10 Lebesgue Measure

Carathéodory criterion

Let $(X,d)$ be a metric space. Consider an outer measure ${\mu }^{\ast }:\mathcal{P}(X)\to [0,\infty ]$. Let $\mathcal{A}\left({\mu }^{\ast }\right)$ be the $\sigma$-algebra consisting of ${\mu }^{\ast }$-measurable subsets of $X$ and $\mathcal{B}(X)\subset \mathcal{P}(X)$ be the Borel $\sigma$-algebra of $(X,d)$. Then the following statements are equivalent:

• $\mathcal{B}(X)\subset \mathcal{A}\left({\mu }^{\ast }\right)$;
• if $A,B\subset X$ satisfy $d(A,B):={\inf }_{a\in A,b\in B}d(a,b)>0$; then ${\mu }^{\ast }(A\cup B)=$ ${\mu }^{\ast }(A)+{\mu }^{\ast }(B)$.

\begin{proof} "→": Let $A,B\subseteq X$ with $d(A,B)>0$. Take open $U$ of $X$ such that $A\subseteq U$ and $U\cap B=\varnothing$. Since $U\in \mathcal{A}({\mu }^{\ast })$, we have

$${\mu }^{\ast }(A\cup B)={\mu }^{\ast }(A\cup B\cap U)+{\mu }^{\ast }(A\cup B\cap U^c)={\mu }^{\ast }(A)+{\mu }^{\ast }(B).$$

"←": Fix a closed set $A\subseteq X$, then we aim to show for any $D\subseteq X$ there is

$${\mu }^{\ast }(D)⩾{\mu }^{\ast }(D\cap A)+{\mu }^{\ast }(D\cap A^c).$$

WLOG we may assume ${\mu }^{\ast }(D)<\infty$. Define $U_k={\cup }_{a\in A}B(a,\frac{1}{k})$, then $d(D\cap A,D-U_k)⩾\frac{1}{k}>0$ and so

$${\mu }^{\ast }(D)⩾{\mu }^{\ast }((D\cap A)\cup (D-U_k))={\mu }^{\ast }(D\cap A)+{\mu }^{\ast }(D-U_k).$$

Claim that ${\lim }_{k\to \infty }{\mu }^{\ast }(D-U_k)={\mu }^{\ast }(D\setminus A)$. Since $A={\cap }_{k=1}^{\infty }{U}_k$, we have that $D-A=(D-U_k)\cup ((U_k-A)\cap D)$. It follows that ${\mu }^{\ast }(D-A)⩽{\mu }^{\ast }(D-U_k)+{\mu }^{\ast }((U_k-A)\cap D)$. Note that $U_k-A=U_k-{\cap }_{j=1}^{\infty }{U}_k={\cup }_{j=k}^{\infty }(U_j-U_{j+1})$ and $D\cap (U_k-A)={\cup }_{j=k}^{\infty }(D\cap U_j-U_{j+1})={\cup }_{j=k}^{\infty }{E}_j$ where $E_j=D\cap U_j-U_{j+1}$.

Sub-claim that ${\sum }_{j=1}^{\infty }{\mu }^{\ast }(E_j)<\infty$. If $d(E_i,E_j)>0$ for any $j⩾i+2$, then we have ${\mu }^{\ast }(D)⩾{\mu }^{\ast }({\cup }_{i=1}^{\ell }{E}_{2i})=\sum {\mu }^{\ast }(E_{2i})$, ${\mu }^{\ast }(D)⩾{\mu }^{\ast }({\cup }_{i=1}^{\ell }{E}_{2i-1})=\sum {\mu }^{\ast }(E_{2i-1})$ and so ${\sum }_{k=1}^{\infty }{\mu }^{\ast }(E_i)⩽2{\mu }^{\ast }(D)<\infty$. Thus, we only need to show $d(E_i,E_j)>0$ if $j⩾i+2$. If $x\in E_i$, $y\in X$ and $d(x,y)<\frac{1}{(i+1)(i+2)}$, then $y\notin E_j$ by the definition of $U_j$. Now we finish the proof. \end{proof}

