Sheet 6

Exercise 1 (Continuous Indicator).

Consider a metric space $(E,d)$. For $x\in E$ and $B\subseteq E$, define

$$d(x,B):=inf\{d(x,y):y\in B\}.$$

1. If $d(x,B)=0$, then $x\in \overline{B}$, the closure of $B$.

   Solution.

   Assume that $x\in {\overline{B}}^c$. By definition

   $\overline{B}={\displaystyle \bigcap _{A\supseteq B\text{ closed}}}A$

   we have

   $x\in {\overline{B}}^c={\displaystyle \bigcup _{A\supseteq B\text{ closed}}}{A}^c={\displaystyle \bigcup _{O\subseteq B^c\text{ open}}}O.$

   So by the definition of open sets, there exists $ϵ>0$, such that $B_{ϵ}(x)\subseteq O\subseteq B^c$ for some open set $O$. So we have for any $y\in B$, $d(x,y)>ϵ$. Therefore $d(x,B)>ϵ>0$. So for all $d(x,B)=0$ we necessarily have $x\in \overline{B}$. ∎

2. Suppose that $B\ne \mathrm{\varnothing }$. Prove that

   $$|d(x,B)-d(y,B)|\le d(x,y).$$

   This is to say, the function $x\mapsto d(x,B)$ is $1$-Lipschitz.

   Solution.

   To remove the infimum choose some $\epsilon >0$. Then we can choose $z\in B$ such that we are $\epsilon$-close to the infimum

   $d(y,z)\le d(y,B)+\epsilon ⟹-d(y,B)\le \epsilon -d(y,z).$

   For $z\in B$ we always have $d(x,B)\le d(x,z)$. Together this results in

   $d(x,B)-d(y,B)$ $\le d(x,z)-d(y,z)+\epsilon$

   $\stackrel{\mathrm{\Delta }}{\le }d(x,y)+\epsilon .$

   Using that $\epsilon$ was arbitrary and symmetry of $x,y$ we conclude

   $|d(x,B)-d(y,B)|$ $\le d(x,y)\mathit{∎}$

3. Let $A\subseteq E$ be closed. Show that, for any $ϵ>0$, the function

   $$f_A^{ϵ}(x)={\left(1-d(x,A)/ϵ\right)}^{+}$$

   satisfies the following properties:

   • $\forall x\in A$, $f_A^{ϵ}(x)=1$

     Solution.

     For $x\in A=\overline{A}$ we have by (i) $d(x,A)=0$ and thus $f_A^{\epsilon }(x)=1$. ∎

   • if $d(x,A)\ge ϵ$ then $f_A^{ϵ}(x)=0$,

     Solution.

     if $d(x,A)\ge ϵ$, then by definition

     $f_A^{\epsilon }(x)$ $={\left(\underset{\le 0}{\underbrace{1-{\displaystyle \frac{d(x,A))}{\epsilon }}}}\right)}^{+}=0.\mathit{∎}$

   • $f_A^{ϵ}\to {\mathrm{𝟙}}_A$ pointwise for $\epsilon \to 0$,

     Solution.

     For all $x\in \overline{A}=A$ we have

     $f_A^{\epsilon }(x)=1={\mathrm{𝟙}}_A$

     For all $x\in A^c$ we have by (i) $d(x,A)>0$. Therefore there exists $\epsilon \le d(x,A)$ such that

     $f_A^{\epsilon }(x)$ $=0={\mathrm{𝟙}}_A.\mathit{∎}$

   • $f_A^{ϵ}\le 1$

     Solution.

     Follows from positiveness of metrics $d(x,B)\ge 0$ and $\epsilon >0$. ∎

   • $f_A^{ϵ}\in {\mathrm{Lip}}_{1/ϵ}(E)$ and $f_A^{ϵ}\in C_b(E)$

     Solution.

     We know from (ii) that $g_1:x\mapsto d(x,A)$ is $1$-Lipschitz. The same is true for $g_3:x\mapsto {(1-x)}^{+}$. Because $g_2:x\to \frac{x}{\epsilon }$ is $\frac{1}{\epsilon }$-Lipschitz we have by repeated application of the definition that

     $f_A^{ϵ}=g_3\circ g_2\circ g_1$

     is $\frac{1}{\epsilon }$-Lipschitz. $f_A^{\epsilon }\in C_b(E)$ now follows from the previous statement. ∎

   • Why do we have to assume $A$ to be closed?

     Solution.

     We would otherwise have $f_A^{ϵ}\to {\mathrm{𝟙}}_{\overline{A}}$, because $d(x,A)=d(x,\overline{A})$. ∎

Exercise 2 (Totally Bounded Spaces).

1. Show that totally bounded sets are always bounded.

   Solution.

