proving measurability

**Dinkydoe** · April 4th 2012, 04:16 AM

Hoi, I want to show that $\psi(x)=\mu(\overline{B}_x(r))$ is measurable. Here $\overline{B}_x(r)$ is the closed ball with center x and radius r in some banach space X=(S,d), and $\mu$ is a "finite" borel-measure.

In the exercise we had to prove $\lim_{n\to\infty}f_{n,x} = \mathbb{I}_{\overline{B}_x(r)}$ for some function $f_{n,x}$ , point-wise convergence (I shall not define f here since I believe only its properties are relevant). and also for any sequence $x_k\to x$ we have $f_{x_k,n}\to f_{x,n}$
( $\mathbb{I}$ is indicated as the indicator-function ;p)

I'm asked to prove the measurability using these results (not even sure if all the results are necessary). I'm somewhat stuck here. The only thing I can think of is writing the following:

$\psi(x) = \mu(\overline{B}_x(r)) = \int_X \mathbb{I}_{\overline{B}_x(r)}d\mu = \int_X \lim_{n\to\infty}f_{n,x} d\mu$

and using the fact we may interchange limit and integral. I don't quite see yet how I might prove measurability...

Am i missing some heavy machinery here? Also, does the measurability of $\psi$ depend on the measurability of $f_{n,k}$ (i'm thinking of dominated convergence)? Do we need to know first whether $f_{n,x}$ is measurable?

**Alibiki** · April 6th 2012, 12:42 PM

You want to show that no matter what element $A$ in $\mathcal{B}([0,\infty))$ you choose, the object $\psi^{-1}(A)$ will always be an element in $\mathcal{B}(S)\ ,$ where $\mathcal{B}(M)$ denotes the Borel sigma-algebra on the metric space $M$ .

To do this you might need the facts that $\mathcal{B}([0,\infty))$ is generated by half-open intervals and that $\mathcal{B}(S)$ is generated by the collection of all open balls in $S$ .

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