The theorem (from Rudin) states:
For a measure space $(X,\mathfrak M, \mu)$, let $\mathfrak M^*$ be the collection of all $E\subset X$ for which there exists sets $A$ and $B\in\mathfrak M$ such that $A\subset E\subset B$ and $\mu(B-A)=0$, and define $\mu(E)=\mu(A)$ in this situation. Then $\mathfrak M^*$ is a $\sigma$-algebra.
(Here $X$ is the set, $\mathfrak M$ is the $\sigma$-algebra, and $\mu$ is the measure)
The proof states:
We begin by checking that $\mu$ is well defined for every $E\in \mathfrak M^*$. Suppose that $A\subset E\subset B$, $A_1\subset E\subset B_1$, and $\mu(B-A)=\mu(B_1-A_1)=0$. Since$$A-A_1\subset E-A_1\subset B-A_1\,\,\,\,\,\,\,\,\,\,\,\,(1)$$we have $\mu(A-A_1)=0$, hence $\mu(A)=\mu(A\cap A_1)$.
My questions are : What does well-defined mean in this proof? Why is it important? I don't think I understand why equation $(1)$ implies that $\mu(A-A_1)=0$. Nor do I understand why $\mu(A-A_1)=0$ implies that $\mu(A)=\mu(A_1\cap A)$.
It means that this definition provides a UNIQUE value to $\mu(E)$, and it is INDEPENDENT of the choice of the sets $A$ and $B$.