Remark. 学实变的时候我好像管这个证明叫瑞士卷来着。

Theorem

Let ${\mu }^{\ast }$ be the Lebesgue outer measure on $\mathcal{P}({\mathbb{R}}^n)$. Then:

• ${\mu }^{\ast }(A+x)={\mu }^{\ast }(A)$ for any $x\in {\mathbb{R}}^n$;
• if $\forall A,B\subseteq {\mathbb{R}}^n$ with $d(A,B)>0$, then ${\mu }^{\ast }(A\cup B)={\mu }^{\ast }(A)+{\mu }^{\ast }(B)$;
• for any $Q$, ${\mu }^{\ast }(Q)=|Q|={\mu }^{\ast }(Q^0)$.

\begin{proof} i) & ii) are easy, but iii) is tricky. Note that $|Q|={\lim }_{N\to \infty }\frac{1}{N^n}\#(Q\cap \frac{1}{N}{\mathbb{Z}}^n)$. We use it to show for any ${\cup }_{i=1}^{\infty }{Q}_i\supseteq Q$, there is $|Q|⩽\sum |Q_i|$. To make sure RHS converges, take open $U_i\supseteq Q_i$ and consider its open cover of $Q$.

Also see here. \end{proof}

Remark. 把算体积变成算格点数. As a corollary of ii), elements in $\mathcal{B}({\mathbb{R}}^n)$ are Lebesgue measurable.

regularity of the Lebesgue outer measure

The followings hold.

• Let $A\subseteq {\mathbb{R}}^n$. Then ${\mu }^{\ast }(A)=\inf \{{\mu }^{\ast }(U):U\text{mbox}isopen,A\subseteq U\}$.
• If $A\in \mathcal{L}({\mathbb{R}}^n)$, then ${\mu }^{\ast }(A)=\sup \{{\mu }^{\ast }(K):K\subseteq A,K\text{mbox}iscompact\}$.
• If $\{A_i\}\subseteq \mathcal{P}({\mathbb{R}}^n)$ and $A_i\uparrow A$, then ${\mu }^{\ast }(A)={\lim }_{i\to \infty }{\mu }^{\ast }(A_i)$.

\begin{proof} i) and ii) are Easy. See here.

iii) If ${\mu }^{\ast }(A)=\infty$, done. Let ${\mu }^{\ast }(A)<\infty$. For any $ϵ>0$, there is open $U_i\supseteq A_i$ and ${\mu }^{\ast }(U_i)⩽{\mu }^{\ast }(A_i)+ϵ$. Let $V_i={\cap }_{j=i}^{\infty }{U}_j$. Then

$${\mu }^{\ast }(A)⩽{\mu }^{\ast }({\cup }_{i=1}^{\infty }{V}_i)=\underset{i\to \infty }{\lim }{\mu }^{\ast }(V_i)⩽\underset{i\to \infty }{\lim }{\mu }^{\ast }(U_i)⩽\underset{i\to \infty }{\lim }{\mu }^{\ast }(A_i)+ϵ$$

and so we finish the proof. \end{proof}

the Lebesgue measure as a completion

Let ${\mu }^{\ast }:\mathcal{P}\left({\mathbb{R}}^n\right)\to$ $[0,\infty ]$ be the Lebesgue outer measure, let $m={{\mu }^{\ast }|}_{\mathcal{L}\left({\mathbb{R}}^n\right)}:\mathcal{L}\left({\mathbb{R}}^n\right)\to [0,\infty ]$ be the Lebesgue measure, let $\mathcal{B}\subset \mathcal{L}\left({\mathbb{R}}^n\right)$ be the Borel $\sigma$-algebra of ${\mathbb{R}}^n$, and define

$$\nu ={{\mu }^{\ast }|}_{\mathcal{B}}:\mathcal{B}\to [0,\infty ]$$

Then $\left({\mathbb{R}}^n,\mathcal{L}\left({\mathbb{R}}^n\right),m\right)$ is the completion of $\left({\mathbb{R}}^n,\mathcal{B},\nu \right)$.

