Prove that, if a sequence is nested from above and has empty intersection, the probability of its terms tends to zero 
Let $(A_n)_{n\geq 1}$ be a sequence in a probability space. Show that $P(A_n)\to 0$ as $n\to \infty$ if $A_{n+1}\subset A_{n}$ for all $n>1$ and $\bigcap_{n=1}^{\infty} A_n = \emptyset.$

Here's my attempt. I'm having difficulty formalising the last step. 
First of all:
\begin{align*} 0 &= P(\emptyset)\\
&= P(\bigcap_{i=1}^{\infty}A_i)\\
&= 1-P(\bigcup_{i=1}^{\infty}A_i^{\complement}).
\end{align*}
Then, define
$B_1:= A_1^{\complement}$
$B_2:= A_1\backslash A_2$
...
$B_n:= A_{n}\backslash A_{n+1}$
The $B_n$'s are all disjoint, and so, from $\sigma$-additivity, 
\begin{align*} P(\bigcup_{i=1}^{\infty}A_i^{\complement}) &= P(\bigcup_{i=1}^{\infty}B_i) = \sum_{i=1}^{\infty}P(B_i).
\end{align*}
From above, this sum converges to $1$. Hence, its terms converge to $0$, that is to say, $P(B_n)\to 0$ as $n\to\infty$. Intuitively, this should show that $P(A_n)\to 0$ as well. But why should it?
 A: $B_i:=X\setminus A_i$ where $X$ is the whole space. Then 
$$\bigcup_iB_i=\bigcup_i(X\setminus A_i)=X\setminus\Big(\bigcap_i A_i\Big)$$
by de Morgan's laws. If $\bigcap_i A_i=\emptyset$ then $\bigcup_iB_i=X$. Therefore 
$$\Pr\Big(\bigcap_i A_i\Big)=1-\Pr\Big(X\setminus\bigcap_i A_i\Big)=1-\Pr(\bigcup_iB_i\Big)=1-\Pr(X)=1-1=0$$
Since $(A_n)\subset X$ are nested in a decreasing order then for every $n$ we have $$A_n=\bigcap_{1\leqslant i\leqslant n}A_i\Rightarrow A:=\lim_nA_n=\bigcap_{i\in\mathbb{N}}A_i$$
On the other hand $A_{n+1}\subset A_n$ for every $n$ implies $\Pr(A_{n+1})\leqslant\Pr(A_n)$. Also for all $n$ we have $0\leqslant\Pr(A_n)\leqslant \Pr(X)=1$ hence we have a sequence $(\Pr(A_n))$ of real numbers which is bounded and monotonic decreasing. By Bolzano-Weierstrass this sequence converges. Therefore
$$\lim_n\Pr(A_n)=\Pr(\lim_nA_n)=\Pr(A)=\Pr\Big(\bigcap_i A_i\Big)=0$$
where we have used the right continuity of $\Pr(\cdot)$. 
A: A measure $\mu:\Omega\rightarrow\mathbb{R}$ has the decreasing continuity property:
If $A_{n+1}\subset A_n$ and  $\mu(\Omega)<\infty \implies \lim_{n\rightarrow \infty}\mu(A_n)=\mu(\cap_{n\in\mathbb{N}}A_n)$. 
In particular, a probability measure is finite, so in your case, as  $A_{n+1}\subset A_n$ and $\cap_{n\in\mathbb{N}}A_n=\emptyset$,  by decreasing continuity, we have
$$\lim_{n\rightarrow \infty}\mathbb{P}(A_n)=\mathbb{P}(\cap_{n\in\mathbb{N}A_n})=\mathbb{P}(\emptyset)=0$$
