Measurable functions and the lebesgue integral

1. $ω_n≥0⇒μ_ω(E)≥0$. Clearly $μ_ω(∅)=0$. For disjoint subsets $E_i$ of ℕ,\[μ\left(\bigcup_{i=1}^∞E_i\right)=\sum_{i=1}^∞\sum_{n∈E_i} ω_n=\sum_{i=1}^∞μ\left(E_i\right)\]
2. Since ν is countable additive, $ν(E)=\sum_{n∈E}ν(\{n\})=\sum_{n∈E}ω_n=μ_ω(E)$.

If $μ\left(A_1\right)<∞$, $\left(A_1∖A_n\right)$ is an increasing sequence of sets in ℱ,\begin{align*}μ(A_1)-μ\left(\bigcap_{n=1}^∞A_n\right)&=μ\left(A_1∖\bigcap_{n=1}^∞A_n\right)&\text{since $\bigcap_{n=1}^∞A_n$ and $A_1∖\bigcap_{n=1}^∞A_n$ are disjoint}\\&=μ\left(\bigcup_{n=1}^∞(A_1∖A_n)\right)&\text{De Morgan's law}\\&=μ(A_1)-\lim_{n→∞}μ(A_n)&\href{https://courses.maths.ox.ac.uk/pluginfile.php/29962/mod_resource/content/6/integration.pdf#page=10&zoom=auto,-78,283}{\text{Proposition 3.2.2}}\\⇒μ\left(\bigcap_{n=1}^∞A_n\right)&=\lim_{n→∞}μ(A_n)\end{align*} If $μ\left(A_1\right)=∞$ this is false: in (ℝ, ℳ_Bor, m), let $A_n=[n,∞)$, we have $\lim_{n→∞}m(A_n)=∞$ but $m\left(\bigcap_{n=1}^∞A_n\right)=m(∅)=0$.

1. $x∈\bigcup_{n=1}^∞\left(ℝ∖\left(b-\frac1n,∞\right)\right)⇔∃n≥1:x≤b-\frac1n⇔x<b$. So $(-∞,b)=\bigcup_{n=1}^∞\left(ℝ∖\left(b-\frac1n,∞\right)\right)$.
   If ℱ is a σ-algebra on ℝ containing $(a,∞)$ for each $a∈ℝ$, by the above identity, $(-∞,b)∈ℱ$.
   For $a< b$, $(-∞,b)∩(a,∞)=(a,b)∈ℱ$, so ℱ contains all open intervals, and hence all open subsets of ℝ, since they are countable union of open intervals.
2. $f^{-1}(∅)=∅∈ℱ⇒∅∈𝒢$.
   If $G∈𝒢$ then $f^{-1}(ℝ∖G)=ℝ∖f^{-1}(G)∈ℱ⇒ℝ∖G∈𝒢$.
   If $G_i∈𝒢$, then $f^{-1}\left(\bigcup_{n=1}^∞G_i\right)=\bigcup_{n=1}^∞f^{-1}(G_i)∈ℱ⇒\bigcup_{n=1}^∞G_i∈𝒢$.
   Hence 𝒢 is a σ-algebra. By (a) if $(a,∞)∈𝒢$ for every $a$, then $G∈𝒢$ for every $G∈ℳ_\text{Bor}$.
   When do Inverse Images of Measurable Sets Generate the σ-algebra?
   Inverse images and σ-algebras

1. $∀λ∈Λ:∅∈ℱ_λ⇒∅∈⋂_{λ∈Λ}ℱ_λ$.
   If $E∈⋂_{λ∈Λ}ℱ_λ$ then $∀λ∈Λ:E∈ℱ_λ⇒Ω∖E∈ℱ_λ⇒Ω∖E∈⋂_{λ∈Λ}ℱ_λ$.
   If $∀n∈ℕ:E_n∈⋂_{λ∈Λ}ℱ_λ$ then $∀λ∈Λ:E_n∈ℱ_λ⇒∀λ∈Λ:\bigcup_{n∈ℕ}E_n∈ℱ_λ⇒\bigcup_{n∈ℕ}E_n∈⋂_{λ∈Λ}ℱ_λ$.
2. Let 𝒮 be the family of all σ-algebras containing ℬ. By (a) $ℱ_ℬ=⋂_{𝒢∈𝒮}𝒢$ is a σ-algebra.
   If $ℱ_ℬ'$ also satisfies the conditions, then $ℱ_ℬ⊂ℱ_ℬ'⊂ℱ_ℬ$, so $ℱ_ℬ=ℱ_ℬ'$ is unique.

