Lebesgue measure

1. By L'hospital's rule$$\lim_{n→∞}n^2x^n=\lim_{n→∞}\frac{n^2}{x^{-n}}=\lim_{n→∞}\frac{2n}{-nx^{-n-1}}=\lim_{n→∞}\frac{2}{-x^{-n-1}}=0$$
2. $$\int_0^1 f_n(x)\d x=n^2\left(\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\right)\bigg|_0^1=\frac{n^2}{(n+1)(n+2)}→1\text{ as }n→∞$$

$$\int_{y=0}^{y=1}\left(\int_{x=0}^{x=1} \frac{x-y}{(x+y)^{3}} \d x\right)\d y=\int_{y=0}^{y=1}\left(\int_{x=0}^{x=1}\frac1{(x+y)^2}-\frac{2y}{(x+y)^3}\d x\right)\d y=\int_{y=0}^{y=1}\frac{-1}{(1+y)^2}\d y=-\frac12$$ $$\int_{x=0}^{x=1}\left(\int_{y=0}^{y=1} \frac{x-y}{(x+y)^{3}} \d y\right)\d x=\int_{x=0}^{x=1}\left(\int_{y=0}^{y=1}-\frac1{(x+y)^2}+\frac{2x}{(x+y)^3}\d y\right)\d x=\int_{x=0}^{x=1}\frac{1}{(1+x)^2}\d x=\frac12$$ (a) Define relation ∼ in ℝ by $x∼y⇔x-y∈ℚ$. Each equivalence class $ℚ+x$ is countable, but ℝ is uncountable, so number of equivalence classes is uncountable. Take $x_n∈[0,1]$ from distinct equivalence classes, the equivalence classes $ℚ+x_n$ are disjoint, so their subsets $E+x_n$ are disjoint.
Since $E+x_n⊂[0,2]\,∀n≥1$, \[U=\bigcup_{n=1}^kE+x_n⊂[0,2]\] Since $E+x_n$ are disjoint, \[ \sum_{n=1}^k χ_E\left(x-x_n\right)=\sum_{n=1}^k χ_{E+x_n}(x)=χ_U(x)≤χ_{[0,2]}(x) \] (b) By 1, $ϕ\left(χ_{[0,2]}\right)=2$. By 2, $ϕ(χ_E)≥0$. By 3, $ϕ(χ_E(x-x_n))=ϕ(χ_E)$. By linearity,\[kϕ(χ_E)=\sum_{n=1}^kϕ(χ_E(x-x_n))≤2⇒ϕ(χ_E)≤\frac{2}k\] Let $k→∞$ we get $ϕ(χ_E)=0$.

1. $\liminf _{n→∞} a_n=\exp (-1),\limsup _{n→∞} a_n=\exp (1)$
   By uniform continuity it suffices to show that $0,π$ are limit points of $b_n=n\bmod2π$
   Claim: $(b_n)$ is dense in $[0,2π]$.
   Proof: Since $π∉ℚ$, $b_n$ are distinct, by compactness, $(b_n)$ has a limit point $p∈[0,2π]$.
   $∀x∈[0,2π],∀δ>0,∃n,k∈ℤ^+$ such that ${|b_n-p|}<δ/2,{|b_{n+k}-p|}<δ/2$. Then $b_{n+k}-b_n=γ,0<{|γ|}<δ$.
   If $γ>0$, since $x≥0≥b_n-2π$,$$∃m∈ℕ,r∈[0,γ): x-(b_n-2π)=mγ+r$$Then $x-b_{n+mk}=r<δ$;
   If $γ<0$, since $x≤2π≤b_n+2π$,$$∃m∈ℕ,r∈[0,-γ): x-(b_n+2π)=mγ+r$$Then $x-b_{n+mk}=r<δ$. So any $x∈[0,1]$ is a limit point of $(b_n)$.
2. $\begin{aligned}[t]n=2k &⟹\sin \left(n \fracπ2\right)=0,\cos \left(n \fracπ2\right)=(-1)^k⟹a_{2 k}=1+\exp \left(\frac1{2 k}(-1)^k\right) \\ n=2 k+1 &⟹\cos \left(n \fracπ2\right)=0,\sin \left(n \fracπ2\right)=(-1)^k⟹a_{2 k+1}=\exp \big((2 k+1)(-1)^k\big)+1\end{aligned}$
   $a_{2k}→1,a_{2k+1}(k\text{ even})→∞,a_{2k+1}(k\text{ odd})→1$, so: $\liminf_{n→∞}a_n=1,\limsup_{n→∞}a_n=∞$
3. Let $b_n = n\sin\left(\frac{n^2+1}n\fracπ2\right)$
   Divide into 4 subsequences: \begin{align*} b_{4k+1}&=(4k+1)\sin\left(\fracπ2+\fracπ{8k+2}\right)→∞&∴a_{4k+1}→∞\\ b_{4k-1}&=(4k-1)\sin\left(-\fracπ2+\fracπ{8k-2}\right)→-∞&∴a_{4k-1}→∞\\ b_{4k}&=4k\sin\fracπ{8k}→\fracπ2&∴a_{4k}→\cosh\fracπ2\\ b_{4k+2}&=(4k+2)\sin\left(π+\fracπ{8k+4}\right)→-\fracπ2&∴a_{4k+2}→\cosh\fracπ2 \end{align*} So, $\liminf_{n→∞} a_n = \cosh\fracπ2,\limsup_{n→∞} a_n = ∞$

