. $f(x)=\frac{\cos x}{x}$ over $(1,∞)$.
      2. $g(x)=\frac{\cos(1/x)}{x}$ over $(0,1)$.
      3. $h(x)=\frac{\sin x}{e^x-1}$ over $(0,∞)$.
   3. Write $ℒ^1(ℝ)=\{f:ℝ→ℝ:f\text{ is integrable}\}$ and define ${‖f‖}_1=∫_ℝ{|f|}$ for $f∈ℒ^1(ℝ)$
      1. Let $\left(f_n\right)_{n=1}^∞$ and $\left(g_n\right)_{n=1}^∞$ be sequences in $ℒ^1(ℝ)$, and suppose that $f,g:ℝ→ℝ$ are such that $f_n→f$ and $g_n→g$ almost everywhere, and that $g∈ℒ^1(ℝ)$. Suppose that $\left|f_n\right|⩽g_n$ for all $n∈ℕ$ and $\left\|g_n\right\|_1→{‖g‖}_1$. Show $f∈ℒ^1(ℝ)$ and $∫_ℝf_n→∫_ℝf$.
      2. Show that $\left\|f_n-f\right\|_1→0$.

Solution

1. 1. 1. $$ m^*(E)=\inf \left\{\sum_{n=1}^∞ m\left(I_n\right): I_n \text { are intervals and } E ⊆ ⋃_{n=1}^∞ I_n\right\}, $$ where $m\left(I_n\right)=b_n-a_n$ when $I_n$ has end points $a_n<b_n$. $E$ is Lebesgue measurable iff $$ m^*(A)=m^*(A ∩ E)+m^*(A ∖ E) $$ for all $A ⊂ ℝ$.
      2. Let $ϵ>0$ and for each $n$ find intervals $\left(I_{n, r}\right)$ such that $E_n ⊂ ⋃_{r=1}^∞ I_{n, r}$ and $$ \sum_{r=1}^∞ m\left(I_{n, r}\right)<m^*\left(E_n\right)+2^{-n} ϵ $$ Then $⋃_{n=1}^∞ E_n ⊆ ⋃_{n, r=1}^∞ I_{n, r}$ so that $$ m^*\left(⋃_{n=1}^∞ E_n\right) ⩽ \sum_{r, n=1}^∞ m\left(I_{n, r}\right)<\sum_{n=1}^∞ m^*\left(E_n\right)+ϵ . $$ Since $ϵ>0$ was arbitrary, $m^*\left(⋃_{n=1}^∞ E_n\right) ⩽ \sum_{n=1}^∞ m^*\left(E_n\right)$.
      3. Set $A_n=F_n ∖ F_{n+1}$ so that $A_n$ are disjoint and $⋃_{n=1}^∞ A_n=F_1 ∖$ $⋂_{n=1}^∞ F_n$. Then by countable additivity $$ m\left(F_1 ∖ ⋂_{n=1}^∞ F_n\right)=\sum_{n=1}^∞\left(m\left(F_n ∖ F_{n+1}\right)\right) . $$Also (finite) addivity gives $m\left(F_1\right)=m\left(F_1 ∖ ⋂_{n=1}^∞ F_n\right)+m\left(⋂_{n=1}^∞ F_n\right)$ and $m\left(F_{n+1}\right)+$ $m\left(F_n ∖ F_{n+1}\right)=m\left(F_n\right)$ So $$ \sum_{n=1}^∞\left(m\left(F_n\right)-m\left(F_{n+1}\right)\right)=m\left(F_1\right)-\lim _{n→∞} m\left(F_n\right) $$ Since $m\left(F_1\right)<∞$, the result follows.
         This solution argues directly from countable additivity. I would expect many students to first prove the bookwork result $m\left(⋃_{n=1}^∞ E_n\right)=\lim m\left(E_n\right)$ when $E_1 ⊆ E_2 ⊆ …$ and then set $E_n=F_1 ∖ F_n$.
      4. We note that $m\left(⋃_{n=1}^∞ E_n\right) ⩽ \sum_n m\left(E_n\right)<∞$ (using (ii)). Hence we can use (iii) to obtain $$ m(E)=\lim _n m\left(⋃_{r ⩾ n} E_r\right) ⩽ \lim _n \sum_{r ⩾ n} m\left(E_r\right)=0 $$
   2. 1. Fix $n ∈ ℕ$, and let $F_r=\left\{x ∈[0,1]:\left|f_n(x)\right|>r / n\right\}$. Since $F_1 ⊃ F_2 ⊃…$ and $⋂_r F_r=∅$, there exists some $r$ with $m\left(F_r\right)<2^{-n}$ (by (a)(iii)). Take $a_n=r$.
