real analysis - Show that: $muleft(bigcup_N bigcap_{n=N}^{infty} A_n right) leq lim inf mu(A_n)$

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.

Answer

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 $$\mu(B_N)\leq\inf\limits_{n\geq N}\mu(A_n)$$