   Let $A\subseteq E$ be totally bounded, where $(E,d)$ is a metric space. By definition there are $x_1,\mathrm{\dots },x_n\in A$ such that

   $A\subseteq {\displaystyle \bigcup _{k=1}^n}{B}_{ϵ}(x_k).$

   Choose some $x\in A$ and define $m:={\max }_{k=1,\mathrm{\dots },n}d(x,x_k)$. Then for all $y\in A$ we have some $k$ such that $y\in B_{ϵ}(x_k)$ and therefore

   $d(x,y)\le d(x,x_k)+d(y,x_k)\le ϵ+m$

   This implies $A\subseteq B_{m+ϵ}(x)$. Therefore $A$ is bounded. ∎

2. Prove that total boundedness and boundedness are equivalent on ${ℝ}^d$ for any metric induced by a norm.

   Solution.

   As the other statement holds in general, we only need to prove boundedness implies total boundedness. Without loss of generality we are going to use the sup-norm (by norm equivalence boundedness in one implies the other). And total boundedness in the sup-norm can be converted back to any other norm with norm equivalence using smaller epsilon.

   Assume there is some bounded set $A\subseteq B_m(x)$ with $m>0$. Using $\tilde{m}=m+d(x,0)$ we can assume without loss of generality that $x=0$. By selecting the finite set

   $\mathrm{\{}{x}_1,\mathrm{\dots },x_n\}:=\{(k_1ϵ,\mathrm{\dots },k_dϵ)\in {ℝ}^d:k_1,\mathrm{\dots },k_d\in \{-\frac{m}{ϵ},\mathrm{\dots },\frac{m}{ϵ}\}\}$

   we get

   $A\subseteq {\displaystyle \bigcup _{k=1}^n}{B}_{ϵ}(x_k).$

   Because for any $y\in A$ we have

   $m\ge {\parallel y\parallel }_{\mathrm{\infty }}=\underset{i=1,\mathrm{\dots },d}{sup}|y^{(i)}|,$

   and for every $i$ there is ${\widehat{k}}_i\in \{-\frac{m}{ϵ},\mathrm{\dots },\frac{m}{ϵ}\}$ such that $|y^{(i)}-{\widehat{k}}_iϵ|\le ϵ$. Therefore

   $y$ $\in B_{ϵ}(({\widehat{k}}_1ϵ,\mathrm{\dots },{\widehat{k}}_dϵ)).\mathit{∎}$

3. Prove that no infinite set is totally bounded in the discrete metric

   Solution.

   Assume $A$ is totally bounded and let $x_1,\mathrm{\dots },x_n$ be the points in the definition for $ϵ=\frac{1}{2}$. Then we have

   $A\subseteq {\displaystyle \bigcup _{k=1}^n}{B}_{\frac{1}{2}}(x_k)={\displaystyle \bigcup _{k=1}^n}\{x_k\}=\{x_1,\mathrm{\dots },x_n\}$

   Therefore $A$ is finite. ∎

4. Prove that the closure of a totally bounded set is totally bounded.

   Remark.

   In a complete metric space totally bounded is therefore by Lemma 7.1.8 equivalent to relatively compact.

   Solution.

   Let $A$ be totally bounded, and select $ϵ>0$. Because

   $A\subseteq {\displaystyle \bigcup _{k=1}^n}{B}_{\frac{ϵ}{2}}(x_k)$

   for some $x_1,\mathrm{\dots },x_n\in A$, we have with $M_1\subseteq M_2$ implies ${\overline{M}}_1\subseteq {\overline{M}}_2$

   $\overline{A}\subseteq \overline{{\displaystyle \bigcup _{k=1}^n}{B}_{\frac{ϵ}{2}}(x_k)}\stackrel{(*)}{\subseteq }{\displaystyle \bigcup _{k=1}^n}\overline{B_{\frac{ϵ}{2}}(x_k)}\subseteq {\displaystyle \bigcup _{k=1}^n}{B}_{ϵ}(x_k)$

   where $(*)$ is due to the fact that finite unions of closed sets are closed and the closure is the smallest closed set which contains the original. ∎

Exercise 3 (Non-separable Space).

Consider $ℝ$ endowed with the Lebesgue measure $\lambda$. Let $L_{\mathrm{\infty }}$ be the space of measurable functions that are bounded almost everywhere, with the essential supremum of its absolute value as a norm: for any measurable function $f$,

$$\parallel f{\parallel }_{\mathrm{\infty }}=inf\{C\ge 0:\lambda (f>C)=0\}.$$

Show that the metric space $L^{\mathrm{\infty }}$ (with metric induced by the norm $\parallel \cdot {\parallel }_{\mathrm{\infty }}$) is not separable.

Hint.

consider all the indicator functions ${\mathrm{𝟙}}_{\mathrm{[}\mathrm{0}\mathrm{,}r\mathrm{)}}$, $r\mathrm{>}\mathrm{0}$

Solution.

