I found this problem while going through some old question papers of a test I'm preparing for:
Let $f: [0,1] \to \Bbb R$ be a bounded measurable function such that $\int_0^1 f(t) e^{nt} dt = 0$ for every $n=0,1,2,\ldots$ Prove that $f(t)=0$ for almost every $t \in [0,1]$.
I've seen similar questions before where $e^{nt}$ was replaced by $t^n$ or $t^{2n}$. They could be solved by using the Stone-Weierstrass approximations of $\cos(2 \pi nt),\sin(2 \pi nt)$ and then looking at the Fourier series of $f$. Again, if we had $e^{int}$ then we could've used use Parseval's identity as $f \in L^2$.
But I've no idea how to continue with this particular problem. Any help is appreciated.