\begin{proof} Use ^7b1c20 to construct union of compact sets $B_0$ and intersection of open sets $B_1$ such that $B_0\subseteq A\subseteq B_1$ with ${\mu }^{\ast }(B_1-B_0)=0$. Also see here. \end{proof}

Theorem

Let $C_0({\mathbb{R}}^n)$ be the set of continuous functions with compact support. Then $C_0({\mathbb{R}}^n)$ is dense in $L^1({\mathbb{R}}^n)$.

\begin{proof} It suffices to show, for any $f\in L^1({\mathbb{R}}^n)$, there exists $g\in C_0({\mathbb{R}}^n)$ with $||g-f|{|}_1<ϵ$. On the other word, there exists $\{f_k\}\subseteq C_0({\mathbb{R}}^n)$ with $||f_k-f|{|}_1\to 0$.

Let $\mathcal{J}=\{f\in L^1(\mathbb{R}):\exists \{f_k\}\subseteq C_0({\mathbb{R}}^n),||f_k-f|{|}_1\to 0\}$. We can show that

• $\mathcal{J}$ is closed w.r.t. finite linear combinations.
• If $\{f_i\}\subseteq \mathcal{J}$ and $||f_k-f|{|}_1\to 0$, then $f\in \mathcal{J}$.

Then we approximate $f$ with simple functions, and it is enough to show ${\chi }_E\in \mathcal{J}$ with ${\mu }^{\ast }(E)<\infty$ and so for ${\chi }_I$ with half open cube $I$. The last one is easy to prove.

Also see here. \end{proof}

Theorem

The set of continuous functions with compact support is dense in $L^p\left({\mathbb{R}}^n\right)$ for $1⩽p<\infty$.

\begin{proof} Suppose $f\in L^p$. We have ${\int }_{{\mathbb{R}}^n}{\left|f-f{\chi }_{B(0,n)}\right|}^p\to 0$ as $n\to \infty$ by DCT. Hence it suffices to approximate functions in $L^p$ that have compact support. By writing $f=f^{+}-f^{-}$ we may suppose $f⩾0$.

Similarly as ^110128, it suffices to approximate characteristic functions of any partly open cube. For any given partly open cube $I$, there exists $g$ continuous with compact support and with values in $[0,1]$ such that ${\int }_{{\mathbb{R}}^n}\left|g-{\chi }_I\right|<ϵ$. Since $\left|g-{\chi }_I\right|⩽1$, then ${\int }_{{\mathbb{R}}^n}{\left|g-{\chi }_I\right|}^p⩽{\int }_{{\mathbb{R}}^n}\left|g-{\chi }_I\right|<ϵ$. This completes the proof. \end{proof}

Continuity in $L^p$

Let $f\in L^p\left({\mathbb{R}}^n\right),1\le p<\infty$. Then

$$\underset{|h|\to 0}{\lim }\Vert f(\cdot +h)-f(\cdot ){\Vert }_p=0$$

\begin{proof} Define the set $C_p$ as the class of $f\in L^p$ such that $||f(\cdot +h)-f(\cdot )|{|}_p\to 0$ as $|h|\to 0$. Then $f$ is closed under finite linear combination and the convergence with norm $L^p$. Since the characteristic function of a cube is contained in $C_p$, by ^110128 and ^fe0685 we finish the proof.

Also see here. \end{proof}

transformation formula

Suppose $\varphi :U\to V$ is a $C^1$ diffeomorphism between open subsets of ${\mathbb{R}}^n$.

• If $f:V\to [0,\infty ]$ is Lebesgue measurable, then $f\circ \varphi :U\to [0,\infty ]$ is Lebesgue measurable and

$${\int }_U(f\circ \varphi )|\det (d\varphi )|dm={\int }_{V}fdm.$$

• If $E\in {\mathcal{L}}_U$ and $f\in L^1\left(m_V\right)$, then $\varphi (E)\in {\mathcal{L}}_V,(f\circ \varphi )|\det (d\varphi )|\in L^1\left(m_U\right)$ and

$${\int }_E(f\circ \varphi )|\det (d\varphi )|dm={\int }_{\varphi (E)}fdm.$$

Remark. The proof is omitted.