The set of indicator functions $I=\{{\mathrm{𝟙}}_{[0,r)}:r>0\}$ is a subset of $L_{\mathrm{\infty }}$ and for any $r_1\ne r_2$ (without loss of generality ) we have

${\parallel {\mathrm{𝟙}}_{[0,r_2)}-{\mathrm{𝟙}}_{[0,r_1)}\parallel }_{\mathrm{\infty }}={\parallel {\mathrm{𝟙}}_{[r_1,r_2)}\parallel }_{\mathrm{\infty }}=1.$

So we have an uncountable set of functions which are all at distance $1$ to each other. This rules out separability. Because if we had

$I\subseteq L_{\mathrm{\infty }}\subseteq {\displaystyle \bigcup _{n\in \mathbb{N}}}{B}_{\epsilon }(f_n),$

we could find $r_1\ne r_2$ with ${\mathrm{𝟙}}_{[0,r_1)},{\mathrm{𝟙}}_{[0,r_2)}\in B_{\epsilon }(f_n)$, since $I$ is uncountable and the number of $f_n$ is just countable. But then

$1={\parallel {\mathrm{𝟙}}_{[0,r_1)}-{\mathrm{𝟙}}_{[0,r_2)}\parallel }_{\mathrm{\infty }}\le {\parallel {\mathrm{𝟙}}_{[0,r_1)}-f_n\parallel }_{\mathrm{\infty }}+{\parallel f_n-{\mathrm{𝟙}}_{[0,r_2)}\parallel }_{\mathrm{\infty }}\le 2\epsilon$

which is a contradiction for small epsilon. ∎

Exercise 4 (Weak Limits).

Find the weak limit, if exists, of the following sequence.

1. Assume $x_n\to x$ in some metric space $(E,d)$. Prove that ${\delta }_{x_n}\to {\delta }_x$.

   Solution.

   We have for any $f\in C_b(E)$ by continuity

   ${\displaystyle \int fd{\delta }_{x_n}}$ $=f(x_n)\to f(x)={\displaystyle \int fd{\delta }_x}.\mathit{∎}$

2. ${\mathbb{P}}_n=𝒩(0,1/n)$, the normal distribution.

   Hint.

   What does weak convergence have to do with random variables?

   Solution.

   Let $X_n\sim {\mathbb{P}}_n$, then

   $𝔼[{|X_n-0|}^2]=\text{Var}(X_n)=\frac{1}{n}\to 0$

   therefore $X_n\to 0$ in $L^2$. This implies convergence in distribution, i.e.

   $𝔼[f(X_n)]\to 𝔼[f(0)]$

   for all $f\in C_b$. Therefore we have ${\mathbb{P}}_n\to {\delta }_0$ weakly. ∎

Exercise 5 (Borel $\sigma$-Algebra).

Let $(E,d)$ be a metric space with Borel $\sigma$-Algebra

$ℬ(E):=\sigma (\{O\subseteq E:O\text{ open})\}$

1. Assume there exists a countable base $\pi$ of the topology $\tau$ of $E$. Prove that $\sigma (\pi )=ℬ(E)$.

   Solution.

   For any set $O\in \tau$ we have $O={\bigcup }_n{O}_n$ with $O_n\in \pi$. This implies $O\in \sigma (\pi )$ and therefore $\tau \subseteq \sigma (\pi )$. But since $\pi \subseteq \tau$ we have

   $ℬ(E)$ $=\sigma (\tau )=\sigma (\pi ).\mathit{∎}$

2. Assume that $E$ is separable (has a countable dense subset $A$). Prove that

   $\pi :=\{B_q(x):x\in A,q\in \mathbb{Q}\}$

   is a (countable) base of $\tau$.

   Remark.

   In case of $ℝ$, this base $\pi$ is the set of all intervals with rational endpoints.

   Hint.

   $\bigcup \{B\in \pi :B\subseteq O\}=O$

   Solution.

   Let $O\in \tau$. We claim that

   $U:={\displaystyle \bigcup \{B\in \pi :B\subseteq O\}}=O.$

   By definition $U\subseteq O$ holds. So consider $y\in O$. Since $O$ is open, there exists an $ϵ>0$ (without loss of generality in $\mathbb{Q}$) such that $B_{ϵ}(y)\subseteq O$. Since $A$ is dense, there exists $x\in A$ with (otherwise $B_{\frac{ϵ}{2}}(y)\subseteq {\overline{A}}^c=E^c=\mathrm{\varnothing }$). This implies

   $y\in \underset{\in \pi }{\underbrace{B_{\frac{ϵ}{2}}(x)}}\subseteq B_{ϵ}(y)\subseteq O.$

   Since $y$ was arbitrary we have $U\supseteq O$. ∎

3. Prove that $\sigma (\{B_{ϵ}(x),x\in E,ϵ>0\})=ℬ(E)$ if $E$ is separable.

   Solution.

   Using $\pi$ from (ii), we have

   $ℬ(E)$ $\stackrel{\text{<a href="\#S0.I4.i1" title="item (i) ‣ Exercise 5 (Borel σ -Algebra). ‣ Sheet 6" class="ltx\_ref" style="font-size:70\%;"><span class="ltx\_text ltx\_ref\_tag">(i)</span></a>}}{=}\sigma (\pi )\subseteq \sigma (\{B_{ϵ}(x),x\in E,ϵ>0\})\subseteq ℬ(E).\mathit{∎}$