8. Lebesgue measure in \(\mathbb R^d\)  PDF TEX

Metric outer measures

Proof \((\Longrightarrow)\). Let us assume that \(\mu^*\) is a metric outer measure.

Step 1. Since the closed sets generate the Borel \(\sigma\)-algebra \({\rm Bor}(X)\), it suffices to show that every closed set \(F\subseteq X\) is \(\mu^*\) measurable. Thus, given \(A\subseteq X\) with \(\mu^*(A)<\infty\), we have to prove that \[\mu^*(A)\ge \mu^*(A\cap F)+\mu^*(A\cap F^c).\]

Step 2. For each \(n\in \mathbb N\) let \(B_n=\{x\in A\cap F^c: \rho(x, F)\ge n^{-1}\}\). Then \(B_n\subseteq B_{n+1}\) and \[\bigcup_{n\in \mathbb N}B_n=A\cap F^c.\] Indeed, since each \(B_n\subseteq A\cap F^c\) then \(\bigcup_{n\in \mathbb N}B_n\subseteq A\cap F^c.\) Conversely, if \(x\in A\cap F^c\) then \(\rho(x, F)>0\) since \(F\) is closed. Thus there is \(n\in\mathbb N\) such that \(\rho(x, F)\ge n^{-1}\) and consequently \(x\in B_n\) yielding \(A\cap F^c\subseteq \bigcup_{n\in \mathbb N}B_n\).

By the construction of \(B_n\) we also see that \(\rho(B_n, F)\ge n^{-1}\), giving \[\mu^*(A)\ge \mu^*((A\cap F)\cup B_n)\ge \mu^*(A\cap F)+\mu^*(B_n),\] since \(\mu^*\) is a metric outer measure.

It suffices to prove that \[\lim_{n\to \infty}\mu^*(B_n)=\mu^*(A\cap F^c).\]

Step 3. For each \(n\in \mathbb N\) let \(C_n=B_{n+1}\setminus B_{n}\). We claim that \[\rho(C_{n+1}, B_n)\ge \frac{1}{n(n+1)}.\] Indeed, if for some \(x\in C_{n+1}\) and \(y\in B_{n}\) we have \(\rho(x, y)< \frac{1}{n(n+1)}\) then \[\frac{1}{n}\le \rho(y, F)\le \rho(x, y)+\rho(x, F)<\frac{1}{n(n+1)} + \frac{1}{n+1}=\frac{1}{n},\] which is impossible and the claim follows.

A simple induction therefore shows that \[\begin{aligned} \mu^*(B_{2n+1})&\ge \mu^*(C_{2n}\cup B_{2n-1})=\mu^*(C_{2n})+ \mu^*(B_{2n-1})\\ &\ge \mu^*(C_{2n})+ \mu^*(C_{2n-2}\cup B_{2n-3})\\ &=\mu^*(C_{2n})+\mu^*(C_{2n-2})+ \mu^*(B_{2n-3})\ge \sum_{j=1}^n\mu^*(C_{2j}), \end{aligned}\] and similarly one can prove that \(\mu^*(B_{2n})\ge \sum_{j=1}^n\mu^*(C_{2j-1}).\)

Step 4. We have shown that \[\mu^*(B_{2n+1})\ge \sum_{j=1}^n\mu^*(C_{2j}),\qquad \text{ and } \qquad \mu^*(B_{2n})\ge \sum_{j=1}^n\mu^*(C_{2j-1}).\] Since \(\mu^*(B_n)\le \mu^*(A)<\infty\) then \[\sum_{j=1}^{\infty}\mu^*(C_{2j})<\infty, \qquad \text{ and } \qquad \sum_{j=1}^{\infty}\mu^*(C_{2j-1})<\infty.\] By subadditivity, for each \(n\in\mathbb N\), we have that \[\mu^*(A\cap F^c)\le \mu^*(B_n)+\sum_{j=n+1}^{\infty}\mu^*(C_j).\] Passing with \(n\to\infty\) the last sum vanishes and we conclude that \[\mu^*(A\cap F^c)\le \liminf_{n\to\infty}\mu^*(B_n)\le \limsup_{n\to\infty}\mu^*(B_n)\le \mu^*(A\cap F^c).\] This completes the proof of one implication.$$\tag*{$\blacksquare$}$$

Proof \((\Longleftarrow)\). Let us assume that \({\rm Bor}(X)\subseteq \mathcal M(\mu^*)\) and \(\mu^*\) is an outer measure on \(X\). We will show that \(\mu^*\) is a metric outer measure.

