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Show $A \subseteq \mathbb{R}$ is a Lebesgue Measurable set $\iff$ there exists $F_\sigma$ set $F \subseteq \mathbb{R}$ contained in $A$ and $G_\delta$ set $G \subseteq \mathbb{R}$ containing $A$ s.t. $m(G \setminus A) = m(A \setminus F) = 0$


Note: Up to now, I have proven inner and outer regularity of $m$. Definitions below:

Inner Regular: If $A$ is Lebesgue Measurable, then $m(A)=sup\{m(K) | K$ is an closed set contained in $A \}$

Outer Regular: If $A$ is Lebesgue Measurable, then $m(A)=inf\{m(U) | U$ is an open set containing $A \}$

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  • $\begingroup$ Would you mind elaborate the so-called "inner and outer regularity"? Different text might use different terminologies. $\endgroup$
    – xbh
    Sep 5, 2018 at 16:02
  • $\begingroup$ Thank you, editted the question $\endgroup$
    – NazimJ
    Sep 5, 2018 at 16:45

2 Answers 2

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Different authors define Lebesgue measurability differently. One common definition is that $A \subset \mathbb R$ is measurable if and only if for every $\epsilon > 0$ there exists an open set $G \supset A$ satisfying $m^*(G \setminus A) < \epsilon$ where $m^*$ is the Lebesgue outer measure. A simple consequence of this definition is that any set $E$ with $m^*(E) = 0$ is measurable.

Suppose $A$ is measurable. For every $k \ge 1$ there exists $G_k \supset A$ so that $m^*(G_k \setminus A) < \frac 1k$. Define $G = \cap G_k$. Then $G$ is a $G_\delta$ set, $G \subset G_k$ for all $k$ and $$m^*(G \setminus A) \le m^*(G_k \setminus A) < \frac 1k$$ for all $k$, forcing $m^*(G \setminus A) = 0$.

Likewise, $A$ measurable implies $\mathbb R \setminus A$ measurable, so by the above reasoning there exists a $G_\delta$ set $H \supset \mathbb R \setminus A$ with $m^*(H \setminus (\mathbb R \setminus A)) = 0$. If you let $F = \mathbb R \setminus H$ you find that $F$ is an $F_\sigma$ set, $F \subset A$, and $$m^*(A \setminus F) = m^*(H \setminus (\mathbb R \setminus A)) = 0.$$

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  • $\begingroup$ I see you use the fact the the compliment of a 'G_\delta' set is an 'F_\sigma' set. But one piece of logic I can't seem to connect is the last line, 'm^{*} (A \setminus F) = m^{*} (H \setminus (\mathbb{R} \setminus A))'. Why is their measure the same? $\endgroup$
    – NazimJ
    Sep 6, 2018 at 14:15
  • $\begingroup$ Write set difference using intersection and complement: $A \setminus F = A \cap F^c$. Since $F^c = H$ you have $$A \cap F^c = A \cap H = H \cap A^{cc} = H \cap (\mathbb R \setminus A)^c.$$ $\endgroup$
    – Umberto P.
    Sep 6, 2018 at 15:33
  • $\begingroup$ That is, $A \setminus F$ and $H \setminus (\mathbb R \setminus A)$ are the same set. $\endgroup$
    – Umberto P.
    Sep 6, 2018 at 15:33
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If $m(A)<\infty$ we have the following: For each $j\in \mathbb{N}$ there exist by inner and outer regularity a compact set $K_j\subset A$ and an open set $U_j\supset A$ s.t $$ m(U_j)+1/j \leq m(A)\leq m(K_j)-1/j. $$

Put $G=\cap_{j}A_j$ and $F=\cup_j(K_j)$. Then $G$ and $F$ are the desired $G_\delta$ and $F_{\sigma}$ sets respectively.

If $m(A)=\infty$, then decompose $A$ as $A=\cup_j (A\cap(j,j+1])$ then apply the above exposed to each $A_j=A\cap(j,j+1]$.

On the other hand if there exists a $G$ a $G_\delta$ set satisfying $m(G\setminus A)=0$ and $m(A\setminus F)=0$, we have for each test set $E$

$$ m(E)=m(E\cap G)+m(E\cap G^c)\geq m(E\cap A)+m(E\cap A^c\cap(G\setminus A)^c)\geq m(E\cap A)+m(E\cap A^c) $$

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  • $\begingroup$ What if $m(A) = \infty$? $\endgroup$
    – Umberto P.
    Sep 5, 2018 at 16:59
  • $\begingroup$ Thanks for the warning $\endgroup$
    – Eduardo
    Sep 5, 2018 at 17:03

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