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Fatous's lemma states that:

Let $f_1,f_2, \ldots, f $ be Borel Measurable and $f_n \leq f$ for all n, where $\int_\Omega f \;d\mu < \infty$. Then

$$ \limsup_{n\rightarrow\infty} \int_\Omega f_n \; d\mu \leq \int_\Omega \left(\limsup_{n\rightarrow\infty} f_n\right) \; d\mu $$

Now, the hypothesis requires that there exist an $f$ which is integrable.

What I want to know is a counter-example (if it exists, or a proof if it doesn't) for the case when $\int_\Omega f \;d\mu = \infty$ for a probability measure $\mu$ i.e. $f_n, f$ such that $$ \limsup_{n\rightarrow\infty} \int_\Omega f_n \; d\mu > \int_\Omega \left(\limsup_{n\rightarrow\infty} f_n\right) \; d\mu $$


Here's is a counter-example for the case when $\mu$ is not a finite measure.

Let $0<x<1$, $\Omega = \mathbb{N}$, and $\mu(\{k\}) = \frac{1}{k}$, $\forall k \in \mathbb{N}$.

$f_n(k) = x^n k$. Thus,

$$ \limsup_n \sum_k p_k f_n(k) = \limsup_n x^n \sum_k 1 = \infty $$

But,

$$ \sum_k p_k \left(\limsup_n f_n(k)\right) = 0 $$

Reference: Problem 5 in section 1.6 in Robert Ash.

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up vote 3 down vote accepted

The standard example on $[0,1]$ with the Lebesgue measure is $f_n=n\chi_{[1/(2n),1/n]}$. Here $f_n(x)\le f(x)=1/x$ (which is not integrable), and $$\lim \int f_n = \frac12 >0 =\int \lim f_n$$

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Is there any example of a probability measure? –  UnadulteratedImagination Jan 27 '13 at 0:43
    
@UnadulteratedImagination The Lebesgue measure on $[0,1]$ is a probability measure, as far as I'm concerned. –  user53153 Jan 27 '13 at 0:44
    
ohh yes. I forgot $[0,1]$. Thank you. –  UnadulteratedImagination Jan 27 '13 at 0:47
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