# Let $f: \mathbb{R} \to \mathbb{R}$ be continuous and let $g: \mathbb{R} \to \mathbb{R}$ Lebesgue measurable. Is $f \circ g$ Borel measurable?

Let $$f: \mathbb{R} \to \mathbb{R}$$ be continuous and $$g: \mathbb{R} \to \mathbb{R}$$ Lebesgue measurable. Is $$f \circ g$$ Borel measurable?

I think that the answer to this question is no. I know that the hypothesis implies that $$f \circ g$$ is Lebesgue measurable, but I am sure that there is a counter-example to show that $$f \circ g$$ is not Borel measurable, since the Borel sigma algebra is a proper subset of the Lebesgue sigma algebra.

Note: In my textbook, a function is measurable with respect to a sigma algebra $$\mathcal{E}$$ if and only if $$[f > \alpha] = \{x : f(x) > \alpha\} \in \mathcal{E}$$ for all $$\alpha \in \mathbb{R}$$.

My current idea is to find a Lebesgue measurable set $$D$$ that is not Borel measurable. Then define $$g(x) = \chi_{_{D}}$$, which is Lebesgue measurable, as $$D$$ is Lebesgue measurable. Then $$f(x) = x$$ is continuous and $$f \circ g = g$$, but $$g$$ is not Borel measurable.

• Does $X=\mathbb{R}$? – confused_wallet Mar 10 at 3:02
• Yes, that was a typo. Fixed. – johnny133253 Mar 10 at 3:05
• I think your idea is a good one! – confused_wallet Mar 10 at 3:07
• Your current idea is correct. – Ramiro Mar 10 at 15:31