Caratheodory: Inner vs. Outer Problem
Given a plain space $\Omega$ and a ring $\mathcal{R}$.
(In fact, a semiring would do the job, too.)
Consider a premeasure $\mu:\mathcal{R}\to\overline{\mathbb{R}}_+$.
For simplicity, preextend this to:
$$\quad D=\bigsqcup_kR_k:\quad\mu_\infty(D):=\sum_k\mu(R_k)$$
(This one is well-defined and countably additive.)
Introduce inner and outer measure as:
$$\mu_+(A):=\inf_{A\subseteq D}\mu_\infty(D)$$
$$\mu_-(A):=\sup_{D\subseteq A}\mu_\infty(D)$$
(CAUTION: There are various definitions of inner measure floating around.)
Meanwhile, perform Caratheodory's construction:
$$E\in\mathcal{M}:\iff\mu_+(A)=\mu_+(A\cap E)+\mu_+(A\cap E^c)\quad(A\subseteq\Omega)$$
Now, are there measurable sets whos inner measure differs from its outer measure, here:
$$E\in\mathcal{M}:\quad\mu_-(E)<\mu_+(E)$$
and are there nonmeasurable sets whos inner measure agrees with its outer measure, here:
$$V\notin\mathcal{M}:\quad\mu_-(V)=\mu_+(V)$$
Attempt
Given the ring $\mathcal{R}:=\{\varnothing,(1)\}$ over the space $\Omega=\{1,2\}$.
Consider the premeasure $\mu(\varnothing)=0$ and $\mu(1)=1$.
Then for the space itself one has:
$$\Omega\in\mathcal{M}:\quad\mu_-(\Omega)=1<\infty=\mu_+(\Omega)$$
Now, the second problem is still open...
 A: So, if we allow the non-measurable set to have infinite measure, this is easy:
Take the standard construction of Lebesgue measure (i.e. $R$ is a suitable semiring of intervals) and let $V = [10,\infty) \cup \Bbb{V}$, where $\Bbb{V} \subset (0,1)$ is nonmeasurable. Then $V$ is not measurable, but the inner and outer measure both equal $\infty$.
If we require that both measures (inner/outer) are finite and equal, this implies measurability:
By definition, there are sequences $D_n \subset V \subset E_n$ (where $D_n, E_n$ are countable (disjoint) unions of elements of $R$, hence in $\mathcal{M}$, because each element of $R$ is measurable by Caratheodory's theorem) with $\mu_\infty (D_n) \to \mu_- (V) = \mu_+ (V) =\lim \mu_\infty (E_n)$.
It is easy to see that we can assume $(D_n)_n, (E_n)_n$ to be increasing/decreasing. Let us denote the Caratheodory extension of $\mu$ (or $\mu_\infty$) by $\nu$.
Then the above yields $\nu(\bigcup D_n) = \mu_- (V)=\nu(\bigcap E_n)<\infty$, so that $V = \bigcup D_n \cup V\setminus \bigcup D_n$ is measurable, because $V\setminus \bigcup D_n \subset \bigcap E_n \setminus \bigcup D_n$ is a ($\nu$)null-set as a subset of a null-set and hence measurable (the Caratheodory construction yields a complete measure).
