Show that: $\mu\left(\bigcup_N \bigcap_{n=N}^{\infty} A_n \right) \leq \lim \inf \mu(A_n)$ Let $(X,\mathcal{M},\mu)$ be a measure space. Let $A_1, A_2, \ldots \in \mathcal{M}$.
Then, I want to show that: $$\mu\left(\bigcup_N \bigcap_{n=N}^{\infty} A_n \right) \leq \lim \inf \mu(A_n)$$
There is a solution in lecture notes:
Let $B_N = \bigcap_{n=N}^{\infty} A_n$. $B_N$ form an increasing sequence of elements, then by continuity from below:
$$\mu\left(\bigcup_N \bigcap_{n=N}^{\infty} A_n \right) = \mu\left(\bigcup_N B_N\right) = \lim_{N \to \infty} \mu(B_N) \leq \lim_{N \to \infty} \inf_{n \geq N} \mu(A_n) = \lim \inf \mu(A_n)$$
OK. I do not understand how $\mu\left(\bigcup_N B_N\right) = \lim_{N \to \infty} \mu(B_N)$. And then following inequality and equality. Can somebody give a detailed explanation? Thank you very much.
 A: As for the equality
$$
\mu\left(\bigcup\limits_{N=1}^\infty B_N\right)=\lim\limits_{N\to\infty}\mu(B_N)
$$
the proof is the following. Since $\{B_N:N\in\mathbb{N}\}$ is the increasing sequence of sets we have 
$$
\bigcup\limits_{N=1}^\infty B_N=\coprod\limits_{N=1}^\infty C_N
$$
where $C_1=B_1$ and $C_N=B_{n+1}\setminus B_N$ for $N>1$. Moreover for $N>1$
$$
\mu(C_N)=\mu(B_{N+1})-\mu(B_N)
$$
hence from $\sigma$-additivity of measure we have
$$
\begin{align}
\mu\left(\bigcup\limits_{N=1}^\infty B_N\right)
=\mu\left(\coprod\limits_{N=1}^\infty C_N\right)
&=\mu(C_1)+\sum\limits_{N=2}^\infty\mu(C_N)\\
&=\mu(B_1)+\lim\limits_{M\to\infty}\sum\limits_{N=2}^M\mu(C_N)=\\
&=\mu(B_1)+\lim\limits_{M\to\infty}\sum\limits_{N=2}^M(\mu(B_{N+1})-\mu(B_N))\\
&=\mu(B_1)+\lim\limits_{M\to\infty}(\mu(B_{M+1})-\mu(B_1))\\
&=\lim\limits_{M\to\infty}\mu(B_{M+1})\\
&=\lim\limits_{M\to\infty}\mu(B_M)
\end{align}
$$
As for the inequality $\mu(B_N)\leq\inf\limits_{n\geq N}\mu(A_n)$ you need to note that 
$$
B_N=\bigcap\limits_{k=N}^\infty A_k\subset A_n
$$ for all $n\geq N$, hence for the same $n$ we have $\mu(B_N)\leq \mu(A_n)$. Therefore
$$
\mu(B_N)\leq\inf\limits_{n\geq N}\mu(A_n)
$$
A: In general: If $(A_n)_{n\geq 1}$ is an increasing sequence of sets, i.e. $A_1\subseteq A_2\subseteq \cdots$, then
$$
\mu\left(\bigcup_{n\geq 1}A_n\right)=\lim_{n\to\infty}\mu(A_n).
$$
This should be present in your lecture notes.
Since $B_N\subseteq A_n$ for every $n\geq N$, then $\mu(B_N)\leq \mu(A_n)$ for all $n\geq N$ and so
$$
\mu(B_N)\leq \inf_{n\geq N}\mu(A_n).
$$
Now take $\lim_{N\to\infty}$ on both sides. The last equality is the definition of $\liminf$.
A: For the first part, set $B_N'= B_N \setminus \bigcup_{i=1}^{N-1} B_i$, 
Then $\bigcup B_N' = \bigcup B_N$ and so $\mu(\bigcup B_N) = \mu(\bigcup B_N')$, which by additivity of measure gives us $$\mu(\bigcup B_N) = \sum \mu(B_N') = \lim_{N\rightarrow\infty} \sum_1^N \mu(B_i') = \lim_{N\rightarrow\infty}\mu(B_N)$$ where the last equality follows from how we constructed the $B_N'$'s. 
Then the inequality follows by monotonicity of measure because $B_N \subseteq A_n$ for all $n\geq N$.
