Let $X := \mathbb R^d$ and $h: X \to \mathbb R$ be strictly convex. Then
- $\partial h (x)$ is non-empty, convex, and compact for all $x \in X$.
- $\partial h (x) \cap \partial h (y) \neq \emptyset$ if and only if $x=y$.
Let $D := \bigcup_{x\in X} \partial h (x)$. Then the inverse $(\partial h)^{-1} : D \to X$ of $\partial h$ is well-defined. Let $g:=(\partial h)^{-1}$ and $d_H$ be the Hausdorff distance on the collection of compact subsets of $X$.
For $x, x_n \in X$, if $d_H(\partial h(x_n),\partial h(x)) \to 0$ then $x_n \to x$. This implies $g$ is continuous w.r.t. the subspace topology of $D$. Hence $g$ is Borel measurable w.r.t. the Borel $\sigma$-algebra $\mathcal B(D)$ induced from the subspace topology of $D$. It's not necessarily true that $D \in \mathcal B(X)$.
Now we assume $f: S\to X$ is Borel measurable for some $S \in \mathcal B(X)$ such that $f(S) \subset D$. I would like to prove that
Theorem $g \circ f:S \to X$ is Borel measurable.
Could you confirm if my above understanding and below proof are fine?
My attempt Fix $A \in \mathcal B(X)$. Then $(g \circ f)^{-1} (A) = f^{-1} (g^{-1} (A))$. We have $C := g^{-1} (A) \in \mathcal B (D)$. It suffices to prove that $f^{-1} (C) \in \mathcal B(X)$. There is $C' \in \mathcal B(X)$ such that $C = C' \cap D$. Then $$ f^{-1} (C) = f^{-1} (C' \cap D) = f^{-1} (C') \cap f^{-1} (D) = f^{-1} (C') \cap S = f^{-1} (C') \in \mathcal B(S). $$
This completes the proof.
