In my proof of below result, I'm surprised that $f$ has no effect on the measurability of $f^c$.
Could you have a check on my attempt?
Is there stronger assumption that makes $f^c$ continuous?
Let $X,Y$ be topological spaces, $c:X \times Y \to \mathbb R \cup \{+\infty\}$, and $f:X \to \mathbb R \cup \{-\infty\}$. We define the $c$-transform $f^c:Y \to \mathbb R \cup \{\pm \infty\}$ of $f$ by $$ f^c (y) := \inf_{x \in X} [c(x, y) - f(x)] \quad \forall y \in Y. $$
Theorem: If $c$ is upper semi-continuous (u.s.c.), then $f^c$ is u.s.c. and thus Borel measurable.
My attempt: For each $x \in X$, we define a map $g_x:Y \to \mathbb R \cup \{+\infty\}$ by $$ g_x (y) := c(x, y) - f(x) \quad \forall y \in Y. $$
Then $g_x$ is u.s.c. for all $x \in X$. The pointwise infimum of an arbitrary collection of u.s.c. functions is again u.s.c. Then $$ f^c = \inf_{x \in X} g_x $$ is u.s.c. and thus Borel measurable.