      2. Let $E_n=\left\{x ∈[0,1]:\left|f_n(x)\right|>a_n / n\right\}$ so that $\sum_n m\left(E_n\right)<∞$. Therefore for $E=⋂_{n=1}^∞ ⋃_{m ⩾ n} E_m$, part (a)(iv) gives $m(E)=0$. If $x ∉ E$, then there exists $n$ such that $x ∉ ⋃_{m ⩾ n} E_m$, i.e. $$ \left|\frac{f_m(x)}{a_m}\right| ⩽ \frac{1}{m} \text { for all } m ⩾ n . $$ Therefore $f_m(x) / a_m → 0$ as $m →∞$ for $x ∉ E$.
         Sourced from Stein and Shakarchi chapter 1.
2. 1. Fubini: Let $f: ℝ^2 → ℝ$ be integrable. Then, for almost all $y, x ↦ f(x, y)$ is integrable, and if $F(y)$ is defined (for those $y$) by $$ F(y)=∫_{ℝ} f(x, y) d x $$ then $F(y)$ is integrable and $$ ∫_{ℝ^2} f(x, y) d(x, y)=∫_{ℝ} F(y) d y $$ Similarly $$ ∫_{ℝ^2} f(x, y) d(x, y)=∫_{ℝ}\left(∫_{ℝ} f(x, y) d y\right) d x $$ Tonelli: Such a function is integrable on $ℝ^2$ if either of the repeated integrals $$ ∫_{ℝ}\left(∫_{ℝ}{|f(x, y)|}d x\right) d y \text { or } ∫_{ℝ}\left(∫_{ℝ}{|f(x, y)|}d y\right) d x $$ is finite (and hence Fubini's theorem applies to $|f|$ and $f$).
   2. 1. Note $e^{-x y} ⩾ 0$ on this range. For any $1 ⩽ y ⩽ a$, $$ ∫_0^∞ e^{-x y} d x=\lim _{n →∞} ∫_0^n e^{-x y} d x=\lim _{n →∞}\left(\frac{e^{-n y}}{y}-\frac{1}{y}\right)=\frac{1}{y} $$ using the monotone convergence theorem, and the fundamental theorem of calculus. Then $$ ∫_1^a ∫_0^∞ e^{-x y} d x d y=\log a $$ by the fundamental theorem of calculus. By Tonelli's theorem, $e^{-x y}$ is integrable on the specified domain.
         [Note that the calculation of the integral as $\log a$ really belongs to the next subpart.]
      2. By Tonelli's theorem $$ \log a=∫_0^∞ ∫_1^a e^{-x y} d y d x=∫_0^∞ \frac{e^{-x}-e^{-a x}}{x} d x . $$ where the second integral is calculated by the fundamental theorem of calculus.
      3. Firstly, for any $y ∈[1,∞)$ $$ ∫_0^1\left(e^{-x y}-a e^{-a x y}\right) d x=\frac{e^{-a y}-e^{-y}}{y} $$ by the FTC. So $$ ∫_1^∞ ∫_0^1\left(e^{-x y}-a e^{-a x y}\right) d x d y=∫_1^∞ \frac{e^{-a y}-e^{-y}}{y} d y $$ On the other hand, for $x ∈(0,1)$, \begin{aligned} ∫_1^∞\left(e^{-x y}-a e^{-a x y}\right) d y & =\lim _{n →∞} ∫_1^n\left(e^{-x y}-a e^{-a x y}\right) d y \\ & =\lim _{n →∞}\left(\frac{-e^{-n x}+e^{-a n x}}{x}-\frac{-e^{-x}+e^{-a x}}{x}\right) \\ & =\frac{e^{-x}-e^{-a x}}{x} \end{aligned} by the MCT (applied separately to each term), and the FTC. Thus $$ ∫_0^1 ∫_1^∞\left(e^{-x y}-a e^{-a x y}\right) d y d x=∫_0^1 \frac{e^{-x}-e^{-a x}}{x} d x $$ If $g(x, y)$ is integrable on the specified domain, then $$ ∫_0^1 \frac{e^{-x}-e^{-a x}}{x} d x=∫_1^∞ \frac{e^{-a y}-e^{-y}}{y} d y $$ by Fubini's theorem. Replacing $y$ by $x$ in the right hand integral, this gives $$ ∫_0^∞ \frac{e^{-x}-e^{-a x}}{x} d x=0 $$ contradicting the previous part. Thus $g$ is not integrable.
         This exercise is sourced from Priestley, where it's an omitted example.
   3. By Fubini, for almost all $y, x ↦{|f(x)-f(y)|}$ is integrable. Fix such a $y_0$ so $$ ∫_0^1\left|f(x)-f\left(y_0\right)\right| d x<∞ . $$ Since for all $x ∈[0,1]$, we have $$ {|f(x)|}⩽\left|f(x)-f\left(y_0\right)\right|+\left|f\left(y_0\right)\right|, $$comparison gives $∫^1_0 {|f (x)|} < ∞$. Therefore $f$ is integrable on $[0, 1]$.
