Let $\mu$ be a regular outer measure on $X$, $\mu(X) < \infty$. If $A \subseteq X $ is set such that $\mu(A)+\mu(A^c)=\mu(X)$ then $A$ is measurable

Question

A set $A \subset X$ is measurable if $\forall T \subset X$ $$ \mu(T) = \mu(A \cap T)+\mu(A^c \cap T) $$ By a regular measure I mean a measure $\mu$ such that for every $A \subset X$ there exists $B$ measurable such that $\mu(A) = \mu(B)$.

Edit: Forgot to add that $\mu(X) < \infty$

Answer 1

I think you meant:

A set $B \subseteq X$ is measurable if $\forall T \subseteq X$ $$ \mu(T) = \mu(B \cap T)+\mu(B^c \cap T) $$ By a regular outer measure it is meant an outer measure $\mu$ such that for every $A \subseteq X$ there exists $B$ measurable such that $A \subseteq B$ and $\mu(A) = \mu(B)$.

The result to prove is

Let $\mu$ be a regular outer measure on $X$, $\mu(X) < \infty$. If $A \subseteq X $ is set such that $\mu(A)+\mu(A^c)=\mu(X)$ then $A$ is measurable.

Proof: Let $A \subseteq X $ such that $\mu(A)+\mu(A^c)=\mu(X)$.

For any measurable subset $B$ of $X$, we have $\mu(A)=\mu(A \cap B) + \mu(A \cap B^c)$
and $\mu(A^c)=\mu(A^c \cap B) + \mu(A^c \cap B^c)$. And we also have, by sub-additivity: $\mu(B) \leqslant \mu(A \cap B)+ \mu(A^c \cap B) $ and $\mu(B^c) \leqslant \mu(A \cap B^c)+ \mu(A^c \cap B^c) $.

Since $B$ is measurable, we also have $\mu(X)= \mu(B) + \mu(B^c)$. Since $\mu(X)< \infty$, we have $\mu(B^c) < \infty$ and so we have $\mu(B)= \mu(X) - \mu(B^c)$.

Now, using those equalities, we have:

\begin{align*} \mu(B) & \leqslant \mu(A \cap B)+ \mu(A^c \cap B) \leqslant \\ & \leqslant \mu(A \cap B)+ \mu(A^c \cap B) + ( \mu(A \cap B^c)+ \mu(A^c \cap B^c) -\mu(B^c) ) =\\ & = (\mu(A \cap B) + \mu(A \cap B^c)) + ( \mu(A^c \cap B) + \mu(A^c \cap B^c)) - \mu(B^c) = \\ & = \mu(A) + \mu(A^c) - \mu(B^c) = \\ & = \mu(X) - \mu(B^c) = \\ & = \mu(B) \end{align*} So, any measurable subset $B$ of $X$, we have $$ \mu(B) = \mu(A \cap B)+ \mu(A^c \cap B)$$

Now, given any $T \subseteq X$, since $\mu$ is a regular outer measure, there is $B$ measurable such that $T \subseteq B$ and $\mu(T) = \mu(B)$. So, we have \begin{align*} \mu(T) & \leqslant \mu(A \cap T)+ \mu(A^c \cap T )\leqslant \mu(A \cap B)+ \mu(A^c \cap B) = \mu(B) = \mu(T) \end{align*} So, given any $T \subseteq X$, we have $$ \mu(T) = \mu(A \cap T)+ \mu(A^c \cap T ) $$ It means, $A$ is measurable.