• Let \(A, B\subseteq X\) be such that \(\alpha=\rho(A, B)>0\). Consider an open set \(G_x=\{y\in X: \rho(x, y)<\alpha/2\}\) for every \(x\in X\). Then \[A\subseteq G=\bigcup_{x\in A}G_x.\] Moreover, \(G\) is open and \(G\cap B=\varnothing\).

• Since \(G\in {\rm Bor}(X)\subseteq \mathcal M(\mu^*)\) we have \[\mu^*(A\cup B)=\mu^*((A\cup B)\cap G)+\mu^*((A\cup B)\cap G^c),\] which implies \[\mu^*(A\cup B)=\mu^*(A)+\mu^*(B).\] since \((A\cup B)\cap G=A\) and \((A\cup B)\cap G^c=B\).$$\tag*{$\blacksquare$}$$

Lebesgue outer measure is a metric outer measure

Lebesgue outer measure is a metric outer measure

Proof. Let \(E_1, E_2\subseteq \mathbb R^d\) be such that \(\delta=\rho(E_1, E_2)>0\). We have to show that \[\lambda_d^*(E_1\cup E_2)=\lambda_d^*(E_1)+ \lambda_d^*(E_2).\]

• In fact, by subadditivity we only need to show that \[\lambda_d^*(E_1\cup E_2)\ge \lambda_d^*(E_1)+ \lambda_d^*(E_2).\]

• It suffices to show that for every sequence \((R_n)_{k\in\mathbb N}\subseteq J_{co}^d\) such that \(\bigcup_{k\in\mathbb N}R_k\supseteq E_1\cup E_2\) we have \[\sum_{k\in \mathbb N}{\rm v}_d(R_k)\ge \lambda_d^*(E_1)+ \lambda_d^*(E_2).\] If \({\rm v}_d(R_k)=\infty\) for some \(k\in\mathbb N\) then there is nothing to prove.

• We may assume that \({\rm v}_d(R_k)<\infty\) for every \(k\in\mathbb N\). Now for each \(k\in \mathbb N\) each rectangle \(R_k\) can be decomposed into finitely many closed-open rectangles \(\{R_{k, j}:j\in\{1,\ldots, m_k\}\}\subset J_{co}^d\) with the property that

  • \(\{R_{k, j}:j\in\{1,\ldots, m_k\}\}\) is a disjoint collection and \(R_k=\bigcup_{j=1}^{m_k}R_{k, j}\);

  • the lengths of the edges of \(R_{k, 1}, \ldots, R_{k, m_k}\) are all bounded above by \[\delta(2\sqrt{d})^{-1}.\]

• Note that \[{\rm v}_d(R_k)=\sum_{j=1}^{m_k}{\rm v}_d(R_{k, j})\] and let \(\mathcal Q= \{Q_n:n\in\mathbb N\}\) be an arbitrary enumeration of the set \(\{R_{k, j}: k\in\mathbb N \text{ and } j\in\{1,\ldots, m_k\} \}\). Then \[\sum_{n\in\mathbb N}{\rm v}_d(Q_n)=\sum_{k\in\mathbb N}\sum_{j_k=1}^{m_k}{\rm v}_d(R_{k, j_k})=\sum_{k\in\mathbb N}{\rm v}_d(R_{k}).\]

• Our claim is that no member of \(\mathcal Q\) intersects both \(E_1\) and \(E_2\). Suppose for a contradiction that there is \(n\in\mathbb N\) such that \[Q_n\cap E_1\neq\varnothing \quad \text{ and } \quad Q_n\cap E_2\neq\varnothing.\]

• Let \(x_1\in Q_n\cap E_1\) and \(x_2\in Q_n\cap E_2\) and note that \[\delta=\rho(E_1, E_2)\le \rho(x_1, x_2)\le \Big(d\big(\delta(2\sqrt{d})^{-1}\big)^2\Big)^{1/2}=\frac{\delta}{2},\] which is impossible!

• Let \(\mathcal Q_1= \{Q\in \mathcal Q:Q\cap E_1\neq\varnothing\}\) and \(\mathcal Q_2= \{Q\in \mathcal Q:Q\cap E_2\neq\varnothing\}\). Consequently, we obtain that \[\begin{aligned} \sum_{k\in\mathbb N}{\rm v}_d(R_{k})&=\sum_{n\in\mathbb N}{\rm v}_d(Q_n)\\ &\ge \sum_{Q\in \mathcal Q_1}{\rm v}_d(Q)+ \sum_{Q\in \mathcal Q_2}{\rm v}_d(Q)\\ &\ge \lambda_d^*(E_1)+ \lambda_d^*(E_2). \end{aligned}\]

This competes the proof. $$\tag*{$\blacksquare$}$$