Let $f \in L^1([0,1])$ such that $$\int_0^1 f(x) \exp(2 \pi i n x) dx=0$$ for all $n \in \mathbb Z$. Does this imply $f=0$?
If we would additionally know that $f \in L^2([0,1])$ this would be obvious since the functions $\exp(2 \pi i n x)$ are a Hilbert basis of the Hilbert space $L^2$. But since we only know $f \in L^1([0,1])$ I'm not sure anymore.
The easiest way to do this is the way Lebesgue did it a long time ago; one first shows the result for continuous functions (eg using the $L^2$ arguments or Feijer theorem - both may be actually post-Lebesgue and there are direct proofs too).
Then consider $F(x)=\int_{0}^xf(t)dt$; by general theory $F$ is absolutely continuous (hence differentiable ae) and $F'=f$ ae.
But $F(1)=F(0)=0$ by the hypothesis with $n=0$ and then for $|n| \ge 1$, integrating by parts one has
$\int_0^1 F(x) \exp(2 \pi i n x) dx=(F(1)-F(0))/(2\pi in)-\frac{1}{2\pi i n}\int_0^1 f(x) \exp(2 \pi i n x) dx=0$, so if $A=\int_0^1F(x)dx$ one has by the above result for continuous functions that $F(x)-A=0$ and by differentiating one gets that $f(x)=0$ ae