# 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 $\mathbb X$.

1. If $(E_n)$ is an increasing sequence in $\mathbb X$, then $$\mu\left(\bigcup_{n=1}^\infty E_n\right)=\lim\mu(E_n).$$
2. 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.

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## 2 Answers

Note that $\mu\left(\bigcap_{n=i}^\infty E_n\right)\leq\inf_{n\geq i}\mu(E_n)$ and by this definition you get $$\lim_i\mu\left(\bigcap_{n=i}^\infty E_n\right)\leq\lim_i\inf_{n\geq i}\mu(E_n) = \liminf_n\mu(E_n)$$

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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 $\liminf x_n\leq \liminf y_n$.

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