Does $\int_{0}^{1}x^nf(x)\, dx=0$ imply that $f=0$ a.e. without assuming $f \in C[0,1]$? Suppose that $f \in L^{1}[0,1]$ and $\int_{0}^{1}x^nf(x)\, dx=0$ for $n=0,1,2,\dots$
Does that imply that $f=0$ a.e.?
I think that there will be a counterexample but it is hard to find out.
 A: Your assumptions imply that $\int p(x) f(x) \, dx = 0$ for all polynomials $p$.
Now, by the (Stone)-Weierstraß Theorem, every continuous function $g \in C([0,1])$ is a uniform limit of polynomials. This yields (how exactly?) $\int g(x) f(x) \, dx = 0$ for all continuous functions $g \in C([0,1])$.
Now, every $g \in L^\infty ([0,1]) \subset L^1([0,1])$ can be approximated (in the $L^1$-norm) by a sequence of continuous functions $(g_n)_n$ (see e.g. Compact support functions dense in $L_1$). For a suitable subsequence (again denoted by $(g_n)_n$), this implies convergence almost everywhere (see e.g. $L^1$ convergence gives a pointwise convergent subsequence).
By considering $h_n := \min \{\Vert g \Vert_\infty, \max\{ - \Vert g \Vert_\infty, g_n\} \}$, you can assume that the sequence $(h_n)_n)$ is uniformily bounded and converges to $g$ almost everywhere.
By dominated convergence (what is the dominating function?), this yields
$$
\int f(x) g(x) \, dx = \lim_n \int f(x) h_n(x) \, dx = 0.
$$
But if we now take $g = \chi_{\{x \mid f(x) \geq 0 \}}$, this implies (how?) $f \leq 0$ almost everywhere. An analogous argument yields $f \geq 0$ almost everywhere and hence $f \equiv 0 $ almost everywhere.
