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?