# Integrable functions with non-integrable limit

$$(*)$$ Let $$M > 0$$ be a constant. Assume that $${f_n}$$ is a sequence of lebesgue integrable functions such that $$0 < f_n < M$$ a.e., and $$\lim_{n \rightarrow \infty } f_n = f$$ almost everywhere for some function $$f : X \rightarrow \mathbb{R}$$.

Find an example of $${f_n}$$ as stated in $$(*)$$ such that $$f_n \uparrow f$$ but $$f$$ is not lebesgue integrable. Show in your example that $$f$$ is not lebesgue integrable.

Now I was thinking about a function that has a lot of gaps in it. Therefore, I was thinking about a function similar to the Dirichlet Function. Am I on the right track here?

• Dirichlet function (i.e. $1_{\Bbb Q}$) is Lebesgue integrable, though. Specifically, its integral is $0$.
– user239203
Apr 16, 2020 at 14:28

You seem to be thinking that you need $$f$$ to be non-measurable. But that's not the case, as limits of measurable are measurable. "Not integrable" here means that $$\int f=\infty$$.
An easy example would be to take $$f(x)=\tfrac1x\,1_{[1,\infty)}$$ and then make $$f_n=1_{[1,n]}$$. The problem with this is that you need $$f_n>0$$. So we may modify the example like this: $$f=1_{(-1,1)}+\tfrac1x\,1_{|x|≥1},\ \ \ \ f_n=f\,1_{[-n,n]}+\tfrac1{1+x^2}\,1_{|x|>n}.$$
A standard result in measure theory gives that of measurable functions are measurable. You should be more worried about the integrable part, which means that it has a finite integral in absolute value. Consider the sequence of functions $$1_{[0,n]}$$, i.e. the indicator function on $$[0,n]$$.