3. 1. 1. A simple function is a measurable function which takes only finitely many values. Such a function ϕ can be written in the form $ϕ=∑^n_{i=1}α_iχ_{B_i}$ with $α_i>0$ and measureable sets $B_i$. Then $∫_ℝ ϕ=∑^n_{i=1}α_im(B_i)$.
      2. $∫_ℝf=\sup\left\{∫_ℝϕ:0⩽ϕ⩽f,ϕ\text { simple}\right\}$
      3. Let $f^{+}(x)=\max(f(x),0)$ and $f^{-}(x)=\max(-f(x),0). f$ is integrable iff $∫_ℝf^{+}<∞$ and $∫_ℝf^{-}<∞$. In this case we define $∫_ℝf=∫_ℝf^{+}-∫_ℝf^{-}$.
      4. Firstly assume that $0⩽f⩽g$. If $ϕ:ℝ→[0,∞]$ is simple and $ϕ⩽f$, then $ϕ⩽g$. Therefore $\sup\{∫_ℝϕ:0⩽ϕ⩽f,ϕ\text{ simple}\}⩽\sup\{∫_ℝϕ:0⩽ϕ⩽g,ϕ\text{ simple}\}$,and so $$ ∫_ℝf⩽∫_ℝg . $$ Now in general, we have $0⩽f^{+}⩽g^{+}$ so $∫_ℝf^{+}⩽∫_ℝg^{+}$ and $0⩽g^{-}⩽f^{-}$ so $∫_ℝg^{-}⩽∫_ℝf^{-}$. Assembling this gives $∫_ℝf⩽∫_ℝg$.
   2. 1. We have $$ ∫_{n π}^{(n+1)π}\left|\frac{\cos x}{x}\right|⩾\frac{1}{(n+1)π}∫_{n π}^{(n+1)π}{|\cos x|}=\frac{2}{(n+1)π}. $$ Since $\sum_{n=1}^∞\frac{2}{(n+1)π}=∞,f$ is not integrable over $[π,∞)$, by the baby MCT (and so not integrable over $(1,∞)$ either).
      2. We substitute using the monotonic bijection $h:(0,1)→(1,∞)$, $h(x)=1/x$ which has continuous derivative $h'(x)=-x^{-2}$. Then as the $f$ from the previous subpart is not integrable on $(1,∞)$, it follows that $(f∘h)⋅h'$ is not integrable on $(0,1)$. But $f(h(x))h'(x)=-\frac{\cos(1/x)}{x}=-g(x)$. Thus $g$ is not integrable.
         [We do similar substitutions on sheet 3.]
      3. For $x⩾1$, we have $\left|\frac{\sin x}{e^x-1}\right|⩽\frac{2}{e^x}$. We have $∫_1^n\frac{2}{e^x}=2\left(1-e^{-n}\right)→2$. Therefore by the baby MCT, and comparison $h$ is integrable on $[1,∞)$. Noting that $\lim _{x→0}f(x)=1$ by L'Hopital, if we define $h(0)=1$, $h$ restricts to a continuous function on $[0,1]$, which is therefore bounded and integrable on this domain. Hence $h$ is integrable on $[0,∞)$.
   3. 1. $f$ is an a.e. limit of measurable functions so measurable. As $\left|f_n\right|⩽g_n$, we have ${|f|}⩽g$ a.e. and hence $f∈ℒ^1(ℝ)$. We have $g_n±f_n⩾0$ and $g_n±f_n→g±f$ a.e. Applying Fatou: $$ ∫(g±f)⩽\liminf∫\left(g_n±f_n\right) $$ As $g_n⩾0$, and $\left\|g_n\right\|_1→{‖g‖}_1$, we have $∫ g_n→∫ g$. Therefore [sum of liminf] $$ ±∫ f⩽\liminf∫\left(±f_n\right) $$ i.e. $$ ∫ f⩽\liminf∫ f_n⩽\limsup∫ f_n⩽∫ f $$ Therefore $\lim∫ f_n=∫ f$.
         Potentially hard, but based on proof of the DCT so reworked bookwork in form too. This could be made easier by adding a hint to use Fatou
      2. We have $$ \left|f_n-f\right| ⩽\left|f_n\right|+{|f|} ⩽ g_n+{|f|} → g+{|f|} $$ almost everywhere. As $g+{|f|}$ is integrable, and $\left\|g_n+{|f|}\right\|_1=∫\left(g_n+{|f|}\right) →∫(g+{|f|})={\|g+{|f|}\|}_1$, we can apply the previous part to $\left|f_n-f\right| → 0$ a.e. Therefore $$ \left\|f_n-f\right\|_1=\lim ∫\left|f_n-f\right|=0, $$ as required.