# What's wrong with the proof using Fatou's Lemma?

Fatou's Lemma states that given any measure space $$(\Omega,\Sigma,\mu)$$ and $$X\in\Sigma,$$ let $$(f_n)_{n=1}^\infty$$ be a sequence of real-valued nonnegative measurable functions $$f_n:X\to [0,+\infty]$$ converges to $$f$$ point wise. Then $$\int_X f d\mu\leq \liminf \int_X f_n\,d\mu.$$

If we assume that $$g_n$$ is integrable for each $$n,$$ then its pointwise limit $$g$$ may not be integrable.

However, what goes wrong with the 'proof' below?

Since each $$g_n$$ is integrable, so $$\int_X |g_n| \,d\mu<\infty.$$ Since $$(|g_n|)$$ is a sequence of nonnegative measurable functions, by Fatou's Lemma, we have $$\int_X |g|\,d\mu\leq \liminf \int_X |g_n|\,d\mu<\infty.$$ Therefore, $$g$$ is integrable.

• Why is it true that $\liminf \int |g_n|$ is finite? – D. Brogan Jan 9 '19 at 4:35
• $x_n = n < \infty$ and $\liminf x_n = \infty$. – RRL Jan 9 '19 at 4:38

Take $$X=\mathbb{R}$$ equipped with the Lebesgue Measure. Set $$g_n(x)=\chi_{[-n,n]}(x)$$ i.e. $$g_n(x)= \begin{cases} 1&x\in [-n,n]\\ 0&x\not\in[-n,n]. \end{cases}$$ $$g_n\in L^1(\mathbb{R})$$ for all $$n$$. Moreover, $$\int_\mathbb{R} g_n(x)dx=2n.$$ The pointwise limit $$g(x)$$ exists, and is the constant $$1$$ function. However, if we check our bounds, we see that $$\liminf_{n\to\infty} \int_{\mathbb{R}} \lvert g_n(x)\rvert dx=\liminf_{n\to\infty} \:2n=\infty$$ and as such provides no "bound" on $$\int g(x)dx$$. So, your argument was flawed because you assumed that $$\liminf \int \lvert g_n\rvert<\infty$$.