# Counter-example for Fatou's lemma using a probability measure

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.

-

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 $$\lim \int f_n = \frac12 >0 =\int \lim f_n$$
@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