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 $ℝ$, $$ m^*\left(\bigcup_{n=1}^∞ E_n\right) ⩽ \sum_{n=1}^∞ m^*\left(E_n\right) $$
      3. Let $F_1 ⊇ F_2 ⊇ ⋯$ be Lebesgue measurable sets with $m\left(F_1\right)<∞$. Use countable additivity to show that $$ m\left(\bigcap_{n=1}^∞ F_n\right)=\lim_{n →∞} m\left(F_n\right) . $$
      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 $E=\bigcap_{n=1}^∞ \bigcup_{r ⩾ n} E_r$. 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 Frullani integral $$ ∫_0^∞ \frac{e^{-x}-e^{-a x}}{x} \mathrm{~d} x=\log a $$
      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 $\left\|f_n-f\right\|_1→0$.
         Scheffé's Lemma

Solution

1. 1. 1. For any interval $I$ with endpoints $a$ and $b$, define ${|I|}=b-a$.
         For $A ⊂ ℝ$, we define the Lebesgue outer measure $$ m^*(A)=\inf \left\{\sum_{n=1}^∞ \left|I_n\right|: I_n \text { are intervals}, A ⊆ \bigcup_{n=1}^∞ I_n\right\}, $$For $E⊂ℝ$, we say $E∈ℳ_\text{Leb}$ if for all $A ⊂ ℝ$$$m^*(A)=m^*(A ∩ E)+m^*(A ∖ E)$$
      2. If $m^*(E_n)=∞$ for some $n$, the inequality clearly holds. Assume $m^*(E_n)<∞\ ∀n$.
         For any $ε>0$, let $I_{1, n}, I_{2, n}, …$ be a sequence of intervals cover $E_n$ such that \[ \sum_{j=1}^∞ m\left(I_{j, n}\right)≤\fracε{2^n}+m^*(E_n) \] Summing over $n$ \[ \sum_{n=1}^∞ \sum_{j=1}^∞m\left(I_{j, n}\right) ≤ ε+\sum_{n=1}^∞m^*(E_n) \] Since $\left\{I_{j, n}: j, n∈ℤ^+\right\}$ cover $\bigcup_{n=1}^∞ E_n$ we have $m^*\left(\bigcup_{n=1}^∞ E_n\right)≤ε+\sum_{n=1}^∞m^*(E_n)$.
         Let $ϵ→0$ we get $m^*\left(\bigcup_{n=1}^∞ E_n\right)≤\sum_{n=1}^∞m^*(E_n)$.
      3. Set $E_n=F_1 ∖ F_n$, then $E_1 ⊆ E_2 ⊆ …$
         Set $A_1=E_1;∀n>1,A_n=E_n∖E_{n-1}$. Then $A_1,…$ are disjoint. By countable additivity $$m\left(\bigcup_{n=1}^∞ E_n\right)=m\left(\bigcup_{n=1}^∞ A_n\right)=\lim_{n →∞}\sum_{k=1}^nm\left(A_k\right)=\lim_{n →∞}m\left(E_n\right)$$ By additivity $m\left(E_n\right)=m(F_1)-m\left(F_n\right),m\left(\bigcap_{n=1}^∞ F_n\right)=m(F_1)-m\left(\bigcup_{n=1}^∞ E_n\right),$
         we get $ m\left(\bigcap_{n=1}^∞ F_n\right)=\lim_{n →∞} m\left(F_n\right) .$
      4. By (ii) $m\left(\bigcup_{r ⩾ n} E_r\right)⩽\sum_{r ⩾ n} m\left(E_r\right)$. Since $\sum_{n=1}^∞ m\left(E_n\right)<∞$$$\lim_{n→∞} m\left(\bigcup_{r ⩾ n} E_r\right)⩽\lim_{n→∞} \sum_{r ⩾ n} m\left(E_r\right)=0$$Apply (iii) to the decreasing sequence $F_n=\bigcup_{r⩾n}E_r$ $$ m(E)=\lim_{n→∞} m\left(\bigcup_{r ⩾ n} E_r\right)=0 $$
   2. 1. Apply a(iii) to $F_r=\left\{x∈[0,1]:\left|f_n(x)\right|>r / n\right\}$, since $F_1 ⊃ F_2 ⊃…$ and $m(F_1)⩽1$, \[m\left(\bigcap_{r=1}^∞ F_r\right)=\lim_{r→∞}m(F_r)\] But $\bigcap_{r=1}^∞ F_r=∅$, so $\lim_{r→∞}m(F_r)=0$, so ∃$a_n:m\left(F_{a_n}\right)<2^{-n}$
      2. Fix $x$, if $\frac{f_n(x)}{a_n} ↛ 0$, $∀n∃m⩾n:\left|f_m(x)\over a_m\right| > \frac1n$, so $\left|f_m(x)\right| > \frac{a_m}{m}$, so $x∈E=\bigcap_{n=1}^∞ \bigcup_{m ⩾ n} E_m$.
