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Suppose that $\int_0^1 x^nf(x)=0$ for all nonnegative integers $n$, where $f$ is a Lebesgue measurable function that is bounded. How do you prove that $f(x)=0$ a.e. on $[0,1]$.

I've seen this problem before, except with the condition that $f$ is continuous. In this case, we can use the Weierstrauss Approximation Theorem to approximate $f$. But in this problem we are not given that $f$ is continuous. How to proceed without assuming $f$ is continuous?

ruben
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  • Hint: Use Weierstrass approximation theorem and the fact that continuous functions are dense in $L^1$ to show that $\int_0^1 g(x)f(x) , dx = 0$ for all $g \in L^1([0, 1])$. Then use the fact that $L^\infty([0, 1])$ is the dual space of $L^1([0, 1])$ to finish the proof. – David Gao Mar 31 '24 at 15:57

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