An inequality about sequences in a $\sigma$-algebra Let $(X,\mathbb X,\mu)$ be a measure space and let $(E_n)$ be a sequence in $\mathbb X$. Show that $$\mu(\lim\inf E_n)\leq\lim\inf\mu(E_n).$$
I am quite sure I need to use the following lemma.
Lemma. Let $\mu$ be a measure defined on a $\sigma$-algebra $\mathbb X$.


*

*If $(E_n)$ is an increasing sequence in $\mathbb X$, then 
$$\mu\left(\bigcup_{n=1}^\infty E_n\right)=\lim\mu(E_n).$$

*If $(F_n)$ is a decreasing sequence in $\mathbb X$ and if $\mu(F_1)<+\infty$, then 
$$\mu\left(\bigcap_{n=1}^\infty F_n\right)=\lim\mu(F_n).$$


I know that
$$\mu(\liminf_n E_n)=\mu\left(\bigcup_{i=1}^\infty\bigcap_{n=i}^\infty E_n\right)=
\lim_i\mu\left(\bigcap_{n=i}^\infty E_n\right).$$
The first equality follows from the definition of $\lim\inf$ and the second from point 1 of the lemma above. Here is where I am stuck.

For every $i\geq 1$ we have that 
$$
\bigcap_{n=i}^\infty E_n\subseteq E_i
$$
and so for all $i\geq 1$
$$
\mu\left(\bigcap_{n=i}^\infty E_n\right)\leq \mu(E_i).
$$
Then
$$
\mu(\liminf_n E_n)=\lim_{i\to\infty}\mu\left(\bigcap_{n=i}^\infty E_n\right)\leq \liminf \mu(E_i),
$$
using the fact that if $(x_n)_{n\geq 1}$ and $(y_n)_{n\geq 1}$ are sequences with $x_n\leq y_n$ for all $n$, then $\liminf x_n\leq \liminf y_n$.
