Integrable functions with non-integrable limit

Question

$(*)$ 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?

Answer 1

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: $$ f=1_{(-1,1)}+\tfrac1x\,1_{|x|≥1},\ \ \ \ f_n=f\,1_{[-n,n]}+\tfrac1{1+x^2}\,1_{|x|>n}. $$

Answer 2

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 $[0,n]$.

You should also be able to convince yourself that this can't happen on a finite measure space; you are fundamentally using that the measure of the whole space is infinite.