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I am trying to show that for $A$, $B \; \subset \mathbb{R}$ $A \cup B$ is Lebesgue measurable such that \begin{align} \infty > \mu(A \cup B) = \mu^{*}(A) + \mu^{*}(B), \end{align} Then $A$ and $B$ are Lebesgue measurable.

Thus far, I have tried reversing the directions in the proof of Caratheodory. I seem to want to do the following:

If $A \cup B$ is Lebesgue measurable, then, for any $E \subset \mathbb{R}$, \begin{align} \mu^{*}(E) \geq \mu^{*}(E \cap (A \cup B)) + \mu^{*}(E \cap (A \cup B)^{c}). \end{align}

I want to use this and work backwards from here adding and subtracting sets to get

\begin{align} \mu^{*}(E) \geq \mu^{*}(E \cap A) + \mu^{*}(E \cap A^{c}). \end{align} yet I am having trouble.

DRich
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  • I'm not sure this is true. Take some nonmeasurable set $X$ with infinite outer measure. Then $\mathbb R - X$ and $X$ would have a union of $\mathbb R$, which is measurable. – Trevor Norton Mar 21 '16 at 01:45
  • Yes. See edit. I meant for finite measure. – DRich Mar 21 '16 at 01:48

2 Answers2

2

To prove measurability of $A$, it is sufficient to consider only $E \subset A\cup B$. (Otherwise split $E\subset \mathbb R$ in $E\cap (A\cup B)$ plus $E\cap (A\cup B)^c$ and apply the measurability of $A\cup B$).

For $E \subset A\cup B$, take measurable $G \subset A\cup B$, such that $E\subset G$ and $\mu(G) = \mu^*(E)$. (This is only part, where we use that $\mu$ is the Lebesgue measure.) Therefore \begin{align} \mu^*(A) &\geqslant \mu^*(A\cap G) + \mu^*(A\cap G^c), \tag{1}\\ \mu^*(B) \geqslant \mu^*((A\cup B)\cap A^c) &\geqslant \mu^*(A^c\cap G) + \mu^*((B\setminus A)\cap G^c), \tag{2}\\ \mu(A\cup B) &= \mu(G) + \mu((A\cup B) \cap G^c). \tag{3} \end{align}

Summing up $(1)$ and $(2)$, we have $$ \mu(A\cup B) = \mu^*(A) + \mu^*(B) \geqslant \mu^*(A \cap G) + \mu^*(A^c \cap G) + \mu^*((A\cup B)\cap G^c) \geqslant \mu(A\cup B), $$ and the last inequality is indeed equality. Subtracting it by $(3)$ (this is part where we need finiteness) and using monotonicity of $\mu^*$, we finally get $$ \mu^*(E) = \mu (G) = \mu^*(G \cap A) + \mu^*(G \cap A^c) \geqslant \mu^*(E\cap A) + \mu^*(E\cap A^c) $$

mishmish-dev
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Firstly, fixes a measurable set $X$, given any set $A\subseteq X$, we have a standard method to construct a measurable set $\widehat{A}$, such that $A\subseteq \widehat{A}\subseteq X$ and $\mu^*(A)=\mu(\widehat{A})$. By the definition of outer measure, for any $n\in \mathbb N$, we have a covering $\displaystyle A\subseteq \bigcup_{k=1}^\infty I_k=G_n$, such that every $I_k$ is an open interval in $\mathbb R^1$, and $\displaystyle\mu^*(A)\leqslant\mu^*(G_n)\leqslant \mu^*(A)+\frac{1}{n}$. Trivially $G_n$ is measurable, and let $\displaystyle \widehat{A}=\bigcap_{n=1}^\infty G_n\cap X$, and we will get $\displaystyle\mu(\widehat{A})\leqslant \mu(G_n)\leqslant \mu^*(A)+\frac{1}{n}$. As $A$ is subset of all $G_n$ and $X$, we can get $\mu(\widehat{A})\geqslant \mu^*(A)$, then $\mu(\widehat{A}) = \mu^*(A)$ since $n$ is arbitrary.

Now we return to the original problem. Let $X=A\cup B$, then $\mu(\widehat{A})-\mu(X-\widehat{X\backslash A}) = \mu(\widehat{A})+\mu(\widehat{X\backslash A})-\mu(X)=\mu^*(A)+\mu^*(X\backslash A)-\mu(X)\geqslant 0$ (all these sets are measurable, and the final inequality is sub-additivity). However, $\mu^*(A)+\mu^*(X\backslash A)-\mu(X)\leqslant \mu^*(A)+\mu^*(B)-\mu(X) =0$, so it can only happen that $\mu(\widehat{A})=\mu(X-\widehat{X\backslash A})$. By $\Phi=X-\widehat{X\backslash A}\subseteq A$, we can get $\mu^*(\widehat{A}-A)\leqslant \mu(\widehat{A}-\Phi)= \mu(\widehat{A})-\mu(\Phi)=0$. A set of reals with outer measure 0 is measurable, then $\widehat{A}-A$ is measurable, by $\widehat{A}$ is measurable, we have $A$ is measurable.

Finally, the result can be generalized to countably infinity unions. Let $X$ be a measurable set, and $X_1, X_2, \cdots X_n\subseteq X$, $X=\bigcup_{k=1}^\infty X_k$, $\displaystyle\sum_{k=1}^{\infty}\mu^*(X_k)=\mu(X) < \infty$. Given any $k \in \mathbb N$, we can get $\displaystyle\mu^*(X_k)+\sum_{j\neq k}\mu^*(X_j)=\mu(X)$, by sub-additivity we have $\displaystyle\mu^*\left(\bigcup_{j\neq k} X_j\right)\leqslant \sum_{j\neq k}\mu^*(X_j)$, so we get $\displaystyle\mu(X)\geqslant \mu^*(X_k)+\mu^*\left(\bigcup_{j\neq k} X_j\right)$. By sub-additivity, we can get $\displaystyle\mu(X) = \mu^*(X_k)+\mu^*\left(\bigcup_{j\neq k} X_j\right)$, then $X_k$ is measurable.

user779130
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