Outer measurability

Question

Let $m$ be a finite measure on $X$ and $A$ a sigma-algebra in $\mathcal{P}(X)$.
Now $m^*$ is the outer measure on $X$ and $m_*(Y)=m(X)-m^*(Y^\complement)$.

How do I prove that $Y$ is $m^*$-measurable iff $m_*(Y)=m^*(Y)?$

What I know:
Definition of $Y$ being $m ^*$-measurable is $$m^*(Q)=m^*(Q\cap Y)+m^*(Q\cap Y^\complement).$$ So if $Y$ is $m^*$-measurable, then we can take $X$ as $Q$ in the equation above and conclude $m_*(Y)=m^*(Y)$.
But how do I show now that $m_*(Y)=m^*(Y)\implies$ $Y$ is $m^*$-measurable?

Answer 1

Find $m$-measurable

$$\begin{array} {lcl} A \supseteq Y & & m(A) = m^*(Y) \\ B \supseteq Y^{\complement} & \text{such that} & m(B) = m^*( Y^{\complement} ) \\ C \supseteq Q & & m(C) = m^*(Q). \end{array}$$

(We can have equalities since $\mathcal{A}$ is a $\sigma$-algebra. But we could go with picking those measurable sets to approximate their subsets up to measure $\varepsilon$ and it would work out too.)

Now since $A^{\complement} \subseteq Y^{\complement} \subseteq B$:

$$\begin{align*} m(X) & = m(C \cap A) + m(C \cap A^{\complement}) + m(C^{\complement} \cap A) + m(C^{\complement} \cap A^{\complement}) \\ & \leqslant m(C \cap A) + m(C \cap B^{\phantom{\complement}}) + m(C^{\complement} \cap A) + m(C^{\complement} \cap B^{\phantom{\complement}}) \\ & = m(A) + m(B) = m^*(Y) + m^*(Y^{\complement}) = m(X) \end{align*}$$

so we have equalities everywhere between the corresponding terms, in particular

$$m(C \cap A) + m(C \cap B) = m(C \cap A) + m(C \cap A^{\complement}) = m(C).$$

Since $Q \cap Y \subseteq C \cap A$ and $Q \cap Y^{\complement} \subseteq C \cap B$:

$$m^*(Q \cap Y) + m^*(Q \cap Y^{\complement}) \leqslant m(C \cap A) + m(C \cap B) = m(C) = m^*(Q)$$

so the sides are equal, because the reverse inequality holds as usual.

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Answers to questions in the comments:

1. From the definition of an outer measure, for each $n > 0$ there is an $m$-measurable $C_n \supseteq Q$ such that $m(C_n) \leqslant m^*(Q) + \frac{1}{n}$. Let $\displaystyle C = \bigcap_{n=1}^{\infty} C_n$. Since $\mathcal{A}$ is a $\sigma$-algebra, $C$ is $m$-measurable. Moreover $C \supseteq Q$ so for each $n$:

   $$m^*(Q) \leqslant m(C) \leqslant m(C_n) \leqslant m^*(Q) + \frac{1}{n},$$

   hence $m(C) = m^*(Q)$. Similarly with $A, B$.

2. Clearly

   $$\begin{align*} m(C \cap A) + m(C \cap A^{\complement}) & \leqslant m(C \cap A) + m(C \cap B) \\ & \text{and} \\ m(C^{\complement} \cap A) + m(C^{\complement} \cap A^{\complement}) & \leqslant m(C^{\complement} \cap A) + m(C^{\complement} \cap B). \end{align*}$$

   But if either of the inequalities were strict, so would be the one in

   $$\begin{align*} m(X) & = m(C \cap A) + m(C \cap A^{\complement}) + m(C^{\complement} \cap A) + m(C^{\complement} \cap A^{\complement}) \\ & \color{red}{\leqslant} m(C \cap A) + m(C \cap B^{\phantom{\complement}}) + m(C^{\complement} \cap A) + m(C^{\complement} \cap B^{\phantom{\complement}}) \\ & = m(A) + m(B) = m^*(Y) + m^*(Y^{\complement}) = m(X) \end{align*}$$

   hence $m(X) < m(X)$, which can't be.