Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

$$\lim \limits_{R \rightarrow 0} \frac{1}{m(B_R(x))}\int_{B_R(x)} f \quad \text{ exists} \tag{1}$$ or $$\lim \limits_{R \rightarrow 0} \;\; \sup \limits_{ B, B' \in \mathcal B_R(x)} \left\vert \dfrac{1}{m(B)}\int_{B} f - \dfrac{1}{m(B')}\int_{B'} f \right\vert = 0 \tag{2}$$ 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?

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.

share|cite|improve this question
up vote 1 down vote accepted

It isn't done.

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


share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.