if $g: \mathbb{R}^k \rightarrow \mathbb{R}$ is continuous and

$F_i=X \rightarrow \mathbb{R}, i = 1,2,...,k$ is measurable

Prove that $h(x) = g(f_1(x),f_2(x),...,f_k(x))$ is measurable.

So far I have that if $f$ is measurable and $g$ is Borel Measurable then $g \circ f$ is measurable:-

$f$ is measurable, $(g \circ f)^{-1}=f^{-1} \circ g^{-1}$, so $g \circ f$ is measurable (is this correct?)

Now, I want to use this result in proving that $h$ is measurable since $g$ is continuous implies that $g$ is Borel measurable, but how would I define the inverse of $h$?

Also, any other suggestions on how to prove this?

  • $\begingroup$ In the proof that $g\circ f$ is measurable, we can't write $f^{-1}$: we instead have to consider $f^{-1}(S)$ where $S$ is a set. $\endgroup$ – Davide Giraudo May 18 '14 at 21:34

The map $f\colon \mathbb R\to\mathbb R^k$ (where both spaces are endowed with the Borel-$\sigma$-algebra) defined by $f(x)=(f_1(x),\dots,f_k(x))$, is measurable. Then $g\circ f$ is measurable.

We use the fact that if $(X,\mathcal A)$, $(Y,\mathcal B)$ and $(Z,\mathcal C)$ are measurable spaces and $f\colon X\to Y$, $g\colon Y\to Z$ are measurable then so is $g\circ f$. This follows from the fact that if $C\in\mathcal C$, then $(g\circ f)^{-1}=f^{-1}(g^{-1}(C))$ and $g^{-1}(C)\in \mathcal B$.

  • $\begingroup$ g is only continuous, not measurable. is there a way to work around this? $\endgroup$ – Pablo May 20 '14 at 3:16
  • $\begingroup$ Do you mean Borel or Lebesgue measurable? $\endgroup$ – Davide Giraudo May 20 '14 at 9:40

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