$L^1$ function is bounded almost everywhere

Question

Suppose $f\in L^1(\mathbb{R})$. Is it true that $f$ is bounded (except for a set of measure zero)?

I think it should be true, but can't show it formally. If $f$ is unbounded, why would we have $\int_\mathbb{R}|f|dx=\infty$?

Answer 1

Every measurable function is "approximately continuous" almost everywhere. See https://www.encyclopediaofmath.org/index.php/Approximate_continuity. In view of the examples you are given in the comment, this is the best you can hope for.

Answer 2

Here is a counter example : take the function $f$ defined by $f=0$ on $(-\infty,1]$, $f(n)=n$ for integers $n\ge 2$ , $f\left(n-\frac{1}{n^3}\right)=0=f\left(n+\frac{1}{n^3}\right)$, $f$ is piecewise affine on $\left[ n-\frac{1}{n^3},n+\frac{1}{n^3} \right]$ and $f$ is zero elsewhere. $f$ is clearly unbounded, non-negative and you can easily compute its integral and see (thanks to the fact that $\sum_{n\ge 1} \frac{1}{n^2}<\infty$) that $f\in L^1(\mathbb{R})$.

Answer 3

What about $x^{\alpha}$ on $(0,1]$ for some $\alpha> -1$?