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Consider a locally bounded function $f: X \times W \rightarrow X$, where $X \subseteq \mathbb{R}^n$, $W \subseteq \mathbb{R}^m$, such that

for all $x \in X$ the function $w \mapsto f(x,w)$ is (Borel) measurable;

Consider a locally bounded, (Borel) measurable, function $g: W \rightarrow X$.

Say if the function

$ (w,v) \mapsto f( g(w), v ) $

is (Borel) measurable as well.

Notes: this question differs from both this and that post.

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    Also in $\mathbb{R}^n$, the composition of two Borel mappings is a Borel mapping as well [Bogachev - Measure Theory, pag. 146].2012-06-29

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What you are trying to prove does not hold. Let $V$ be the Vitali Set which is not measurable in $B(\mathcal{R})$. Let $f: R^{2} \mapsto R$, such that:

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

Note that $f$ is locally bounded and for any $x \in R$, $y \mapsto f(x,y)$ is Borel measurable, since it is constant. Let $g$ be the identity function. Observe that $f(g(x),y) = f(x,y)$ which is not measurable in $(R^{2},B(\mathcal{R}^{2}))$.

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    I will edit the answer to make it more explicit.2012-06-29