Property of decreasing sequences in a measure space I want to proof the following theorem regarding decreasing sequences in measure spaces:

Let (S,R,$\mu$) be a  measure space and let $(A_n)_{n\geq1}$ be a sequence in R. Proof that if:
1). $A_n$ decreases to A (i.e. $A_n \supset A_{n+1}$  for all $n \in \mathbb{N}$. and $A = \cap_{n=1}^\infty A_n$) and 
2). $\mu(A_1) < \infty$ 
then $\mu(A_n)$ decreases to $\mu(A)$

So our goal is to show that $\lim_{n\rightarrow\infty} \mu(A_n) = \mu(A)$.
We know that $\mu(A_1)<\infty$ and we know trivially that $\cap_{n=1}^\infty A_n \subset A_1$, therefore $\mu(A_1) \geq \mu(A)$. We also know that $\cap_{n=1}^\infty A_n \subset (A_1 \cap A2)$, therefore $ \mu(A_1) \geq \mu(A_1 \cap A_2) \geq \mu(A)$. 
We note that, since $(A_n)_{n\geq 1}$ is a decreasing sequence, we can write down $B_k = \cap_{n=1}^k A_n$. This is where I can't seem to figure out what to do next. I want to proof that with the boundedness of $B_k$ that $\lim_{k\rightarrow\infty} \mu(B_k) = \mu(A)$ but I don't know how I'd do this.
Is my proof until this point correct, or do I have the wrong idea? I'd appreciate a hint on what to do.
Thanks for reading,
K. Kamal.
 A: Lemma: Let $\mu$ be a measure on a measurable space $(X,\mathcal A)$ and let $(A_n)_n$ be a sequence of measurable sets with: $$A_1\subseteq A_2\subseteq A_3\subseteq\cdots\text{ and }A:=\bigcup_{n=1}^{\infty}A_n$$
then:  $$\lim_{n\to\infty}\mu(A_n)=\mu(A)$$
Proof: Setting $A_0=\varnothing$ and $C_n=A_n-A_{n-1}$ for $n=1,2,\dots$ we have $A=\bigcup_{k=1}^{\infty} C_k$ where the $C_k$ are measurable and disjoint, and also $A_n=\bigcup_{k=1}^nC_k$. 
Then: $$\mu(A)=\sum_{n=1}^{\infty}\mu(C_k)=\lim_{n\to\infty}\sum_{k=1}^n\mu(C_k)=\lim_{n\to\infty}\mu(A_n)$$

Corollary: Let $\mu$ be a measure on a measurable space $(X,\mathcal A)$ with $\mu(A_1)<\infty$  and let $(A_n)_n$ be a sequence of measurable sets with: $$A_1\supseteq A_2\supseteq A_3\supseteq\cdots\text{ and }A:=\bigcap_{n=1}^{\infty}A_n$$
then:  $$\lim_{n\to\infty}\mu(A_n)=\mu(A)$$
Proof: The lemma can be applied on sequence $(A_1-A_n)_n$ and leads to $$\lim_{n\to\infty}\mu(A_1-A_n)=\mu(A_1-A)$$
Then, on base $\mu(A_1-A_n)=\mu(A_1)-\mu(A_n)$ and $\mu(A_1-A)=\mu(A_1)-\mu(A)$  we also find:$$\lim_{n\to\infty}\mu(A_n)=\mu(A)$$