1. Q3b proves $f_*(ℱ)$ is a σ-algebra on $Ω_*$, it remains to prove $f_*μ$ is a measure. Clearly $f_*μ≥0$.
   $\left(f_* μ\right)(∅)=μ\left(f^{-1}(∅)\right)=μ(∅)=0$.
   Let $E_n∈f_*(ℱ)$ be disjoint,\begin{align*}\left(f_* μ\right)\left(\bigcup^∞_{n=1}E_n\right)&=μ\left(f^{-1}\left(\bigcup^∞_{n=1}E_n\right)\right)\\&=μ\left(\bigcup^∞_{n=1}f^{-1}(E_n)\right)&&\text{inverse image of union equals union of inverse image}\\&=\sum^∞_{n=1}μ(f^{-1}(E_n))&&\text{since $f^{-1}(E_n)$ are disjoint}\\&=\sum^∞_{n=1}\left(f_* μ\right)(E_n)\end{align*}
2. 1. Since $f$ is π-periodic, $f_*\left(ℳ_\text{Bor}\right)$ consists of
      if $0∈E$, $\{x+kπ∣x∈E,k∈ℤ\}∪\{π/2+kπ∣k∈ℤ\}$;
      if $0∉E$, $\{x+kπ∣x∈E,k∈ℤ\}$;
      for all $E∈ℳ_\text{Bor},E⊂\left(-\fracπ2,\fracπ2\right)$.
      For $E∈ℳ_\text{Bor}$, $\left(f_* m\right)(E)=m\left(\bigcup_{k∈ℤ}(kπ+\arctan E)\right)=∞⋅m\left(\arctan E\right)$
      so $\left(f_* m\right)(E)=0$ if $m(E)=0$; $\left(f_* m\right)(E)=∞$ if $m(E)>0$.
   2. tan is continuous, so $f(E)∈ℳ_\text{Bor}$ for all $E∈ℳ_\text{Bor}$.
      The range of $f$ is $\left(-\fracπ2,\fracπ2\right)$, so $f_*\left(ℳ_\text{Bor}\right)$ consists of $E∪F$ for all $E∈ℳ_\text{Bor}\left(-\fracπ2,\fracπ2\right)$, $F ⊂\left(∞,-\fracπ2\right]∪\left[\fracπ2,∞\right)$.
      For $E∈f_*(ℳ_\text{Bor})$, if $E∩\left(-\fracπ2,\fracπ2\right)=∅$, then $f^{-1}(E)=∅$, $\left(f_* m\right)(E)=0$; otherwise $E∩\left(-\fracπ2,\fracπ2\right)$ is union of disjoint intervals between $a_i,b_i$ with $-\fracπ2≤a_i<b_i≤\fracπ2$, so $\left(f_* m\right)(E)=\sum\tan b_i-\tan a_i$

1. Assume $f(a)>g(a)$, since $f-g$ is continuous at $a$, $∃r>0∀x∈B(a,r):f(x)>g(x)$, and $m(B(a,r))=2r>0$, contradicting $f(x)=g(x)$ a.e.
2. ℚ, $ℝ∖ℚ$ are dense in ℝ, $∀x∈ℝ∀r>0∃p,q∈B(x,r):χ_ℚ(p)=1,χ_ℚ(q)=0$, so $χ_ℚ$ is discontinuous at any $x$.
   $χ_ℚ(x)=0$ a.e. and 0 is continuous. Warning: equal to a continuous function a.e. doesn't imply continuous a.e.
3. $χ_{(0,∞)}$ is continuous except 0, so it is continuous a.e. Suppose $g$ is a continuous function, $χ_{(0,∞)}=g$ a.e.
   Applying (a) to $I=[0,1]$ and $I=[-1,0]$, we have $g(0)=1$ and $g(0)=0$ respectively, contradiction.

1. $h(x)=x^2$ is continuous, so $f^2=h∘f$ is measurable.
2. By (a), $(f+g)^2,f^2,g^2$ are measurable. so $fg=\frac{(f+g)^2-f^2-g^2}2$ is measurable.
3. $h(x)={|x|}$ is continuous, so ${|f|}=h∘f$ is measurable.
4. By (c), $|f-g|$ is measurable, so $\max(f, g)=\frac{f+g+{|f-g|}}2$ is measurable.

