The following is a question in my midterm exam.
Let $f_n: \mathbb{R} \rightarrow \mathbb{R}$ be a seq. of Lebesgue integrable functions such that
$$
\int_{-\infty}^{\infty} f_n(x) dx =1
$$
and $\sup_n f_n$ is unbounded in every open interval. Find such a sequence of $f_n$'s.
I've tried to think along the lines that for every irrational number my functions are $0$. But I'm unable to proceed along this direction since I don't know how to manipulate the functions at rational numbers to meet the above criteria.
Best Answer
Let $f_0$ be function unbounded in a neighborhood of $0$ with finite integral, e.g. $$f_0(x) = \frac{1}{\sqrt{|x|}} \chi_{[-1,1]}.$$ You may need to rescale to make sure it has integral $1$. If $\{r_n\}$ is an enumeration of the rationals (or any dense countable set) you can define $$f_n(x) = f_0(x-r_n)$$ so that $f_n$ has integral $1$ but is unbounded on every neighborhood of $r_n$. What can you say about $\sup_n f_n$?