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Let $f \in L^1_{loc}$. We call $x \in \mathbb R^n$ an Lebesgue point, if

  1. $\lim \limits_{R \rightarrow 0} \frac{1}{m(B_R(x))}\int_{B_R(x)} f \;\;\;\;$ exists

or

  1. $\lim \limits_{R \rightarrow 0} \;\; \sup \limits_{ B, B' \in \mathcal B_R(x)} \vert \dfrac{1}{m(B)}\int_{B} f - \dfrac{1}{m(B')}\int_{B'} f \vert = 0$, where $\mathcal B_R(x)$ denotes all balls that contain $x$ with radius at most $R$

Most calculus books state only one of these definitions, and whereas they state several conclusions or variations of the theorem, but they usually stick to one of the above definitions.

However, although the equivalence is obvious, and indeed trivial from 2. to 1., I do not manage to prove the implication from 1. to 2. - can someone please explain me, how this is done?

Edit: I should clarify, that I am aware both definitions are in general different for, say, a continuous function with a jump. Nevertheless, for example in Wikipedia I have encountered both defintions entitled as Lebesgue point.

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1 Answer 1

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It isn't done.

0 is a Lebesgue_1 point of signum
0 is not a Lebesgue_2 point of signum

signum

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Well, both definitions exist (cf. "Lebesgue Differentiation theorem" and "Lebesgue point" in Wikipedia). But it would set me somehow aghast if the difference between both definitions would really be swept under the rug. –  shuhalo Jan 19 '11 at 1:17

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