Basic measure theory inequality

Question

Let $(X,\Sigma,\mu)$ be a measure space and $\langle E_n\rangle_{n\in\mathbb{N}}$ a sequence in $\Sigma$. I am trying to show that
$$\mu\left(\bigcup_{n\in\mathbb{N}}\left(\bigcap_{m\geq n} E_m\right)\right)\leq \lim\text{inf}_{n\rightarrow\infty}\mu E_n$$
I know that since $$\bigcap_{m\geq n} E_m$$ is a non-decreasing sequence of sets as $n$ increases, I can conclude that
$$\mu\left(\bigcup_{n\in\mathbb{N}}\left(\bigcap_{m\geq n} E_m\right)\right)=\lim_{n\rightarrow\infty}\mu \bigcap_{m\geq n}E_m$$
However, to replace the right hand side with inf, it seems like I would need to fact that the $E_m$ are a non-increasing sequence of sets. Am I missing something here?

Answer 1

$\mu (\cap_{m \geq n} E_m) \leq \mu (E_n)$. If $a_n \leq b_n$ for all $n$ then $\lim \inf a_n \leq \lim \inf b_n$. Hence $\lim \inf \mu (\cap_{m \geq n} E_m) \leq \lim \inf \mu(E_n)$. Now just note that $\lim \inf$ on the left side is same as the limit.