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Show that if $f\in L^{3/2}([0,1])$ satisfies $ \int_0^1x^nf(x)dx = 0,\:\: n\in\mathbb{N} \cup \{0\}$ then $f = 0$ a.e.

I have not taken a course in integration theory, but my guess is that this apply that $ \int_0^1p(x)f(x)dx \approx \int_0^1|f(x)|^2dx =0 $ and hence we get $f(x) = 0$ a.e.

Im a bit unsure about the $"\approx"$ part? How could I write that more rigorously and does it matter that we are in $L^{3/2}([0,1])$ or could we as well be in any $L^p([0,1])$ space?

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Let $p\in (1,+\infty)$.

  • First, we show that if $A$ is a measurable subset of $[0,1]$ and $\delta>0$, then we can find a continuous function $f$ such that $\lVert f-\chi_A\rVert_p< \delta$. To see that, using inner regularity, we can find a closed subset $F$ of $A$ such that $\mu(A\setminus F)<\delta^p$. The characteristic function of this closed set can be approached pointwise by a monotonic sequence of continuous functions $\{f_n\}$ such that $0\leqslant f_n\leqslant 1$. By monotone convergence, choose $n$ such that $\lVert \chi_F-f_n\rVert_p<\delta$, giving $\lVert \chi_A-f_n\rVert_p<2\delta$.
    • Using the definition of Lebesgue integral, we can show that each $f\in L^p[0,1]$ can be approached in $L^p$ by a continuous function.
    • By Stone-Weierstrass theorem, we can see that we can replace continuous function by polynomials in the last sentence.