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Let $(E, d)$ be a Polish space and $\mu$ a finite Borel measure on $E$. Let $r>0$ and $x, x_n \in E$ such that $x_n \to x$.

Is it true that$$ \mu (B(x_n, r)) \to \mu (B(x, r))? $$

My attempt: Let $A_n := B(x_n, r)$ and $A := B(x, r)$. Then $1_{A_n} \to 1_A$ point-wise on $A$. If $y \notin A$ then $d(y, x) \ge r$. On the other hand, we only have $d(y, x_n) \ge d(y, x) - d(x, x_n) \ge r - d(x, x_n)$. As such, we are not sure if $1_{A_n} \to 1_A$ point-wise on $A^c$. Hence we can not apply DCT.

Akira
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1 Answers1

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Let $E = \mathbb{R}$, $\mu = \delta_1$ (Dirac measure concentrated at $x=1$), $r=1$ and $x_n = {1 \over n}$ (with $x=0$, of course).

Then $\mu B(x_n, r) = 1$ for all $n$ but $\mu B(x,r) = 0$.

copper.hat
  • 172,524