         Apply a(iv) to $E_n=\left\{x ∈[0,1]:\left|f_n(x)\right|>a_n / n\right\}$, since $\sum_{n=1}^∞ m\left(E_n\right)<\sum_{n=1}^∞2^{-n}<∞$
         we have $m(E)=0$, so $\frac{f_n(x)}{a_n} → 0$ for almost every $x∈[0,1]$
2. 1. [Fubini’s Theorem] Let $f:ℝ^2→ℝ$ be integrable. Then, for almost all $y$, the function $x↦f(x,y)$ is integrable. Define $F(y)=∫_ℝf(x,y)dx$, then $F$ is integrable, and$$\int_{ℝ^2} f(x, y) d(x, y)=\int_ℝF(y) d y$$[Tonelli’s Theorem] Let $f:ℝ^2→ℝ$ be measurable. Suppose that either of$$\int_ℝ\left(\int_ℝ{|f(x, y)|} d x\right) d y, \quad \int_ℝ\left(\int_ℝ{|f(x, y)|} d y\right) d x$$ is finite, then $f$ is integrable.
   2. 1. $e^{-x y}>0$. For $y∈[1,a]$, by MCT, $∫_0^∞ e^{-x y} d x=\lim_{n →∞}\frac{1-e^{-n y}}{y}=\frac{1}{y}$$$∫_1^a∫_0^∞ e^{-x y} d xdy=\log a<∞$$By Tonelli's theorem, $e^{-x y}$ is integrable on $[0,∞)×[1, a]$.
      2. By Fubini's theorem $$ ∫_0^∞ \frac{e^{-x}-e^{-a x}}{x} d x=∫_0^∞ ∫_1^a e^{-x y} d y d x=\log a $$
      3. [Sheet 5 Q3a] $g(x, y) = f(xy)-af(axy)$. By linearity of the integral $$ \int_0^1 g(x, y) d x=\int_0^1 f(x y) d x-a \int_0^1 f(a x y) d x $$ Since $f$ is integrable, by changes of variable $y↦x y$ and $y↦ay$\[\frac{1}{x} ∫_0^∞ f(y) d y=∫_0^∞ f(x y) d y=a ∫_0^∞ f(a x y) d y\]Since $f>0$, Baby MCT ensures that $$ \lim _{n → ∞} ∫_0^n g(x, y) d y=\lim _{n → ∞}\left(∫_0^n f(x y) d y-a ∫_0^n f(a x y) d y\right)=0 . $$ $|g|$ is dominated in $y$ by the integrable function $f(x y)+a f(a x y)$, by DCT $$ ∫_0^∞ g(x, y) d y=0 $$For $y>0$ we have $$ ∫_0^1g(x,y)d x=\left[e^{-axy}-e^{-xy}\over y\right]_0^1=\frac{e^{-a y}-e^{-y}}{y}<0 $$and $g(x, y)=g(y, x)$, so $∫_0^1 g(x, y) d y<0$. Thus $∫_1^∞ g(x, y) d y>0$, whenever $x>0$.
         If $g$ is integrable over $(0,1) ×(1, ∞)$, then Fubini's theorem would imply that $$ 0<∫_0^1 ∫_1^∞ g(x, y) d y d x=∫_1^∞ ∫_0^1 g(x, y) d x d y<0 . $$
   3. By Fubini $∃x_0∈[0,1]$ such that $\int_0^1{|f(x_0)-f(y)|}dy<∞$. Then$$\int_0^1{|f(y)|}dy≤\int_0^1{|f(y)-f(x_0)|}dy+\int_0^1{|f(x_0)|}dx<∞$$Therefore $f$ is integrable on $[0,1]$
3. 1. 1. A simple function $ϕ:ℝ→[0,∞]$ has the form $ϕ=\sum^n_{i=1}α_iχ_{B_i}$ with $α_i>0,B_i∈ℳ_\text{Leb}$. Then $∫_ℝ ϕ=\sum^n_{i=1}α_im(B_i)$.
