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

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

($\displaystyle \mathbb{I}$ 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:

$\displaystyle \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 $\displaystyle \psi$ depend on the measurability of $\displaystyle f_{n,k}$ (i'm thinking of dominated convergence)? Do we need to know first whether $\displaystyle f_{n,x}$ is measurable?