If $f(x)$ is integrable as well as differentiable on $\mathbb{R}$, can we conclude that $f$ is bounded on $\mathbb{R}$ 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! 
 A: 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.
A: No, $f(x) = x$ is integrable, differentiable and unbounded.
