Limit inferior/superior of sequence of sets

Let $(\Omega, \mathcal{A}, \mu)$ be a measure space, where $\mu(\Omega)< \infty$. Further $(A_n)_{n \in \mathbb{N}}$ is a a sequence of $\mathcal{A}$-measurable sets. I want to prove, that

$$\mu ( \liminf_{n \rightarrow \infty} A_n) \leq \liminf_{n \rightarrow \infty} \mu (A_n) \leq \limsup_{n \rightarrow \infty} \mu (A_n) \leq \mu (\limsup_{n \rightarrow \infty} A_n)$$ holds for any sequence $(A_n)_{n \in \mathbb{N}}$. I have no experience working with the limit superior/inferior. Clearly $$\mu ( \liminf_{n \rightarrow \infty} A_n) \leq \mu (\limsup_{n \rightarrow \infty} A_n)$$ holds, since it is easy to prove that the one is a superset of the other. Also $$\liminf_{n \rightarrow \infty} \mu (A_n) \leq \limsup_{n \rightarrow \infty} \mu (A_n)$$ holds, for any sqeuence. But I am stuck how to show the connection. I could use the Definitions, then I get $$\mu (\bigcup_n^\infty \bigcap_{k=n}^\infty A_n) \leq \lim_{n \rightarrow \infty} \inf_{k \geq n} \mu(A_n) \leq \inf_{n \geq 0} \sup_{k \geq n} \mu(A_n) \leq \mu (\bigcap_n^\infty \bigcup_{k=n}^\infty A_n)$$ But I don't know if this helps. Anyone got a hint how to go on?

• Since each measurable subset has a finite measure, what can you say about $\mu(\bigcap_{k\geq n}A_k)$? Anyway, +1 for showing your work. Jul 16, 2012 at 12:05

Consider sets $$B_n=\bigcap\limits_{k=n}^\infty A_k$$ Obviously, $B_n\subset A_k$ for all $k\geq n$, so $\mu(B_n)\leq \mu(A_k)$ for all $k\geq n$ After taking infimum we get $\mu(B_n)\leq\inf_{k\geq n}\mu(A_k)$. Since $B_n\subset B_{n+1}$ for all $n\in\mathbb{N}$ then the sequence $\{\mu(B_n):n\in\mathbb{N}\}$ is non-decreasing, so there exist $\lim_{n\to\infty}\mu(B_n)$. Similarly the sequence $\{\inf_{k\geq n}\mu(A_k):n\in\mathbb{N}\}$ is non-decreasing hence there exist $\lim_{n\to\infty}\inf_{k\geq n}\mu(A_k)$. Since existence of limits is justified we can say $$\lim\limits_{n\to\infty}\mu(B_n)\leq \lim\limits_{n\to\infty}\inf\limits_{k\geq n}\mu(A_k)=\liminf\limits_{n\to\infty}\mu(A_n)\tag{1}$$ Again recall that $B_n\subset B_{n+1}$ for all $n\in\mathbb{N}$, so $$\mu\left(\bigcup\limits_{n=1}^\infty B_n\right)=\lim\limits_{n\to\infty}\mu(B_n)\tag{2}$$ It is remains to note that $$\liminf\limits_{n\to\infty}A_n= \bigcup\limits_{n=1}^\infty \bigcap\limits_{k=n}^\infty A_k= \bigcup\limits_{n=1}^\infty B_n\tag{3}$$ From $(1)$, $(2)$ and $(3)$ we have $$\mu\left(\liminf\limits_{n\to\infty}A_n\right)\leq \liminf\limits_{n\to\infty}\mu(A_n)$$ Now try to prove in the similar way the second inequality.