Measure theory, prove the countable additivity of measure

Question

I am reading Cohn's Measure Theory, here is an exercise from Chapter 1, Section 2.

Let ($X,\mathcal A,\mu$) be a measure space, and define $\mu^{\star}:\mathcal A\rightarrow[0,\infty)$ by \begin{equation} \mu^{\star}(A)＝sup\{\mu(B):B\subset A,B\in\mathcal A , and \,\,\mu(B)\leq+\infty\} \end{equation} (a) Show that $\mu^{\star}$ is a measure on ($X,\mathcal A$)

(b) Show that if $\mu$ is $\sigma$-finite, then $\mu^{\star}＝\mu$

In terms of the countable additivity of $\mu^{\star}$, here is my approach.

Let $\{A_{k}\}$ be a sequence of disjoint of sets that belong to $\mathcal A$, let $B$ $\subset \bigcup_{k＝1}^{\infty} A_{k}$, thus, $B＝\bigcup_{k＝1}^{\infty}(B\cap A_{k})$, it follows that $\mu (B)＝\sum_{k＝1}^{\infty}\mu(B\cap A_{k})\leq \mu^{\star}(A_{k})$

My question is:

1. How to prove the reverse inequality holds？
2. And how to prove (b)?