$f_n \to 0$ $ a.e.$ and $\lim \int f_n d\mu =0$ but $\sup_n f_n$ is not in $L^1$

Question

Give an example of a finite measure space $(X,M,\mu)$ and a sequence of functions $f_n:X \to[0, \infty)$ such that $f_n \to 0$ $a.e.$ and $\lim \int f_n d\mu=0$ but $\sup_n f_n$ is not in $L^1$

I have trouble coming up with such an example and prove it. Can someone help me with this problem?

Answer 1

Hint: Try the measure space $X = \mathbb{N}$ with a measure $\mu$ that assigns measure $a_n$ to the integer $n$, where $\sum a_n < \infty$. Let $f_n$ be a function with $f_n(n) = b_n$ and $f_n(k) = 0$ for $k \ne n$. Observe that:

• $f_n \to 0$ everywhere

• $\int f_n \,d\mu = a_n b_n$

• $\int \sup_n f_n = \sum_k a_k b_k$.

Now choose $a_n, b_n$ appropriately.

Answer 2

$X=(0,1)$, with Lebesgue measure, $f_1:=1/x$, $f_n\equiv 0$ for $n\geq 2$.

Answer 3

Take $X=[0,1]$, $f_n(x) = {1 \over x} \cdot 1_{({1 \over n+1}, {1 \over n}]}(x)$.

Then $f_n(x) \to 0$, $\int f_n = \ln(1+ {1 \over n}) \to 0$, for $x \neq 0$, $\sup_n f_n(x) = {1 \over x}$, hence $\sup_n f_n$ is not integrable.