## mean value theorem for integrals [closed]

Suppose I have an integral of $f_\delta:\Omega_\delta \rightarrow \mathbb{R}$, where $\Omega_\delta \in \mathbb{R}^N ~ \forall~ \delta$ is a box with maximum side-length $\delta$. Is it true that that

$\lim_{\delta \rightarrow 0} \frac{1}{|\Omega_{\delta}|}\int_{\Omega_\delta} f(y) dy = f(x)$ ?

(${|\Omega_{\delta}|}$ is the integral with $f=1$). Can't find this, but seems to be true. Maybe not called MVT? Thanks for your help!

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 What is $x$? If it is the center of $\Omega_\delta$ and $f$ is continuous at this point write the integrand as $f(y)=f(x)+f(y)-f(x) \cong f(x)$. – Jochen Wengenroth Feb 7 2012 at 10:20 what conditions are you putting on $f$? and in what context did this question arise? – Yemon Choi Feb 7 2012 at 10:48 Yes, sorry, $x$ is the center of $\Omega_\delta$. What will the re-arrangement buy me? I know $f$ is smooth, but that just means $f(y)-f(x) = \nabla f(x) \cdot (y-x) + \Bigo (|y-x|^2)$, and the latter term could be as large as $\delta^2$, while the denominator is $\Bigo (\delta^N)$. Maybe polar coordinates? – confused Feb 7 2012 at 10:53 I believe this is called the Lebesgue differentiation theorem and the requirement is that the family $\Omega_\delta$ are of bounded eccentricity. – confused Feb 7 2012 at 11:18 For the next person who happens upon this posting and doesn't know the answer, it is both the Lebesgue differentiation theorem, and, more specifically, for continuous $f$, as Jochen alluded to: $f(y) \leq f(x) + C \delta$ Therefor, the integral is equal to $f(x) + C \delta$. Thanks. – confused Feb 7 2012 at 11:49