# Continuity from below for Lebesgue outer measure

Let $\{E_n\}$ be an increasing sequence of subsets of $\mathbb{R}^n$, measurable or not. Then $$m^* \bigg( \bigcup_{n=1}^{\infty}E_n \bigg)=\lim_{n\rightarrow\infty}m^*E_n$$

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Okay, I believe you. What is your question? – Henning Makholm Mar 5 '12 at 23:01
Presumably $m^*$ is supposed to be outer Lebesgue measure? – ShawnD Mar 5 '12 at 23:04
Sorry @Henning Makholm, My question is how to prove it. – Ace Mar 6 '12 at 0:17
@ShawnD, yes that's Lebegues outer measure. – Ace Mar 6 '12 at 0:20
You might rename this "Continuity from below for Lebesgue outer measure" or something like that. – Quinn Culver Mar 6 '12 at 1:51

## 2 Answers

I got it.

Since $\bigcup_{n=1}^{\infty}E_n \supset E_n$ for all $n$, We have $m^* \big( \bigcup_{n=1}^{\infty}E_n \big) \geqslant m^*E_n$. Therefore $m^* \big( \bigcup_{n=1}^{\infty}E_n \big)$ $\geqslant$ $\lim_{n\to\infty}m^*E_n$. So it's clear when $\lim_{n\to\infty}m^*E_n=\infty$. Then we assume $\lim_{n\to\infty}m^*E_n<\infty$. For all $\varepsilon>0$ and $n$, there exists $\{I_{n,i}\}_{i\in N_+}$, a sequence of open intervals in $\mathbb R^n$, covering $E_n$, s.t.

$$m\bigg(\bigcup_{i=1}^{\infty}I_{n,i}\bigg) \leqslant\sum_{i=1}^{\infty}m(I_{n,i}) =\sum_{i=1}^{\infty}|I_{n,i}| <m^*E_n+\frac{\varepsilon}{2^n}$$

by definition and properties of Lebesgue outer measure, and the L-measurability of open intervals. Then every $G_n:=\bigcup_{i=1}^{\infty}I_{n,i}\supset E_n$ is an open set, and $mG_1< m^*E_1+\varepsilon/2$. Assuming $m\big(\bigcup_{k\leqslant n}G_k\big)$ $<$ $m^*E_n+(1-1/2^n)\varepsilon$, we have

\begin{align*} m\Big(\bigcup_{k\leqslant n+1}G_k\Big) &=m\Big(\bigcup_{k\leqslant n}G_k\Big)+mG_{n+1} -m\bigg(\Big(\bigcup_{k\leqslant n}G_k\Big)\bigcap G_{n+1}\bigg) \\ &<\Big[m^*E_n+\Big(1-\frac{1}{2^n}\Big)\varepsilon\Big] +\Big(m^*E_{n+1}+\frac{\varepsilon}{2^{n+1}}\Big)-m^*E_{n} \\ &=m^*E_{n+1}+\Big(1-\frac{1}{2^{n+1}}\Big)\varepsilon. \end{align*}

So $m\big(\bigcup_{k\leqslant n}G_k\big) <m^*E_n+(1-\frac{1}{2^n})\varepsilon$ for all $n.$ That means, since $\varepsilon$ can be arbitrarily small,

$$m\Big(\bigcup_{k\leqslant n}G_k\Big)\leqslant m^*E_n$$ for all $n$. Finally,

$$m^* \bigg( \bigcup_{n=1}^{\infty}E_n \bigg) \leqslant m\bigg( \bigcup_{n=1}^{\infty}G_n \bigg) =m\bigg( \bigcup_{n=1}^{\infty}\bigcup_{k=1}^n G_k \bigg) =\lim_{n\to\infty}m\bigg(\bigcup_{k=1}^n G_k \bigg) \leqslant \lim_{n\to\infty}m^*E_n .\quad\square$$

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Perhaps I am stating the trivial thing, but the Borel sets should be closed under countable union, etc. And if I remember correctly the popular definition of Lesbegue measure allows us to use Borel sets to approach measurable sets. To prove this for Borel set should be more or less automatic, like making standard set manipulations, etc.

You may be interested to read about the limit superior and limit inferior.

Edit: sorry I ignored your condition that $E_{i}$ are not necessarily measurable. Under such condition you may be interested in cutting $\bigcup E_{i}$ into $\bigcup (E_{i}-E_{j})$s. Since the two sets are the same and the second set is a union of separated subsets, you should be able to use subaddivity to prove this.

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I've tried to prove that by cutting $\bigcup E_{i}$ as you suggested, and found that the subaddivity doesn't help. – Ace Mar 6 '12 at 7:58
I agree, the result does not follow directly, we need more assumptions. – checkmath Mar 6 '12 at 12:20
@Ace: By subaddiviy we have $m^{*}(U(A_{i}))=\sum m{*}(A_{i})$. Now use $A_{i}=E_{i}-E_{i-1}$. We have $m^{*}(U(A_{i}))=\sum m{*}(A_{i})=\sum m^{*}(E_{i}-E_{i-1})$. But we can prove $m^{*}E_{i}$ to be $\sum^{i}_{j=k} m^{*}(E_{j}-E_{j-1})$. Assume $\lim m^{*}E_{n}$ is finite (or else there is nothing to prove), then the result should easily follows. – Kerry Mar 6 '12 at 17:30
uh...your format isn't displayed correctly. – Ace Mar 6 '12 at 22:17
@Changwei Zhou: I've copied your comment and edited it so it can be displayed correctly. Here is it: By subaddiviy we have $m^{}(U(A_{i}))=\sum m{}(A_{i})$. Now use $A_{i}=E_{i}-E_{i-1}$. We have $m^{}(U(A_{i}))=\sum m{}(A_{i})=\sum m^{}(E_{i}-E_{i-1})$. But we can prove $m^{}E_{i}$ to be $\sum^{i}_{j=k} m^{}(E_{j}-E_{j-1})$. Assume $\lim m^{}E_{n}$ is finite (or else there is nothing to prove), then the result should easily follows. – Ace Mar 6 '12 at 22:28