1. $∀i≥m:a_i≤b_i≤\sup_{n≥m}b_n$. So $\sup_{n≥m}b_n$ is an upper bound for $a_i$. So $\sup_{n≥m} a_n≤\sup_{n≥m} b_n$.
   Let $m→∞$ we get $\limsup _{n→∞} a_n≤\limsup _{n→∞} b_n$
2. For all $i≥k$ we have$$a_i+b_i≤\sup_{n≥k}a_n+\sup_{n≥k}b_n$$Fix $k$, the right hand side is an upper bound for $a_i+b_i$, so $$\sup_{n≥k}(a_n+b_n)≤\sup_{n≥k}a_n+\sup_{n≥k}b_n$$ Let $k→∞$ we get $\limsup _{n→∞} \left(a_n+b_n\right)≤\limsup _{n→∞} a_n+\limsup _{n→∞} b_n$.
3. Let $\limsup_{n→∞}a_n=α$ and $s_k=\sup_{n≥k}a_n$. Then $∃n_k≥k$ such that $\left|a_{n_k}-s_k\right|<\frac1m$.
   Since $\inf_{k≥1}s_k=α,∃k_m>k_{m-1}:\left|s_{k_m}-α\right|<\frac1m$. $$\left|a_{n_{k_m}}-α\right|≤\left|a_{n_{k_m}}-s_{k_m}\right|+\left|s_{k_m}-α\right|<\frac2m⇒\lim_{m→∞}a_{n_{k_m}}=α$$
4. $a_{k_r}≤\sup_{n≥k_r}a_n$, let $r→∞$ we get $\lim _{r→∞} a_{k_r}≤\limsup _{n→∞} a_n$.

By construction \[ C=\bigcap_{k≥1}\left\{\sum_{n=1}^∞ a_n 3^{-n}: a_1=⋯=a_k=0 \text { or } 2\right\} . \] So \[ C=\left\{\sum_{n=1}^∞ a_n 3^{-n}: a_n=0 \text { or } 2\right\} . \] To prove $C$ is uncountable, by 3 methods:

1. Suppose $C=\left\{\sum_{n=1}^∞ a_{r,n} 3^{-n}:r=1,2,⋯\right\}$, but $\sum_{n=1}^∞(2-a_{r,n})3^{-n} ∈ C$ not included, contradiction.
2. $∀x∈C$, define $f:C→[0,1]$ by$$f\left(\sum_{i=1}^∞ r_j 3^{-j}\right)=\sum_{i=1}^∞ \frac{r_j}2 2^{-j}$$ To show that $f$ is a surjection, define $g:[0,1]→C$ by$$g\left(\sum_{i=1}^∞ r_j 2^{-j}\right)=\sum_{i=1}^∞2r_j 3^{-j}$$Then $f∘g=$id
3. $\frac12C=\left\{\sum_{n=1}^∞ a_n 3^{-n}: a_n=0 \text{ or } 1\right\}.∀a,b∈\frac12C:a=\sum_{n=1}^∞ a_n 3^{-n},b=\sum_{n=1}^∞ b_n 3^{-n}$, where $a_n,b_n∈\{0,1\}$.
   Then $a_n+b_n∈\{0,1,2\}$, so $a+b=\sum_{n=1}^∞(a_n+b_n)3^{-n}∈[0,1]$.
   On the other direction, $∀c∈[0,1]$ has a ternary expansion $c=\sum_{n=1}^∞ c_n 3^{-n}$, where $c_n∈\{0,1,2\}$.
   Since 0=0+0, 1=1+0, 2=1+1, there exist $a,b∈\frac12C$ such that $a+b=c$. So $\frac12C+\frac12C=[0,1]$. So $C+C=[0,2]$.
   We constructed a surjection from $C×C$ onto uncountable set $[0,2]$, so $C$ is uncountable.

1. Since the union of countable null sets is null, we only need to consider numbers between 0 and 1.
   Let $A_n = \{x∈[0,1]:\text{first $n$ digits of }x\text{ not }4\}$. For example 0.4 = 0.399⋯ not containing 4.
   Since $A_n=\{x∈A_{n-1}:n\text{th digit of }x\text{ not }4\}$,
   $$ m(A_n)=0.9m(A_{n-1}) $$ $A_1 = [0,0.4]∪[0.5,1]⇒m(A_1)=0.9$ finite. $A_n$ is decreasing, by Proposition 3.2 $m^*(A)=\lim_{n→∞}m(A_n)=0$
2. Since $m^*$ is translation invariant, $m^*(A+b)=m^*(A)=0$.
   $A+B=⋃_{b∈B}(A+b)$ and $A+b$ are disjoint, by countable subadditivity$$m^*(A+B)≤\sum_{b∈B}m^*(A+b)=0$$Then $A+B$ is null.
3. Since countable union of null sets is null and $A=⋃_{n∈ℤ}A∩[n,n+1]$, we can assume $A⊂[0,1]$.
   Since $f'$ is bounded on $[0,1]$, by mean value theorem, $∃C>0∀a,b∈[0,1]:{|f(a)-f(b)|}<C{|a-b|}$
   Since $A$ is null, $∀ϵ>0∃$a sequence of intervals $I_n$ cover $A$ with $∑_{n=1}^∞m(I_n)<ϵ$.
   Since $I_n$ is connected and $f$ is continuous, $f(I_n)$ is connected, so $f(I_n)$ is an interval, $m(f(I_n))<Cm(I_n)$. \[f(A)⊂\bigcup_{n=1}^∞f(I_n),\sum_{n=1}^∞m(f(I_n))<Cϵ⇒m(f(A))<Cϵ\] So $f(A)$ is null.