      2. $∫_ℝf=\sup\{∫_ℝϕ:0⩽ϕ⩽f,ϕ\text{ simple}\}$
      3. $f^+=\max(f,0)$, $f^-=\max(-f,0).$
         $f$ is integrable iff $∫_ℝf^+$ and $∫_ℝf^-$ are both finite, define $∫_ℝf=∫_ℝf^+-∫_ℝf^-$.
      4. Assume $0⩽f$ then $\{∫_ℝϕ:0⩽ϕ⩽f,ϕ\text{ simple}\}⊂\{∫_ℝϕ:0⩽ϕ⩽g,ϕ\text{ simple}\}$,
         so $∫_ℝf=\sup\{∫_ℝϕ:0⩽ϕ⩽f,ϕ\text{ simple}\}⩽\sup\{∫_ℝϕ:0⩽ϕ⩽g,ϕ\text{ simple}\}=∫_ℝg$
         In general, $\left.\begin{array}l0⩽f^+⩽g^+⇒∫_ℝf^+⩽∫_ℝg^+\\0⩽g^-⩽f^-⇒∫_ℝg^-⩽∫_ℝf^-\end{array}\right\}$Subtracting, $∫_ℝf⩽∫_ℝg$.
   2. 1. For $x∈[nπ,(n+1)π]$, $\left|\frac{\cos x}{x}\right|⩾{|\cos x|\over(n+1)π}$ we have $$ ∫_{n π}^{(n+1)π}\left|f(x)\right|⩾\frac{∫_{n π}^{(n+1)π}{|\cos x|}}{(n+1)π}=\frac2{(n+1)π} $$ Since harmonic series diverges, $f$ is not integrable over $(1,∞)$.
      2. Substitute $u=x^{-1}$ $$\int_0^1g=∫_∞^1\frac{\cos(1/x^{-1})}{x^{-1}}(-x^{-2})dx=∫_1^∞f=∞$$
      3. $\lim_{x→0}h(x)=\lim_{x→0}\frac{x+O(x^2)}{x+O(x^2)}=1$, so $h$ is bounded on $[0,1]$, so $h$ is integrable on $[0,1]$.
         For $x⩾1$, we have $\left|\frac{\sin x}{e^x-1}\right|⩽\frac1{e^x-1}$. Substitute $u=e^x$, $$\int_1^∞\frac1{e^x-1}dx=\int_e^∞\frac1{u(u-1)}du=\log\left(u-1\over u\right)|_e^∞=1-\log(e-1)$$ By comparison $h$ is integrable on $[1,∞)$. So $h$ is integrable on $[0,∞)$.
   3. 1. A sequence of measurable functions $f_n→f$ a.e. so $f$ is measurable.
         $∀n∈ℕ:\left|f_n\right|⩽g_n$, $f_n→f,g_n→g$ a.e. So ${|f|}⩽g$ a.e. But $g∈ℒ^1(ℝ)$, so $f∈ℒ^1(ℝ)$.
         Let $B_n=\{x∈ℝ:g_n(x)>g(x)\}$. Then $\left|f_n\right|χ_{ℝ∖B_n}⩽g_nχ_{ℝ∖B_n}⩽g$.
         By DCT $\lim_{n→∞}∫_{ℝ∖B_n}f_n=∫_ℝf$. Again by DCT $\lim_{n→∞}∫_{ℝ∖B_n}g_n=∫_ℝg$.
         $\left.\begin{array}r\lim_{n→∞}∫_{ℝ∖B_n}g_n=∫_ℝg\\\lim_{n→∞}∫_ℝg_n=∫_ℝg\end{array}\right\}$Subtracting,$\lim_{n→∞}∫_{B_n}g_n=0$,by comparison,$\lim_{n→∞}∫_{B_n}f_n=0$.
         Therefore $\lim_{n→∞}∫_ℝf_n=\lim_{n→∞}∫_{B_n}f_n+\lim_{n→∞}∫_{ℝ∖B_n}f_n=∫_ℝf$.
      2. If $f_n→f$ a.e. then ${|f_n-f|}→0$ a.e.
         Let $F_n={|f_n-f|},F=0,G=2g$. Then $F_n→F$ a.e.
         If ${|f_n|}≤g$ and $f_n→f$ a.e. then ${|f|}≤g$ a.e. so ${|f_n-f|}≤2g$ a.e.
         ${|F_n|}≤G$ and $∫_ℝG<∞$, by DCT $∫_ℝF_n→∫_ℝF$.