# Prove that $\mu(\bigcup_{i=1}^{\infty}\bigcap_{n=i}^{\infty}B_{n}) \le \liminf_{n\to \infty} \mu (B_{n})$ [duplicate]

Possible Duplicate:
Fatou's lemma and measurable sets

Let $(X,\Sigma, \mu)$ be a measurable space and $\{B_{i}\}$ be a sequence of sets in $\Sigma$. Then $$\mu\left(\bigcup_{i=1}^{\infty}\bigcap_{n=i}^{\infty}B_{n}\right) \le \liminf_{n\to \infty} \mu (B_{n}).$$

What I know is that $\liminf_nA_n=\bigcup\limits_{n=1}^{\infty}\bigcap\limits_{k=n}^{\infty}A_{k}$ for any sequence $\{A_n\}$. What will happen if I apply measure on both sides?

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## marked as duplicate by t.b., tomasz, sdcvvc, arjafi♦, copper.hatAug 29 '12 at 5:30

We have using property that if $\{A_n\}$ is an increasing sequence of measurable sets, $\mu(A_n)\to \mu\left(\bigcup\limits_{j=1}^{+\infty}A_j\right)$. Hence $$\mu\left(\bigcup_{i=1}^{+\infty}\bigcap_{k=i}^{+\infty}B_k\right)=\lim_{n\to +\infty}\mu\left(\bigcup_{i=1}^n\bigcap_{k=i}^{+\infty}B_k\right)=\lim_{n\to +\infty}\mu\left(\bigcap_{k\geq n}B_k\right).$$ Now, use the fact that $\bigcap\limits_{k\geq n}B_k\subset B_j$ for any $j\geq k$ to get the result.
Note that this inequality doesn't need to be an equality, for example with $\{0,1\}$ with counting measure, $B_{2n}=\{0\}$, $B_{2n+1}=\{1\}$. Then $\liminf_n B_n=\emptyset$ but $\liminf_n\mu(B_n)=1$.
What Davide uses is that $\Big(\bigcup_{i=1}^{n}\bigcap_{k=i}^{\infty}B_{k}\Big)_{n=1}^{\infty}$ is an increasing sequence of sets and applies then convergence of measure. Also, since $\mu(\bigcap_{k\geq n}B_{k})\leq \mu(B_{j})$ for all $j\geq k$, then $\mu(\bigcap_{k\geq n}B_{k})\leq \inf_{k\geq n}\mu(B_{k})$, which yields the result since $\bigcup_{i=1}^{n}\bigcap_{k=i}^{\infty}B_{k}=\bigcap_{k\geq n}B_{k}$ for all $n\in\mathbb{N}$. – T. Eskin May 24 '12 at 7:11