1. $∀ϵ>0∃$a sequence of intervals $I_n$ cover $A$ with $\sum_{n=1}^∞ m\left(I_n\right)≤m^*(A)+ϵ$.
   Then $A+x⊂⋃_{n=1}^∞I_n+x$ and $I_n+x$ are intervals$⇒m^*(A+x)≤\sum_{n=1}^∞ m\left(I_n+x\right)$.
   Since $I_n$ are intervals, $m(I_n+x)=m(I_n)$. So $m^*(A+x)≤m^*(A)+ϵ$.
   Let $ϵ→0$ we get $m^*(A+x)≤m^*(A)$. Since $(A+x)+(-x)=A$, we can prove $m^*(A+x)≥m^*(A)$.
2. For $α=0$ it is trivial. For $α≠0$:
   $∀ϵ>0∃$a sequence of intervals $I_n$ cover $A$ with $\sum_{n=1}^∞ m\left(I_n\right)≤m^*(A)+ϵ$.
   Then $αA⊂⋃_{n=1}^∞αI_n$ and $αI_n$ are intervals$⇒m^*(αA)≤\sum_{n=1}^∞ m\left(αI_n\right)$.
   Since $I_n$ are intervals, $m(αI_n)={|α|}m(I_n)$. So $m^*(αA)≤{|α|}m^*(A)+{|α|}ϵ$.
   Let $ϵ→0$ we get $m^*(αA)≤{|α|}m^*(A)$. Since $α^{-1}(αA)=A$, we can prove $m^*(αA)≥{|α|}m^*(A)$.
3. $∀ϵ>0∃$a sequence of intervals $I_{n,1}$ cover $A$ with $\sum_{n=1}^∞ m\left(I_{n,1}\right)≤m^*(A)+ϵ$.
   ∃a sequence of intervals $I_{n,2}$ cover $B$ with $\sum_{n=1}^∞ m\left(I_{n,2}\right)≤m^*(B)+ϵ$.
   The collection of open intervals $\left\{I_{j, k}: j∈ℤ^+,k=1,2\right\}$ is countable.
   Thus $m^*(A∪B)≤m^*(A)+m^*(B)+2ϵ$.
   Let $ϵ→0$ we get $m^*(A∪B)≤m^*(A)+m^*(B)$.
4. If $m^*(A_k)=∞$ for some $k∈ℤ^+$, the inequality clearly holds. Thus assume $m^*(A_k)<∞$ for all $k∈ℤ^+$.
   Let $ε>0$. For each $k∈ℤ^+$, let $I_{1, k}, I_{2, k}, …$ be a sequence of intervals cover $A_k$ such that \[ \sum_{j=1}^∞ m\left(I_{j, k}\right)≤\fracε{2^k}+m^*(A_k) \] Thus \[\tag1 \sum_{k=1}^∞ \sum_{j=1}^∞m\left(I_{j, k}\right) ≤ ε+\sum_{k=1}^∞m^*(A_k) \] The doubly indexed collection of open intervals $\left\{I_{j, k}: j, k∈ℤ^+\right\}$ is countable.
   Inequality (1) shows that the sum of the lengths of the intervals listed above$≤ε+\sum_{k=1}^∞m^*(A_k)$.
   Thus $m^*\left(\bigcup_{k=1}^∞ A_k\right)≤ε+\sum_{k=1}^∞m^*(A_k)$.
   Let $ϵ→0$ we get $m^*\left(\bigcup_{k=1}^∞ A_k\right)≤\sum_{k=1}^∞m^*(A_k)$.

1. If $N$ is null, then $m^*(N)=0$. So for any $A⊂ℝ$ we have \begin{aligned} m^*(A∩N)≤m^*(N)=0&\text { since } A∩N⊂N\\ m^*\left(A∖N\right)≤m^*(A)&\text { since } A∖N⊂A \end{aligned} and adding together we have proved $m^*(A∩N)+m^*(A∖N)≤m^*(A)$.
2. If $I$ is bounded, $∀A⊂ℝ$, if $m(A)=∞$, then $m(A∖I)=∞$ we have $∞=∞$; if $m(A)<∞$, $∀ε>0,∃$a sequence of intervals $I_n$ cover $A$ with $\sum_{n=1}^∞m\left(I_n\right)≤m^*(A)+ε$……(1)
   The intervals $I_n'=I_n∩I$ cover $A∩I$, so \[ m^*(A∩I)≤\sum_{n=1}^∞m\left(I_n'\right) . \] Let $I⊂[a, b]$. The intervals $I_n''=I_n∩(-∞, a), I_n'''=I_n∩(b,∞)$ cover $A∖I$, so \[ m^*\left(A∖I\right)≤\sum_{n=1}^∞m\left(I_n''\right)+\sum_{n=1}^∞m\left(I_n'''\right) . \] Add these inequalities then use (1) and $m(I_n)=m(I_n')+m(I_n'')+m(I_n''')$,$$m^*(A∩I)+m^*(A∖I)≤m^*(A)+ε$$Let $ε→0$ we get $m^*(A∩I)+m^*(A∖I)≤m^*(A)$.
   If $I$ is unbounded, say $I=[a, ∞)$, it suffices to consider $I_n'=I_n∩[a, ∞)$ and $I_n''=I_n∩(-∞, a)$.
3. By Q8, $m^*(A-x)=m^*(A),m^*((A-x)∩E)=m^*(A∩(E+x)),m^*((A-x)∖E)=m^*(A∖(E+x))$.
   $E$ is measurable$⇒m^*(A-x)=m^*((A-x)∩E)+m^*((A-x)∖E)$
   $⇒m^*(A)=m^*(A∩(E+x))+m^*(A∖(E+x))⇒E+x$ is measurable.