1. Let $E≔\{x∈Ω: f(x)=g(x)\}$, then $Eᶜ$ is null. For $a∈ℝ$, we have \[ \{x∈Ω: f(x)>a\}=\{x∈E: f(x)>a\}∪\left\{x∈Eᶜ: f(x)>a\right\} . \] Moreover, \[ \{x∈E: f(x)>a\}=\{x∈E: g(x)>a\}=\{x∈Ω: g(x)>a\}∖\left\{x∈Eᶜ: g(x)>a\right\} . \] Since μ is a complete measure and $Eᶜ$ is a null set, $\left\{x∈Eᶜ:f(x)>a\right\}$ and $\left\{x∈Eᶜ:g(x)>a\right\}$ are measurable.
   Because $g$ is measurable, $\{x∈Ω:g(x)>a\}$ is measurable. This shows that $\{x∈E: f(x)>a\}$ is measurable, hence $\{x∈Ω:f(x)>a\}$ is measurable for every $a∈ℝ$. Therefore, $f$ is measurable.
   Real Analysis, Stein & Shakarchi, Chapt 1 Theorem 4.1, Property 6, p.30.
   completeness of measure space is necessary
2. First assume $f≥0$ and $f(x)=0$ for $x∉[0,1]$. By Prop 3.9 ∃ a sequence $(g_n)_{n=1}^∞$ of non-negative simple functions such that $g_n(x)→f(x)∀x∈ℝ$. Let $g_n=\sum_{i=1}^ka_{n,i}χ_{E_{n,i}}$ for disjoint measurable sets $(E_{n,i})_{i=1}^k$.
   By Cor 2.7 ∃ open set $O_{n,i}$ such that $E_{n,i}=O_{n,i}∖N_{n,i},μ(N_{n,i})<\frac1{kn}$.
   Let $N_n=\bigcup_{i=1}^kN_{n,i}$, then $μ(N_n)<\frac1n$. Let $N=\bigcap_{n=1}^∞N_n$, then $μ(N)=0$.
   Each $O_{n,i}$ is a finite union of open intervals, so $ϕ_n=\sum_{i=1}^ka_{n,i}χ_{O_{n,i}}$ is a step function. $∀x∉N_n:ϕ_n(x)=g_n(x)$.
   $∀x∉N:f=\lim_{n→∞}ϕ_n$. So $f=\lim_{n→∞}ϕ_n$ a.e. By Prop 3.7 $\lim_{n→∞}ϕ_n$ is measurable, then by part (a) $f$ is measurable.
   Extend to general $f$: apply the above to $fχ_{[n,n+1)}$ then use $f=\sum_{n∈ℤ}fχ_{[n,n+1)}$.
   $f^+$ and $f^-$ can be approximated by step functions $ϕ^+$ and $ϕ^-$, set $ϕ=ϕ^+-ϕ^-$.
   If a function is almost everywhere continuous then it is the limit of step functions almost everywhere.
   Stronger result ‘For every measurable function, ∃ a sequence of step functions’ See
   Real Analysis, Stein & Shakarchi, Theorem 4.3 p.32.
   MSE – Measurable Functions as Limits (a.e) of Step Functions
   Measure Integral & Probability, Capinski & Kopp, Theorem 4.25

1. $ϕ=αχ_{\{x:{|f(x)|}≥α\}}$ is a simple function and $0≤ϕ≤{|f|}$, by definition of $∫{|f|}$, \[∫ϕ≤∫{|f|}⇒m(\{x:{|f(x)|}≥α\})≤\frac1α∫{|f|}\]
2. 1. $m(\{x:{|f(x)|}=∞\})≤m(\{x:{|f(x)|}≥α\})≤\frac1α∫{|f|}$ when $α→∞$ we get $m(\{x:{|f(x)|}=∞\})=0$.
   2. By (a), $E_n=\left\{x:{|f(x)|}≥\frac1n\right\}$ is null, then $\bigcup_{n=1}^∞ E_n=\{x:f(x)>0\}$ is a countable union of null sets, so is null.
      Measure Integral & Probability, Capinski & Kopp, Theorem 4.8

Lebesgue integral of a positive function on a set of positive measure

Let $E$ be a subset of ℝ with positive Lebesgue measure, $\lambda(E)>0$. Let $f$ be a function from ℝ to ℝ which is positive on $E$, that is $f(x)>0$ for all $x\in E$. Prove that $\int_E f\,d\lambda$ is strictly positive.

