Let $f_n : (0,1) \rightarrow \mathbb R$ be a sequence of $L^1([0,1])$ functions continuous almost everywhere and non-decreasing. Let $f (0,1) \rightarrow \mathbb R$ be a function of $L^1([0,1])$ continuous almost everywhere and non-decreasing. Define $$c_{n,m} =\int_0^1 f_n(x) P_m(x) dx $$ $$c_{m} =\int_0^1 f(x) P_m(x) dx $$ where $(P_m)_{m\geq 0}$ are the shifted Legendre polynomials (orthogonal basis of $L^2([0,1])$).
Is it true that $f_n \rightarrow f$ almost everywhere iff $c_{n,m} \rightarrow c_m$ for all $m\geq 0$ ? Left to right is only dominated convergence as the shifted Legendre polynomials are bounded. But what about the other way ? If it is not true, does it become true with $C^0([0,1])$ or with $C^1([0,1])$ instead of continuous almost everywhere in the hypotheses ?
EDIT: I feel this has something to do with the fundamental lemma of variational calculus as we have $|\int_0^1 (f(x)-f_n(x)) P_m(x) dx| < \epsilon$ and we can estimate any continuous function on $[0,1]$ with shifted Legendre polynomials. I am not aware of any measure-theoric formulation of this lemma and I don't see how we would get $\int_0^1 |f(x)-f_n(x)| P_m(x) dx < \epsilon$ instead.