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I think the solution to this question somehow involves Riesz Representation Theorem, but I don't see how to apply it.

Suppose $g$ is a measurable function on $[0,1]$ such that $\int_0^1 fg~dx$ exists for all $f\in L^2[0,1]$. Then $g\in L^2[0,1]$.

How do I use the above mentioned theorem to show it.

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    And yes would have to be $1$ but in the right set. You can argue as follows: let $f_1=\chi_{\{x\in [0,1]:g(x)\gt h(x)\}}$. Then $\int f_1^2\leq \int_0^1 dx=1$, so $f\in L^2$, then $\int f_1(g-h)=0$. Similarly if $f_2=-\chi_{\{x\in [0,1]:g(x)\lt h(x)\}}$ you can see that $f_2\in L^2$ etc. If you consider $f=f_1+f_2$, then $\int f(g-h)=0$ is equivalent to $\int |g-h|=0$, therefore $g=h$ almost everywhere.2012-05-09

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