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Fatou's lemma says that for any sequence $(f_n)$ of positive measurable functions.

$$\int_X \lim \inf f_n du \leq \lim \inf \int_Xf_n du$$

Here is a proof:

Set $g_k = \inf_{n\geq k} f_n$ then $(g_k)$ is an increasing sequence of positive measurable functions such that $g_k \rightarrow \lim \inf_n f_n$ as $k \rightarrow \infty$; and $g_k \leq f_k$ for any $k$ therefore by monotone convergence theorem

$$\int_X \lim \inf_n f_n du = \lim_{n \rightarrow \infty} \int_X g_n du = \lim \inf_n \int_X g_n du \leq \lim \inf_n \int_X f_n du.$$

My questoin is: why can we say

$\lim_{n \rightarrow \infty} \int_X g_n du = \lim \inf_n \int_X g_n du$

Should the equality not be a $\leq$sign? Since the limit may not exist?

How is this proof working?

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  • $\begingroup$ As the answer said, the limit exists (in $[-\infty,\infty]$). However, if it didn't, that wouldn't mean there's a $\le$ sign. If the limit exists (in $[-\infty,\infty]$) then it's equal to the liminf. If it doesn't then there's no comparison to be made. $\endgroup$ – spaceisdarkgreen Jun 25 '17 at 14:55
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Note that the sequence $\underset{X}\int g_n$ is increasing because $g_n$ is increasing, and consequently the limit exists.

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