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Suppose that $f$ is Lebesgue measurable and $g$ is real valued, continuous, and has the property that for any null set $N$, $g^{-1} (N)$ is measurable. Then $f \circ g$ is also Lebesgue measurable.

That seems entirely too strong? Any idea why this is true?

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  • $\begingroup$ Please see here for how to typeset common math expressions with LaTeX, and see here for how to use Markdown formatting. $\endgroup$ – Zev Chonoles May 20 '13 at 20:28
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Any Lebesgue measurable set can be written as the symmetric difference $B \vartriangle N$ of a Borel set $B$ and a null set $N$. Since $g$ is continuous, $g^{-1}(B)$ is then also Borel, and, by assumption, $g^{-1}(N)$ is measurable, so $g^{-1}(B \vartriangle N)=g^{-1}(B) \vartriangle g^{-1}(N)$ is also Lebesgue measurable. This means that applying $g^{-1}$ to a Lebesgue measurable set gives another Lebesgue measurable set. Also, since $f$ is Lebesgue measurable, applying $f^{-1}$ to an open set gives a Lebesgue measurable set. So, if $O$ is open, $$(f\circ g)^{-1}(O)=g^{-1}(f^{-1}(O))$$ must be Lebesgue measurable, meaning that $f\circ g$ is Lebesgue measurable.

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