Let $f$ be Lebesgue measurable, and $\int_a^b x^\alpha f(x) = 0$ for every $\alpha\ge 0$.

How do I show that $f(x)=0 ~ a.e.$

and if the condition change "$\alpha\ge 0$" to "$\alpha\ge k$ for some $k\ge 1$", if the statement is also true?

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Shouldn't there be a $dx$ or a $df(x)$ in your integral? –  draks ... Dec 16 '12 at 12:54

For $f$ a polynomial, both cases should be true. In the first case, by the linearity of integration, we have $\int f p = 0$ for any $p \in \mathbb{R}[x]$, in particular $\int f^2 = 0$ giving that $f^2 \equiv 0$ and the result.
In the second case, we have $\int f p = 0$ for any $p$ with lowest term $x^k$. In particular we have $\int f^2 x^{2n} = 0$, for some $n \gg 0$, so same trick applies.
For general $f$, I think polynomials are dense in $L^1$, from which the result should follow, namely if $f_i \in \mathbb{R}[x]$ $\rightarrow f$, we have $0 = \int (x^{2n})f_i f \rightarrow \int (x^{2n}) f^2$.
One way to phrase this is: the dual of $L^1$ is $L^\infty$ (because the interval is $\sigma$-finite) and the hypothesis is that a dense subspace of $L^\infty$ vanishes on $f$. Since the dual of $L^1$ separates points in $L^1$, $f=0$. –  Mariano Suárez-Alvarez Dec 16 '12 at 11:01
Shouldn't there be a $dx$ or a $df(x)$ in your integral? –  draks ... Dec 16 '12 at 12:54
@draks, the «$dx$» is usually omitted. –  Mariano Suárez-Alvarez Dec 18 '12 at 15:43