Now that the measures are assumed finite I think this is doable. First we wish to show that $\mu_*(A) = \mu(\Omega) - \mu^*(A)$. Let $\epsilon > 0$. Choose $B_i$ so that $\cup_i B_i \subset A$, and $\mu_*(A) \le \mu(\cup_i B_i) + \epsilon$. It follows since $\mu$ is additive that for $C\in \mathcal{A}$, $\mu(C) = \mu(\Omega) - \mu(C^c)$. A picky detail, but additionally there exists $n$ so that $\mu(\cup_i B_i) \le \mu(\cup_{i=1}^n B_i) + \epsilon$. This is needed since $\cup_{i=1}^n B_i \in \mathcal{A}$ ($\mathcal{A}$ is an algebra). With this, and noting that $ A^c \subset \left(\cup_{i=1}^n B_i \right)^c $,
$$
\mu_*(A) \le \mu(\cup_i B_i) + \epsilon \le \mu(\cup_{i=1}^n B_i) + 2\epsilon = \mu(\Omega) - \mu(\left(\cup_{i=1}^n B_i \right)^c) + 2\epsilon \le \mu(\Omega) - \mu^*(A^c) + 2\epsilon.
$$
Starting with a cover $C_i$ such that $A^c \subset \cup_i C_i$ and $\mu(\cup_i C_i) \le \mu^*(A^c) + \epsilon$, one can show very similarly that
$$
\mu^*(A^c) \ge \mu(\Omega) - \mu_*(A) -2\epsilon.
$$
Manipulating a bit and putting the above equations together we have
$$
\mu(\Omega)-\mu^*(A^c) - 2\epsilon \le \mu_*(A) \le \mu(\Omega)-\mu^*(A^c) + 2\epsilon.
$$
Since $\epsilon$ is arbitrary, we have $\mu_*(A) = \mu(\Omega)-\mu^*(A^c)$. Now this implies that the collection of sets $\mathcal{B}$ is closed under the complement. Suppose $A\in \mathcal{B}$ so that $\mu_*(A)= \mu^*(A)$. Then $\mu_*(A^c) =\mu(\Omega)-\mu^*(A)= \mu(\Omega)-\mu_*(A)= \mu^*(A^c)$.
It also came up in the comments to show that $\mathcal{B}$ is closed under intersection. To that end let $\epsilon>0$, and suppose $A_1,A_2 \in \mathcal{B}$. As a result we may choose $B_{i,1},B_{i,2},C_{i,2},C_{i,2}$ with the following properties: 1) $\cup_i B_{i,j} \subset A_j \subset \cup_i C_{i,j}$, $j=1,2$. 2) $\mu( \cup_{i} C_{i,j} / \cup_{i} B_{i,j}) < \epsilon/2$, $j=1,2$. Evidently then $(\cup_i B_{i,1}) \cap (\cup_i B_{i,2}) \subset A_1 \cap A_2 \subset (\cup_i C_{i,1}) \cap (\cup_i C_{i,2})$. Further, it can be shown that $(\cup_{i } C_{i,1} \cap C_{i,2}) /(\cup_{i } B_{i,1} \cap B_{i,2}) \subset (\cup_{i} C_{i,1} / \cup_{i} B_{i,1}) \cup (\cup_{i} C_{i,2} / \cup_{i} B_{i,2}). $ It follows that $\mu ( (\cup_i C_{i,1} \cap \cup_i C_{i,2}) /(\cup_i B_{i,1}) \cap (\cup_i B_{i,2})) < \epsilon$, giving that $\mu^*(A_1 \cap A_2) - \mu_*(A_1\cap A_2) < \epsilon$. Since $\epsilon$ is arbitrary, $\mu^*(A_1 \cap A_2)=\mu_*(A_1\cap A_2)$.
Best Answer
You need to look for weird sets, like a Vitali set.