The following is from Achim Klenke - "Probability Theory" p. 28, Theorem 1.65 (ii):
Let $\mathcal{A}\subset 2^{\Omega}$ be a semiring and let $\mu$ be a measure on $\sigma(\mathcal{A})$ that is $\sigma$-finite on $\mathcal{A}$.
For any $\mathcal{A}\in\sigma(\mathcal{A})$ with $\mu(A)<\infty$ and any $\epsilon>0$ there exists an $n\in\mathbb{N}$ and mutually disjoint sets $A_1, ..., A_n\in \mathcal{A}$ such that $\mu(A\Delta \bigcup_{k=1}^n A_k)<\epsilon$.
The proof starts as follows:
As $\mu$ and the outer measure $\mu^*$ coincide on $\sigma(\mathcal{A})$ and since $\sigma(\mathcal{A})$ is finite, by the very definition of $\mu^*$ there exists a covering $B_1,...\in\mathcal{A}$ of $A$ such that $\mu(A)\le \sum^\infty_{i=1}\mu(B_i) -\epsilon/2$
Here I think we need the construction of the outer measure, which is a lemma in my book, but is equivalent to this: Outer measure construction
So in the proof the epsilon property of the infimum is used and it should start like this
$\mu^*(A)\le \sum^\infty_{i=1}\mu(B_i) -\epsilon/2$
Now if the outer measure conincides on $\sigma(\mathcal{A})$ it does so on its generator. But the problem is, that $A$ is not necessarily in $\mathcal{A}$, hence the outer measures on $\mathcal{A}$ and $\sigma(\mathcal{A})$ don't have to be equal. If I were to take a cover from $\sigma(\mathcal{A})$ the outer measure is smaller or equal than that on just $\mathcal{A}$.
Why can I take a cover in $\mathcal{A}$ which is guaranteed to satisfy the inequality?