I want to prove the equality of inner and outer measures for measurable sets. The hypothesis are these:
Let be $R$ an algebra over $X$, $\mu$ a quasi-measure over $R$ and $R^*$ the set of the measurable subsets of $X$, so we can build the outer and inner measures respectively:
$$\mu^*(E)=\text{inf}\{\mu(A): A\in S(R),E\subseteq A\}$$
$$\mu_{*}(E)=\text{sup}\{\mu(A): A\in S(R),A\subseteq E\}.$$ Then: $E\in R^*$ iff $\mu_{*}(E)=\mu^*(E)$. We can suppose $\mu^*(E)<\infty.$
Best Answer
Step 1: Let $E \subseteq X$ be an arbitrary set. Show that there exists $B \in S(R)$ such that $E \subseteq B$ and $\mu(B) = \mu^*(E)$.
Step 2: Let $E \subseteq X$ be such that $\mu^*(E) < \infty$. Choose $B$ be as in step 1. Then
$$\begin{align*} \mu_*(E) &= \sup \{\mu(A); A \subseteq E, A \in S(R)\} \\ &= \sup\{\mu(A); B \cap A^c \supseteq B \cap E^c, A^c \in S(R)\} \\ &= \sup\{\mu(B)-\mu(B \cap A^c); B \cap A^c \supseteq B \cap E^c, B \cap A^c \in S(R)\} \\ &= \mu(B) - \inf\{\mu(C); C \supseteq B \cap E^c, C \in S(R)\} \\ &= \mu(B)-\mu^*(B \backslash E). \end{align*}$$
Step 3: "$\Rightarrow$" Let $E \in R^*$ be such that $\mu^*(E) < \infty$. By the measurability of $E$, we have
$$\mu^*(E) = \mu^*(B) = \mu^*(B \cap E) + \mu^*(B \backslash E) = \mu^*(E) + \mu^*(B \backslash E).$$
As $\mu^*(E)<\infty$, this implies $\mu^*(B \backslash E)=0$. Combining this with step 2, we get
$$\mu_*(E) = \mu(B) = \mu^*(E).$$
Step 4: "$\Leftarrow$": Suppose that $E \subseteq X$ is such that $\mu^*(E) = \mu_*(E)$. Choose $B$ as in step 1. It follows from step 2 that $\mu^*(B \backslash E)=0$. For an arbitrary set $A \subseteq X$, we have
$$\begin{align*} \mu^*(A \cap E) + \mu^*(A \cap E^c)&\leq \mu^*(A \cap B) + \mu^*\big( (A \cap B^c) \cup (B \backslash E) \big) \\ &\leq \mu^*(A \cap B) + \mu^*(A \cap B^c) + \mu^*( B \backslash E) \end{align*}$$
which implies by the measurability of $B$
$$\mu^*(A \cap E) + \mu^*(A \cap E^c) \leq \mu^*(A) + \mu^*(B \backslash E) = \mu^*(A).$$
Since the inequality "$\geq$" is trivially satisfied (because of the subadditivity of the outer measure), this shows that $E$ is measurable.