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Let $f_i$ be a sequence of smooth functions on $S^2$ such that the measures $\mu_i=f_i \;d\mathrm{vol}_{S^2}$ converge weakly to $d\mathrm{vol}_{S^2}$. Now suppose $\epsilon_i$ is a sequence going to zero. My question now is if for each $x\in S^2$ $$\limsup_i \epsilon_i^{-2}\mu_i(B_{\epsilon_i}(x))=1.$$

(I've chosen to phrase the question on $S^2$ because I want it on a compact set w/o boundary, so that nothing funny can happen at the boundary or "at infinity".)

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Is $\mathrm{vol}_{S^2}$ the 2-dimensional Lebesgue measure on $S^2$ (the surface area of the sphere)? – Ben Derrett Feb 11 '12 at 13:41
Hi alfred, and welcome to math.SE! Users with any number of "reputation points" can comment on their own questions and answers (once you obtain 50 points, you gain the ability to comment anywhere), but you were not able to comment because you were not signed into the account that asked the question. I've now merged the duplicate account into the original. If you register your account that should help to prevent login difficulties in the future. Here is an explanation of reputation points. – Zev Chonoles Feb 11 '12 at 17:01

Unfortunately, weak convergence of measures is not strong enough to imply your conclusion. For a given $x\in S^2$ and sequence $\epsilon_i$ it is easy to define smooth functions $f_i$ with $f_i(y)=0$ for $y\in B(x,\epsilon_i)$, but $f_i(y)=1$ for $y\notin B(x,2\epsilon_i)$. This gives a counterexample to your proposal.

What you want is convergence in total variation.

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Thanks Byron. How about if I only want the $\limsup$ to be $\leq 1$? Can I at least get that? – alfred jodokus Feb 11 '12 at 16:41
@alfredjodokus No. The "zero" in my answer could be set to any other value, e.g. "two". – Byron Schmuland Feb 11 '12 at 16:58

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