# If $f,g:(\Omega, \mathcal{F},P) \to (\mathbb{R}, \mathcal{B}(\mathbb{R})$ are measurable, $f=h\circ g$, then is $h$ Borel-measurable?

If $$f,g:(\Omega, \mathcal{F},P) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$$ are measurable, and $$f=h\circ g$$ where $$h: \mathbb{R}\to \mathbb{R}$$, then is $$h$$ Borel-measurable?

The above question is motivated by the following statement: for any $$x,y\in \Omega$$ if $$g(x) = g(x)$$ implies that $$f(x)=f(y)$$, then $$f$$ is a function of $$g$$, say $$f = h\circ g$$. Here, how can we analyze the measurability of $$h$$? Alternatively, can we claim that there is a $$\tilde h$$ such that $$\tilde h = h$$ almost everywhere?

(I tried to approach this using: for any set with positive outer measure, there is a non-measurable subset $$A$$ of it, therefore $$\mathbb{1}_{A}$$ is non-measurable. Yet it is hard to characterize a non-measurable function in general.)

• Borel-measurable means that pre-image of Borel set is Borel. Assume $X$ is Borel, then $f^{-1}(X) = g^{-1}(h^{-1}(X))$, $h$ is measurable so $h^{-1}(X)$ is Borel, and as $g$ is measurable, $g^{-1}(h^{-1}(X))$ is also Borel. Commented Sep 25, 2022 at 20:55
• Thanks for your answer. Yet I assumed the measurability of $f,g$ here and want to find out the measurability of $h$. The statement you proposed shows that given $g$ measurable and $h$ Borel-measurable, their composition is measurable. Commented Sep 25, 2022 at 21:17
• I think I just figured out the answer: let $\sigma(T')$ and $\sigma(T)$ be the $\sigma$-algebra generated by $T',T$, then $\sigma(T)\subseteq \sigma(T')$, hence there is a Borel-measurable function $h$ such that $T=h\circ T'$. Commented Sep 25, 2022 at 21:27

Let $$h$$ be non-measurable and let $$g$$ be constant. Then so is $$f$$. In particular both $$f$$ and $$g$$ are measurable but $$h$$ is not.