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If $f_n \rightarrow f$ pointwise, then does $$\liminf \int f_n=\lim\int f_n?$$ I know that $\liminf f_n=\lim f_n$ since the sequence converges, but I'm not sure if the $(L)$ integral throws us off.

I'm trying to prove the Fatou's Reverse Lemma, and I got stuck.

EDITED: $f$ is integrable and $f_n\le f$

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3 Answers

up vote 2 down vote accepted

If we had some guarantee that the limit on the right hand side exists, then of course this equality would be true, because for a convergent sequence $a_n = \int f_n d \mu$ we would have $\liminf a_n = \lim a_n$.

But in your case there doesn't seem to be such a guarantee. For example, consider a sequence of functions $f_n: \mathbb{R} \to \mathbb{R}$ like this: for $n$ even, $f_n=0$. And for $n$ odd, $f_n = 1_{[n, n+1]}$. Then $\int f_n d\mu$ is $0$ for n even and $1$ for n odd, so $\liminf$ on the LHS is equal to $0$, and the $\lim$ on the RHS doesn't exist.

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What if $f_n\le f$ and $f$ is integrable? –  cap Nov 11 '12 at 6:32
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I'm going to assume your measure space is $\mathbb{R}$ with Lebesgue measure.

Consider $f_n=n1_{[0,\frac{1}{n}]}$ where $1_A$ is the characteristic function of the set $A\subset \mathbb{R}$. Then $f_n \to f=0$ pointwise but

$$ \liminf\int_{\mathbb{R}} f_n(x)dx=1 > 0= \int_{\mathbb{R}} f(x)dx $$

(You can even take $f_n, f\in C^{\infty}(\mathbb{R})$ so it's not a regularity issue).

In other measure spaces it might still be false: For example in $\mathbb{N}$ with counting measure take $f_n(m)= 1$ if $m=1,n$ and zero otherwise then $f_n \to f$ pointwise, where $f(1)=1$ and is zero otherwise, and

$$ \liminf \int_{\mathbb{N}} f_n(m)dm = 2 > 1 = \int_{\mathbb{N}} f(m)dm $$

So no, in general only one inequality is true in Fatou's lemma.

With the edit it's still not true: Take $g_n=-f_n$ as above. You could put $|f_n|\leq f$ but then this is just the dominated convergence theorem.

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If $f_{n}=n1_{[0,\frac{1}{n}]}$ then $f_{n}\to 0$ almost everywhere, not pointwise. Since $f_{n}(0)=n$ for all $n\in\mathbb{N}$. –  Thomas E. Nov 26 '12 at 9:15
@ThomasE: Use $f_n=n1_{\left(0,\frac1n\right]}$, instead. –  robjohn Nov 26 '12 at 9:23
@robjohn. Yeah, it's just a matter of small modification. –  Thomas E. Nov 26 '12 at 9:25
The question does not ask about $\int_{\mathbb{R}}f(x)\,\mathrm{d}x$; it only asks about the $\liminf$ and $\lim$ of $\int_{\mathbb{R}}f_n(x)\,\mathrm{d}x$. –  robjohn Nov 26 '12 at 9:32
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A slight modification of the usual counterexample works here: $$ f_n=\left\{\begin{array}{}n1_{(0,1/n]}&\mbox{if $n$ even}\\0&\mbox{if $n$ odd}\end{array}\right. $$ Here, $f_n\to0$ pointwise, and $$ \liminf_{n\to\infty}\int_{\mathbb{R}}f_n(x)\,\mathrm{d}x=0 $$ yet $$ \limsup_{n\to\infty}\int_{\mathbb{R}}f_n(x)\,\mathrm{d}x=1 $$ so the limit does not exist.

Of course if $\lim\limits_{n\to\infty}\int_{\mathbb{R}} f_n(x)\,\mathrm{d}x$ exists, then $\liminf\limits_{n\to\infty}\int_{\mathbb{R}} f_n(x)\,\mathrm{d}x=\lim\limits_{n\to\infty}\int_{\mathbb{R}} f_n(x)\,\mathrm{d}x$ by definition.

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