# Proving an extension of Fatou's Lemma

Fatou's Lemma states:

Let $$(f_n) : \Omega \to [0, \infty]$$ be a sequence of measurable functions. Then $$\int \lim \inf f_n \le \lim \inf \int f_n.$$

Later, my professor wrote

Let $$(f_n)$$ be a sequence of measurable functions which is bounded below by an integrable function $$g$$. Then $$\int \lim \inf f_n \le \lim \inf \int f_n.$$

I am having two issues with this extension.

$$1)$$ Don't we need to assume that the $$f_n$$'s are eventually integrable? My reason for thinking this is that the straightforward method of proof that I do below seems to require it.

$$2)$$ What codomain can the $$f_n$$'s have? My proof below uses the linearity of the integral, but I only know that the integral is linear for functions with codomain $$\mathbb{R}$$. Is it possible for both $$f_n$$ and $$g$$ to have codomain $$[-\infty, \infty]$$?

Thank you very much for your help.

My Work:

Assuming that the $$f_n$$'s are eventually integrable, and we have codomain $$\mathbb{R}$$, we have $$f_n \ge g \implies f_n-g \ge 0$$. Applying Fatou's Lemma to $$f_n - g$$, we get

$$\int \lim \inf (f_n-g) \le \lim \inf \int (f_n-g).$$

We know $$g$$ is integrable, we can write

$$\int \lim \inf (f_n) - \int g \le \lim \inf \int (f_n) - \int g$$

and deduce the desired result.

• Why can we write $\int \liminf (f_n -g)=\int \liminf f_n -\int g$?
– Koro
Commented Oct 8, 2022 at 18:42

## 1 Answer

No further assumptions are necessary. Integrability of $$g$$ implies that it takes values in $$\mathbb R$$ except on a set of measure $$0$$. $$f_n$$'s may take infinite values for this proof to work.

• But on the RHS, we used $\int (f_n-g) = \int f_n - \int g$. Don't we need to know that both functions are integrable in order to do this?
– Ovi
Commented May 7, 2020 at 12:25
• $\int (f_n-g)$ surely exists but it may be $+\infty$. Since $g$ is integrable this implies that $\int f_n$ also exists ( in the sense either $\int f_n^{+}$ or $\int f_n^{-}$ is finite); of course $\int f_n$ may be $\infty$ but the integral exists an an extended real number. @Ovi Commented May 7, 2020 at 12:29
• Thanks for the reply. So in general, $\int(f \pm g)$ may not equal $\int f \pm \int g$. But if $g$ is integrable (and no restrictions on $f$), then always $\int (f \pm g) = \int f \pm \int g$?
– Ovi
Commented May 7, 2020 at 14:55
• Yes, that is correct. @Ovi Commented May 7, 2020 at 23:12
• Thank you ${}{}$
– Ovi
Commented May 7, 2020 at 23:16