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I will use the notation of Wheeden & Zygmund's Measure and Integral because the problem come from that book: $|\cdot|_e$ for outer measure and $|\cdot|$ for Lebesgue measure. An exercise says: prove (3.29) and (3.29) is:

Theorem Suppose that $|E|_e<\infty$. Then $E$ is measurable if and only if given $\epsilon>0$, $E=(S\cup N_1)\setminus N_2$, where $S$ is a finite union of non-overlapping intervals and $|N_1|_e,|N_2|_e<\epsilon$.

I did "$\implies$", and I was thinking that "$\impliedby$" would be easier. But is not after all. I appreciate any advice.

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Please let me know if something is unclear. –  leo Sep 24 '11 at 3:30

1 Answer 1

up vote 5 down vote accepted

I think I got it. Now the objective of my question (and my answer) is to know if my proof is acceptable. Please let me know it in the comments. Of course, any other answer is welcome.

Let $\epsilon>0$. Let $S'=\bigcup_{k=1}^mI_k$ a finite union of nonoverlapping intervals; $N_1'$ and $N_2$ such that $$E=(S'\cup N_1')\setminus N_2$$ and $|N_1'|_e,|N_2|_e<\epsilon/4.$ Then $$E=\left[\bigcup_{k=1}^mI_k^\circ \cup \left(\bigcup_{k=1}^m\partial(I_k)\cup N_1'\right)\right]\setminus N_2.$$ Let $$S=\bigcup_{k=1}^mI_k^\circ \text{ and } N_1=\bigcup_{k=1}^m\partial(I_k)\cup N_1'.$$ Therefore $S$ is open and by subadditivity and monotony of the outer meausure $|N_1|_e<\epsilon/4.$ Also $$E=(S\cup N_1)\setminus N_2.$$ Since the outer measure is defined as an inf, there exist a countable collection of open intervals $\{J_k\}_{k\in\Delta}$ that cover $N_1$ and satisfies $$\sum_{k\in\Delta}|J_k|<|N_1|_e+\epsilon/4<\epsilon/2.$$ Let $O=\bigcup_{k\in\Delta} J_k$. Then $O$ is open, $N_1\subseteq O$ and by the subadditivity and monotony of the outer meausure $$|O\setminus N_1|_e\leq |O|\leq \sum_{k\in\Delta}|J_k|<\epsilon/2.$$

In the other hand, is possible to show that $$(S\cup N_1)\setminus E=(S\cup N_1)\cap N_2\subseteq N_2,$$ and therefore $$|(S\cup N_1)\setminus E|_e\leq |N_2|_e<\epsilon/4<\epsilon/2.$$

Finally, $E\subseteq S\cup N_1\subseteq S\cup O$, wich is open, and by the subadditivity and monotony of the outer measure we get

$$\begin{align*} |(S\cup O)\setminus E|_e &= |[S\cup (N_1\cup (O\setminus N_1))]\setminus E|_e\\ &= |[(S\cup N_1)\cup (O\setminus N_1)]\setminus E|_e\\ &\leq |(S\cup N_1)\setminus E|_e+|(O\setminus N_1)\setminus E|_e\\ &\leq |(S\cup N_1)\setminus E|_e+|O\setminus N_1|_e\\ &<\frac{\epsilon}{2} + \frac{\epsilon}{2}=\epsilon. \end{align*}$$

Therefore $E$ is measurable.

Just an observation, I did not use of the hypothesis $|E|_e<\infty$. Am I missing something?

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