Finiteness of $\lim_{h \to 0^+} \frac1{h} \int_x^{x+h} f(t)\mathrm{d}t$ 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. 

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 $0\lt h\lt 1$, $1/h\int_0^h f(t)dt=2/\sqrt{h}$.
