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prove or give a counterexample: for all Lebesgue measurable set $E\subset \mathbb{R}^{n}$ and all continuous function $f:\mathbb{R}^{n} \to \mathbb{R}$, there is a point $\xi_{E}\in\mathbb{R}^{n}$ such that $$\int_{E}{f}dm=f(\xi_{E})m(E)$$ where m is the Lebesgue measure.

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Thisx resembles the mean value theorem for integrals, saying there is $c\in (a,b)$ for which $\int_a^b f(x)\ dx=f(c)(b-a)$. Maybe look at that proof. –  coffeemath Feb 4 '13 at 9:54
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Let $V = \frac{\int_E f dm}{m(E)}$ and suppose without loss it is positive. If $f(x) > V$ for all $x \in E$, then $\int_E f dm > m(E) V$. If on the other hand $f(x) < V$ for all $x \in E$, then $\int_E f dm < m(E) V$. So there must be $x_0, x_1 \in E$ satisfying $f(x_0) \leq V \leq f(x_1)$. Connect $x_0$ and $x_1$ up by a line and use the intermediate value theorem.

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