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   For $α=0$, $αA=\{0\}$ is measurable. For $α≠0$,
   By Q8, $m^*(A)={|α|}m^*(α^{-1}A),m^*(A∩αE)={|α|}m^*(α^{-1}A∩E),m^*(A∖αE)={|α|}m^*(α^{-1}A∖E),$
   $E$ is measurable$⇒m^*(α^{-1}A)=m^*(α^{-1}A∩E)+m^*(α^{-1}A∖E)$
   $⇒m^*(A)=m^*(A∩αE)+m^*(A∖αE)⇒αE$ is measurable.

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   $F$ is measurable$⇒m^*(A)=m^*(A∩F)+m^*(A∩Fᶜ)$
   With $A∩Eᶜ$ in place of $A$ we get \[\tag1 m^*\left(A∩Eᶜ\right)=m^*\left(A∩Eᶜ∩F\right)+m^*\left(A∩Eᶜ∩Fᶜ\right) . \] $E$ is measurable$⇒m^*(A)=m^*(A∩E)+m^*(A∩Eᶜ)$……(2)
   $A∩\left(E∪F\right)$ is disjoint union of $A∩E$ and $A∩Eᶜ∩F$
   We insert (1) into (2) to get \begin{align*} m^*(A)&=m^*\left(A∩E\right)+m^*\left(A∩Eᶜ∩F\right)+m^*\left(A∩Eᶜ∩Fᶜ\right)\\ \text{additivity} &=m^*\left(A∩\left(E∪F\right)\right)+m^*\left(A∩Eᶜ∩Fᶜ\right)\\ \text{de Morgan's law} &=m^*\left(A∩\left(E∪F\right)\right)+m^*\left(A∩\left(E∪F\right)ᶜ\right) \end{align*}Therefore $E∪F$ is measurable.
4. Induct on $n$. For $n=1$, $m^*(A)=m^*(A)$ is true. For $n=2$, we need to prove \[\tag1 m^*\left(E_1∪E_2\right)=m^*(E_2)+m^*(E_1) \] Since $E_2$ is measurable, apply definition with $A=E_1∪E_2$ : \[ m^*\left(E_1∪E_2\right)=m^*\left((E_1∪E_2)∩E_2\right)+m^*\left((E_1∪E_2)∩E_2^{\mathrm{c}}\right) \] Since $E_1,E_2$ are disjoint, we get (1).
   Assume theorem is true for $n-1$, then \[\tag2 m^*\left(\bigcup_{k=1}^{n-1}E_k\right)=\sum_{k=1}^{n-1} m^*(E_k) \] Applying (1) with $E_n$ and $\bigcup_{k=1}^{n-1} E_k$ \[ m^*\left(\bigcup_{k=1}^n E_k\right)=m^*\left(E_n\right)+m^*\left(\bigcup_{k=1}^{n-1} E_k\right), \] Inserting (2) we get \[ m^*\left(\bigcup_{k=1}^n E_k\right)=\sum_{k=1}^n m^*(E_k) \] as required to complete the induction step. It remains true when $n→∞$.

1. If $x∈G$, then $I_{x,x}=\{x\}⊂G$, so $x∼x$
   If $x∼y$, then $I_{y,x}=I_{x,y}⊂G$, so $y∼x$
   If $x∼y,y∼z$, then $I_{x,z}⊂I_{x,y}∪I_{y,z}⊂G$, so $x∼z$
2. Let $A$ be an equivalence class. By metric space Theorem 7.2.1. $A$ is an interval.
   For any $a∈A$, since $G$ is open, $∃r>0:B(a,r)⊂G$. Note that $∀b∈B(a,r):b∼A$. So $B(a,r)⊂A$. So $A$ is open.
3. Each rational number is in one equivalence class and each equivalence class contains some rational numbers. We constructed a surjection from ℚ to $G/$∼ and $G/$∼ is infinite, so $G/$∼ is countable.
4. $G$ is the union of equivalence classes, which are disjoint open intervals. We proved they are countable.