Let $\mathcal{E}$ be a collection of subset of $\Omega$ and $\varnothing \in \mathcal{E},$ let $\mu$ be a countable subadditive set function defined on $\mathcal{E}$ for which $\mu(\varnothing) = 0.$ If $\tau$ is the outer measure that generated by $\mu,$ and $A\subseteq \Omega$ satisfying $$\tau(X) = \tau(X\cap A) + \tau(X\cap A^c), \quad \forall X \in \mathcal{E},$$ then $A$ is $\tau$-measurable.
I immediately got a "solution":
Let $D\subseteq \Omega, B_n \in \mathcal{E}$ such that $$D \subseteq \bigcup_{n=1}^{+\infty}B_n \quad \text{and} \quad \sum_{n=1}^{+\infty}\mu(B_n) < \tau(D) + \varepsilon.$$ Then we have \begin{aligned} \tau(D) &\le \tau(D\cap A) + \tau(D\cap A^c)\\ &\le \tau\left(\bigcup_{n=1}^{+\infty}B_n \cap A\right) + \tau\left(\bigcup_{n=1}^{+\infty}B_n \cap A^c\right) \qquad &\text{since monotonicity}\\ &\le \sum_{n=1}^{+\infty}\tau(B_n \cap A) + \sum_{n=1}^{+\infty}\tau(B_n \cap A^c) \qquad &\text{since subadditive}\\ &= \sum_{n=1}^{+\infty}\tau(B_n) \qquad &\text{follow by the condition}\\ &\le \sum_{n=1}^{+\infty}\mu(B_n) < \tau(D) + \varepsilon. \end{aligned}
However I believed my solution is wrong, since I don't use the countable subadditive of $\mu,$ but I don't know in which step of my proof is fake.