If $f(x)$ is integrable as well as differentiable on $\mathbb{R}$, can we conclude that $f$ is bounded on $\mathbb{R}$, or at least, there exists some $M>0$ so that $m(\{x:|f(x)| > M\}) = 0$?
I feels that this should be true, because at least $f$ is bounded on any compact set.. But, is there a way to prove it or is there any counter example?
Many thanks to any help!
Continuous unbounded but integrable functions is similar. As Daniel Fischer says, if you want smoother functions, mollify.
In essence, for a continuous function, integrability is a global property and depends on the behavior of the function at infinity. Differentiability is a local property, so there isn't really any reason to expect that they should be related.
For an alternate construction, let $\psi$ be your favorite bump function. Let's say $\psi$ is $C^\infty$, nonnegative, compactly supported in $(0,1)$, has $\int_0^1 \psi(x)\,dx = 1$, and $\sup_{[0,1]} \psi = 3$. Then for real numbers $a>0$, $c\in \mathbb{R}$ and $b \ge 1$, the function $a \psi(b(x-c))$ is supported in $(c,c+1)$, takes the value $3a$ at some point in that interval, and has $\int_{\mathbb{R}} a \psi(b(x-c))\,dx = a/b$.
Now let $$f(x) = \sum_{n=1}^\infty 2^n \psi(2^{2n}(x-n)).$$ $f$ is $C^\infty$ since on each interval $(n, n+1)$ only one of the summands is nonzero. And by Tonelli's theorem, since all the summands are nonnegative we can interchange sum and integral to get $$\int_{\mathbb{R}} f(x)\,dx = \sum_{n=1}^\infty \int_{\mathbb{R}} 2^n \psi(2^{2n}(x-n)) \,dx = \sum_{n=1}^\infty 2^n/2^{2n} = \sum_{n=1}^\infty 2^{-n} = 1.$$ So $f$ is integrable. And clearly $f$ is unbounded since for each $n$ there is an $x \in (n,n+1)$ with $f(x) = 3 \cdot 2^n$.
Intuitively, $f$ consists of a series of smooth bumps marching off to infinity. Each bump is twice as high and one-fourth as wide as the previous, hence has half the area.
