Refferring to the Lebesgue decomposition theorem in Lebesgue decomposition theorem and fundamental theorem of calculus there is a corollary when the measure is the Lebesgue measure that states: if $f\in L^p_{loc}(\mathbb{R}^n)$ then $$\lim \frac{1}{|B|}\int_{B}|f-f(x)|^p=0$$ for a.e. $x$, where the limit is taken on balls B each containing $x$ with diameter tending to zero (x is not necessarily the center of the balls). Is there a counter-example of this fact if we choose a general Radon measure (I know taking the limit on balls centered in x holds the same result, but could this not happen if the balls are not centered in $x$?)
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$\begingroup$ Looking at the proof I guess maybe is the measure is not doubling this couldn't work, but I don't even know real examples od non-doubling measures.. $\endgroup$– balestravApr 20, 2012 at 13:03
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Your conjecture is right. The point is that the balls need not be centered at x. This theorem and its proof can be found in Measure Theory and Fine Properties of Functions (Revised Edition) by Lawrence C. Evans and Ronald F. Gariepy. I have posted the theorem and its proof below.
