Determine all integrable functions $f$ defined on $[0,1]$ so that $\int_0^1 f(t) t^n\, dt = 0$ for all $n =0,1,2,\dots$.

Question

Question: Determine all integrable functions $f$ defined on $[0,1]$ so that $\int_0^1 f(t) t^n\, dt = 0$ for all $n =0,1,2,\dots$.

My attempt: Using Stone-Weierstrass theorem, if $g \colon [0,1] \to \mathbb{R}$ satisfy $$ \int_0^1 g(x) x^n \, dx =0$$ for all $ n= 0,1,2,\dots$ then $g\equiv 0$ in $[0,1]$. Let $f$ be such an integrable function and $\epsilon > 0$ be given then there exists continuous $g\colon [0,1] \to \mathbb{R}$ so that $$\int_0^1 |f(x) - g(x)| \, dx < \epsilon.$$ If $f$ is nonnegative for a.e. $x \in [0,1]$ then by triangle inequality we have $$\int_0^1 |g(x)| x^n \, dx \leq \epsilon$$ for all $n = 0,1,2,\dots$ and $\epsilon >0$ can be made arbitrary small then $$\int_0^1 |g(x)| x^n \, dx = 0$$ for all $n =0,1,2,\dots$ so $g\equiv 0$ in $[0,1]$ and $f \equiv 0$ in $[0,1]$. Is there any way to tackle the problem for $f$ be any arbitrary Lebesgue integrable function? Any hints/solutions are appreciated.

Answer 1

Note that $\int f p = 0$ for all polynomials $p$.

Pick some $x \in (0,1)$ and let $s_n$ be the function whose graph consists of the points $(0,1), (x, 1), (x+{1 \over n}, 0), (1,0)$. This is basically a continuous approximation to the step function $s=1_{[0,1]}$.

Let $p_k$ be a sequence of polynomials that converge uniformly to $s_n$, we see that $\int f p_k = 0 \to \int f s_n$ and so $\int f s_n = 0$. The dominated convergence theorem shows that $\int f s_n = 0 \to \int f s$ and so $\int_0^x f = 0$. The Lebesgue differentiation theorem gives $f(x) = 0 $ for ae. $x$.