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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.

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

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