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Consider a locally-bounded function $f: W \rightarrow X$, $X \subseteq \mathbb{R}^n$, $W \subseteq \mathbb{R}^m$.

Assume that $f$ is Borel measurable

(for every open $O \in \Sigma_X$ ($\sigma$-algebra of $X$) we have $f^{-1}(O) \in \Sigma_W$ ($\sigma$-algebra of $W$))

Consider a locally-bounded discontinuous function $g: X \rightarrow Y$, $Y \subseteq \mathbb{R}^p$.

  1. Say if $\ g \circ f: W \rightarrow Y$ is measurable as well.

  2. If not: counterexample?

  3. If not: under which conditions is $g \circ f$ measurable?

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    Could be. I mean the definition of measurability provided in the question. To avoid confusion, let's delete the question on measurability of $g$.2012-06-29

1 Answers 1

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Consider $n=m=p=1$. Take $f:R \mapsto (0,1)$, $f(x) = \frac{1}{1+x^{2}}$, which is locally bounded, continuous and measurable. Let $V$ be the vitali set which is not measurable in $B(\mathcal{R})$. Let $g: (0,1) \mapsto \{0,1\}$,

$g(x) = \left\{ \begin{array}{ll} 1 & \mbox{$x \in V$};\\ 0 & \mbox{$x \notin V$}.\end{array} \right. $

$g \circ f$ is not measurable. If $g$ is continuous, $g \circ f$ is measurable. In general, if $g$ is measurable, $g \circ f$ is measurable.