here is the main body of the question:

suppose $\mu$ is a regular radon measure on $\mathbb R^n$, let $B(x,r)$ represents the closed ball on $\mathbb R^n$.

(1) prove that $$\lim_{y\to x} \sup(\mu(B(y,r))) \le \mu (B(x,r))$$ (2) make an example to illustrate that the "$\le$" can be "$<$" strictly

Here is my confusion. what is regular radon measure? We all know that radon measure is inner regular and local finite, and "regular" means inner and outer regular, does that mean the regular radon measure is actually a regular borel measure? And I don't know how to solve this inequation.

  • $\begingroup$ Radon measures are usually inner regular on open sets and outer regular on all Borel sets, whereas regular means inner and outer regular on all Borel sets. Does that help clear up the confusion? $\endgroup$ – postmortes Jan 11 '18 at 6:13
  • $\begingroup$ @postmortes ok,that helps. $\endgroup$ – user415071 Jan 11 '18 at 6:19

OK,I found a solution to (1).But right now I have something to do.I'll give a hint:use the Fatou lemma.I will finish my answer soon.


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