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?

  • $\begingroup$ 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

Note that the sequence $\underset{X}\int g_n$ is increasing because $g_n$ is increasing, and consequently the limit exists.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.