If there is $B \subset A$ with $\mu(B)=\mu^{\ast}(A)$ then $A$ is measurable Let $A\subset E$ such that $\mu^{\ast}(A)<+\infty$, How can I prove that If there is $B \subset A$ with $\mu(B)=\mu^{\ast}(A)$ then $A$ is measurable.
Is it true if $\mu(A)=+\infty$?
I don't know how attack this problem, could someone help me with this, please.
Thanks for your help and time.
 A: 
Let $A$ be such that $\mu^*(A) < \infty$ and let $B \subset A$ be a measurable set such that $\mu(B) = \mu^*(A).$ Then $$\mu^*(A \setminus B) = \mu^*(A) - \mu^*(B) = 0.$$

Notice that for the previous equalities to be true we need the assumption $\mu^*(A) < \infty$. This is in general not true otherwise, indeed one can consider $A = \mathbb{R}$ and $B = (0,\infty)$. Then $\mu(A) = \mu(B) = \infty$, but clearly $$\mu(A \setminus B) = \mu((-\infty,0)) = \infty.$$

This shows that $A \setminus B$ is measurable since by the Caratheodory condition sets of outer measure $0$ are measurable. To conclude the argument it is enough to notice that $A$ belongs to the $\sigma$-algebra of measurable sets since it can be written as the union of two measurable sets: $$A = B \cup (A \setminus B).$$

The result is not true in general if $\mu^*(A) = \infty$. Indeed, let $C \subset (-1,0)$ be a non-measurable set and consider $B = (0,\infty)$ and $A = B \cup C$.
A: As suggested already, first show $\mu^*(A \cap B^c) = 0$ if $\mu^*(A) < \infty$.
Next, using this result, you can show directly that $\mu^*(T \cap A)=\mu^*(T \cap B)$ and $\mu^*(T \cap A^c) = \mu^*(T \cap B^c)$ for all $T \subset \Omega$. This implies by measurability of $B$ that
\begin{equation}
\mu^*(T \cap A) + \mu^*(T \cap A^c) = \mu^*(T \cap B) + \mu^*(T \cap B^c) = \mu^*(T) .
\end{equation}
Hence, $A$ is measurable.
To prove the claims that $\mu^*(T \cap A)=\mu^*(T \cap B)$ and $\mu^*(T \cap A^c) = \mu^*(T \cap B^c)$ for all $T \subset \Omega$, note that from $B \subset A$ and monotonicity of $\mu^*$ it follows that
\begin{equation}
\mu^*(T \cap A) = \mu^*(T \cap A \cap B) + \mu^*(T \cap A \cap B^c) = \mu^*(T \cap B) + 0 .
\end{equation}
Note that we have $\mu^*(T \cap A \cap B^c) = 0$ by monotonicity. 
Moreover, $\mu^*(T \cap A^c) \leq \mu^*(T \cap B^c)$ by monotonicity and 
\begin{equation}
\mu^*(T \cap B^c) \leq \mu^*(T \cap B^c \cap A) + \mu^*(T \cap B^c \cap A^c) = 0 + \mu^*(T \cap A^c)
\end{equation}
by subadditivity. Hence, $\mu^*(T \cap A^c) = \mu^*(T \cap B^c)$, as claimed.
A: Hint: For the first part, note that $\mu(B)=\mu_*(B)=\mu^*(A)$. Conclude that $\mu^*(A\setminus B)=0$. For the second part, let $A'$ be your favourite non-measurable subset of $[0,1]$ and consider $A=A'\cup (1,\infty)$.
