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My question is based on this post, which I summarize below.
Claim: Let $(X, \textbf{X})$ be a measurable space, $f:X \to \mathbb{R}$ is X-measurable, $g: \mathbb{R} \to \mathbb{R}$ is continuous. Then $g \circ f$ is X-measurable.
Pf: WTS $(g \circ f)^{-1}( \alpha, \infty) \in \textbf{X}$, $\forall \alpha \in \mathbb{R}$. We have $(g \circ f)^{-1}( \alpha, \infty) = f^{-1}(g^{-1}( \alpha, \infty))$. Now $g^{-1}( \alpha, \infty) \in \textbf{B}$ (is a Borel set), since g is cont. Thus, $f^{-1}(g^{-1}( \alpha, \infty)) \in \textbf{X}$, by this result.

Looking at the Pf, it seems the only property of g being cont that we used was that $g^{-1}( \alpha, \infty) \in \textbf{B}$. But if we weaken the hypothesis to g being B-measurable, then it seems that it's still true that $g^{-1}( \alpha, \infty) \in \textbf{B}$, again by the cited result and since $( \alpha, \infty) \in \textbf{B}$.
Is this true? Sorry I'm new to measure theory and unsure. After all, if it was true, why bother with the special case that g is cont at all, since the proof technique would be no different?

Community
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Jason
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1 Answers1

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The composition of measurable functions where we use the same $\sigma$-algebras throughout, is measurable.

However, when speaking of Lebesgue measurable functions $\Bbb R\to\Bbb R$, the definition says that the inverse image of a set belonging to the Borel $\sigma$-algebra is a set that belongs to the Lebesgue $\sigma$-algebra. Hence for Lebesgue measurable $g$ it may happen that for some Borel set $E$, the set $g^{-1}(E)$ is Lebesgue, but not Borel. In that situation we are also no longer guaranteed that the $f^{-1}(g^{-1}(E))$ is Lebesgue.

Hagen von Eitzen
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  • Thx for the confirmation! I have not yet learned about what the Lebesgue $\sigma$-algebra is yet, but your remark motivates me to continue studying measure theory – Jason May 28 '16 at 21:46