Show that there exists a non-measurable set $A\subseteq \mathbb{R}$ such that:
$\mu_{\ast}(A\cap E)=0$ and $\mu^{\ast}(A\cap E)=\mu(E)$
For each measurable set E with $\mu(E)>0$.
So my attempt was trying to get a Vitali set (which is known to be unmeasurable) and constructing a set E that can have inner measure 0 and outer measure equals its Lebesgue measure, but I'm stuck in constructin a good set E that can satisfy these conditions.
I'm sure there's a simpler answer to this question, but this was the best I could come up with.
One can modify the Vitali set construction in $[0,1]$ such that the resulting nonmeasurable set $V$ has outer measure $1$ and inner measure $0$. This problem is discussed at length here. Then put $$A = \bigcup_{n\in\mathbb{Z}} (V+n),$$ which is clearly nonmeasurable.
We claim the inner measure $0$ property is retained. Indeed, given $F$ compact with $F\subset A$, we have $(F\cap [n,n+1])-n\subset V$, so we find $\mu(F\cap[n,n+1]) = 0$ and thus $\mu(F)=0$. This proves $\mu_*(A\cap E)\leq\mu_*(A) = 0$ for any measurable set $E$.
To prove the outer measure property, we first assume the measurable set $E$ is bounded, so there is positive integer $N$ large enough that $E\subset [-N,N]$. Also put $A_N: = A\cap [-N,N]$, which is easily seen to have outer measure $2N$. Applying Carathéodory criterion gives $$\mu^*(A_N) = \mu^*(E\cap A_N)+\mu^*(E^c\cap A_N).$$ Since $E^c\cap A_N\subset [-N,N]\setminus E$, we have $$\mu^*(E^c\cap A_N)\leq \mu^*([-N,N]\setminus E) = 2N-\mu(E).$$ Thus our inequality becomes $2N\leq \mu^*(E\cap A_N)+2N-\mu(E)$. Rearranging, $$\mu(E) \leq \mu^*(E\cap A_N)\leq \mu^*(E) =\mu(E),$$ so $\mu(E)= \mu^*(E\cap A_N) = \mu^*(E\cap A)$ in this case.
In the general case, we use inner regularity of measure and continuity of outer measure from below. Measurability of $E$ implies $E=F\cup Z$, where $F$ is $F_\sigma$ and $Z$ is measure $0$. It is sufficient to prove $\mu^*(F\cap A) = \mu(E)$ (exercise). Now, there is a sequence of compact sets $F_n\nearrow F$ such that $\mu(F_n)\nearrow\mu(F)=\mu(E)$ as $n\to\infty$. Each $F_n$ satisfies $\mu(F_n) = \mu^*(F_n\cap A)$. Yet $F_n\cap A\nearrow F\cap A$, so $$\mu^*(F\cap A) = \lim \mu^*(F_n\cap A) = \lim \mu(F_n) = \mu(E).$$ This completes the proof.
