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Let $\def\AAA{\mathcal{A}} \def\BBB{\mathcal{B}}$ $f:(X,\AAA)\to (Y,\BBB_Y)$ be a function from a measurable space $(X,\AAA)$ to a Borel space $(Y,\BBB_Y)$, meaning $\BBB_Y$ is the Borel $\sigma$-algebra generated by a fixed metrizable topology on $Y$. I've been wondering if the following result is true:

Theorem (?): $f$ is measurable if and only if it is the limit of simple measurable functions.

The result is true if the word 'simple' is removed, as shown here.

Sam
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1 Answers1

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If $\langle f_n\rangle$ is a sequence of simple functions converging to $f$, then the values of must lie in the closure of the union of the ranges of the $f_n$. In particular, $f$ must have a separable range.

So, a simple counterexample is the identity function on a nonseparable metric space (i.e., an uncountable discrete metric space).

Michael Greinecker
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