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If f is in $L^1(\mathbb{R})$, is it true that $\lim_{h \to 0^+} \frac1{h} \int_x^{x+h} f(t)\mathrm{d}t$ exists and is finite for every x in $\mathbb{R}$?

Would it be possible to use something along the lines of the following argument: The Lebesgue Differentiation Theorem says that this integral is equal to f(x) a.e., which is finite a.e. if f is in $L^1(\mathbb{R})$. And since the integral doesn't change on a set of measure 0, then the limit itself must be finite a.e.

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    Simplest example: $f(x)=1$ on $[0,1]$ and $f(x)=2$ on $(1,2]$. What happens at the jump discontinuity?2011-07-20
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    @N.S.: In that case $\lim\limits_{h\to0^+}\frac{1}{h}\int_1^{1+h}f(t)dt=2$, so it doesn't give a counterexample.2012-01-05

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No. As you said, it is true almost everywhere, but for example if $f(x)=x^{-1/2}$ on $(0,1)$ (and whatever you like elsewhere), then for $0\lt h\lt 1$, $1/h\int_0^h f(t)dt=2/\sqrt{h}$.