Since $f$ is strictly positive on $E$, we have $$ E = \bigcup_{n \geq 1} E_n, \quad \mbox{ where } E_n = \left\{x \in E: f(x) > \frac{1}{n}\right\}. $$ Since $\lambda(E) > 0$ there is some $n$ for which $\lambda(E_n)$ is positive (otherwise $E$ would be the countable union of measure $0$ sets, implying $\lambda(E)=0$). We then have $$ \int_E f \, d\lambda \geq \int_{E_n} f \, d\lambda > \int_{E_n} \frac1n \, d\lambda = \frac{\lambda(E_n)}{n} > 0, $$ as desired.

1. $f(x)=0$ a.e. By Proposition 5.1.5 $∫_I f=0$.
2. $\lim_{ϵ→0}∫_0^{π/2-ϵ}f=-\lim_{ϵ→0}\log\sinϵ=∞$. So $f$ isn't integrable on $I$.
3. $∫_1^{n+1}{|f|}=\sum^n_{k=1}\frac1k→∞$. So $f$ isn't integrable on $I$.
   But $f$ is R-integrable, $∫_1^∞f=\ln2$.
4. $f$ is continuous so measurable and ${|f|}≤1$ on finite interval $I$, by Corollary 5.2.3, $f$ is integrable on $I$.
   $\int_0^1 \sin(1/x)\,dx = \sin(1)-\operatorname{Ci}(1)≈0.504067$
5. $\lim_{x→∞}{f(x)\over x^{-2}}=0$ and $x^{-2}$ is integrable on $(1,∞)$, by Comparison Test, $f$ is integrable on $(1,∞)$.
   $f$ is bounded on $[0,1]$, so $f$ is integrable on $[0,1]$. So $f$ is integrable on $[0,∞)$.

   ---

   Set $p=1$ in $\int_0^∞x^ne^{-px}=ℒ\{x^n\}(p)=\frac{n!}{p^{n+1}}$ we get $\int_0^∞x^ne^{-x}=n!$
   Or use the definition of Gamma function.
6. $\lim_{x→0}{f(x)\over x^{-1/2}}=0$ and $x^{-1/2}$ is integrable on $(0,1)$, so $f$ is integrable on $(0,1)$.
   $\lim_{x→∞}{f(x)\over x^{-2}}=0$ and $x^{-2}$ is integrable on $(1,∞)$, so $f$ is integrable on $(1,∞)$.
   So $f$ is integrable on $(0,∞)$.
   Let $s=1$ in $ℒ\left\{ \log x\right\}(s)=-\frac{\log s+γ}s$, we get $\int_0^∞(\log x)e^{-x}=-γ$. 拉普拉斯变换的问题
7. Integrating by parts,$$\int_1^A x^α\log x=\left.\frac{x^{α+1}}{α+1}\left(\log x-\frac1{α+1}\right)\right|_1^A=\frac1{(α+1)^2}+L(A)$$ where$$\lim_{A→∞}L(A)=\lim_{x→∞}\frac{x^{α+1}}{α+1}\left(\log x-\frac1{α+1}\right)=\begin{cases}∞&α≥-1\\0&α<-1\end{cases}$$ For $α≥-1$, we also can use comparison test: $\int_1^∞ x^α\log x≥\int_1^∞ x^{-1}\log x=\left.\frac{\log^2x}2\right|_1^∞=∞$.
   
8. Since $\sin x>\frac2πx$ for $x∈\left(0,\fracπ2\right)$, we have $\operatorname{cosec}x<\fracπ{2x}$ for $x∈\left(0,\fracπ2\right)$. But $x^{-1/2}$ is integrable on $\left(0,\fracπ2\right)$, by Comparison Test $f(x)=(\operatorname{cosec}x)^{1/2}$ is integrable on $\left(0,\fracπ2\right)$, by symmetry, $f$ is integrable on $(0,π)$.
9. $∫_{rπ}^{(r+1)π}{|f|}≥∫_{rπ}^{(r+1)π}\frac{|\cos x|}{(r+1)π}dx=\frac2{(r+1)π}$ and $\sum_{r=0}^∞\frac2{(r+1)π}=∞⇒f$ is not integrable on $I$.
10. Since $\sin x>x\sin1$ for $x∈(0,1)$, we have $\sin x^{-1}>x^{-1}\sin1$ for $x∈I$.
    By Comparison Test, $x^{-1}$ is not integrable on $I⇒f$ is not integrable on $I$.