# Pointwise a.e. convergence implies $\int_E f d\mu \leq \lim (\int f_n d\mu)$.

I'm trying to prove / disprove the following:

If $f_n \geq 0$ is a sequence of integrable functions and $f_n \to f$ a.e., then $\lim \int_E f_n d\mu$ exists and $\int_E fd\mu \leq \lim \int_E f_n d\mu$. All limits here are as $n \to \infty$.

What I have so far: As $f_n \to f$, we have that $\liminf f_n = f$. If $(\int_Ef_nd\mu)$ converges, it is equal to $\liminf (\int_E f_n d\mu)$, and so by Fatou's Lemma, $$\int_E fd\mu = \int_E\left(\liminf f_n\right)d\mu \leq \liminf \left(\int_E f_nd\mu\right) = \lim\left(\int_Efd\mu\right).$$

However, I'm stuck on showing that $\lim (\int_E f_n d\mu)$ exists, if at all.

• $\lim(\int_E f_n d\mu)$ need not exist. Consider e.g. $E = \ \mathbb R$, $\mu$ = Lebesgue measure, and $f_n = n^2 g_n$ where $g_n$ is the characteristic function of the interval $(0,1/n)$. Then $f_n \to 0$ pointwise but $\int f_n = n$, so $\int f_n \to \infty$. With a simple modification you can also obtain an example where $\int f_n$ oscillates between two (or more) finite values, so there's no limit even in the extended reals. – Bungo Sep 17 '17 at 20:14

## 1 Answer

Upgrading my comment to an answer:

$\lim(\int_E f_n d\mu)$ need not exist. For a counterexample, let $E = \mathbb R$ and $\mu$ = Lebesgue measure. Define $f_n = n^2 g_n$, where $g_n$ is the characteristic function of the interval $(0,1/n)$. Then $f_n \to 0$ pointwise, but $\int f_n = n$, so $\int f_n \to \infty$.

We can easily modify the above example so that $f_n$ oscillates between two values, e.g. $$f_n = \begin{cases} n g_n & \text{n even} \\ 0 & \text{n odd} \\ \end{cases}$$ in which case $\int f_n$ is $1$ for even $n$ and $0$ for odd $n$, hence the sequence $\int f_n$ has no limit even in the extended reals. (Its $\liminf$ is zero, so Fatou's lemma holds with equality.)

• Perfect - thank you. If the limit of the integral of $f_n$ does indeed exist, was my argument using Fatou's lemma correct? – user381606 Sep 17 '17 at 20:35
• Yes, if the limit exists then your argument is correct. – Bungo Sep 17 '17 at 21:03