Finite at every point but unbounded on every interval Is is possible that a function $f$ is finite at every point but unbounded on every interval?
What if f is measurable?
 A: Yes; Conway base 13 function is an example for such function. It even satisfies the intermediate value property!
Granted the function itself is defined only for the interval $(0,1)$ but by agreeing to send integers to themselves, and taking translations of the function we can easily extend it to the entire real line.
One can also precompose this function with any homeomorphism of $\Bbb R$ with $(0,1)$ actually.
A: Asaf's example is very nice due to its additional property. An easier example (without worrying about the intermediate value property) is to let $f(x)$ be whatever you want if $x$ is irrational, it doesn't matter, while if $x=p/q$ is rational, with $\gcd(p,q)=1$ and $q>0$, then we set $f(x)=q$.
A: If you insist on continuity, then I don't think so.  But there are functions whose graphs are dense in the plane, so that does it.  For each rational number $m$ pick an irrational number $a_n$ and let
$$
f(x) = \begin{cases} mx & \text{if $x$ is a rational multiple of $a_m$,} \\
0 & \text{if for every rational $m$, $x$ is not a rational multiple of $a_m$.}  \end{cases}
$$
I think that does it.
