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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
    
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1 Answer 1

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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