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.
Best Answer
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 $$ \mu^*(E) = \mu (G) = \mu^*(G \cap A) + \mu^*(G \cap A^c) \geqslant \mu^*(E\cap A) + \mu^*(E\cap A^c) $$