Proof of continuity from above using continuity from below

I'm reading a proof of continuity from above using continuity from below.
Let $(X,\mathscr{A},\mu)$ be a measure space. Suppose $A_{1} \supseteq A_{2} \supseteq \cdots $ is a sequence of decreasing sets in $\mathscr{A}$. Suppose $\mu(A_{1}) < \infty$, we want to show that $$\lim_{k \to \infty} \mu(A_{k}) = \mu(\bigcap_{i=1}^{\infty} A_{i}).$$Proof: We let $E_{i} = A_{1}\setminus A_{i}$ for $i = 1,2,3,\ldots$. We then have $E_{1} \subseteq E_{2} \subseteq \cdots$. By continuity from below, we have $\lim_{k \to \infty} \mu(E_{k}) = \mu(\bigcup_{i=1}^{\infty} E_{i})$. Then the book just says note the fact that $\mu(E_{i}) = \mu(A_{1}) - \mu(A_{i})$. Then we're done. However, I plug it in. I get $\lim_{k \to \infty} \mu(E_{k}) = \lim_{k \to \infty} (\mu(A_{1})-\mu(A_{k}))$. So I have $$\lim_{k \to \infty} (\mu(A_{1}) - \mu(A_{k})) = \mu(\bigcup_{i=1}^{\infty} (A_{1} \setminus A_{i})) .$$However, I don't see clearly how to get from here to the equality we want.