Below we use Bochner measurability and Bochner integral. Let $T>0$ and $p \in [1, \infty)$. Let $X :=[0, T]$ and $Y:= \mathbb R^d$. Let $L^p_{\text{loc}} (Y)$ be the space of measurable functions $f:Y \to \mathbb R$ such that $$ \|f\|_{L^p_{\text{loc}}} := \sup_{y \in Y} \|1_{B(y, 1)} f\|_{L^p} < \infty, $$ where $B(y, 1)$ is the open unit ball centered at $y$. Then $(L^p_{\text{loc}} (Y), \|\cdot\|_{L^p_{\text{loc}}})$ is complete but not separable. Let $(y_m)$ be a countable dense subset of $Y$. The sphere has Lebesgue measure $0$. Then by dominated convergence theorem (DCT), $$ \|f\|_{L^p_{\text{loc}}} = \sup_{m \in \mathbb N} \|1_{B(y_m, 1)} f\|_{L^p} \quad \forall f \in L^p_{\text{loc}} (Y). $$
I'm trying to prove a statement in a thread, i.e.,
(S2) Let $f:X \times Y \to \mathbb R$ be measurable such that $f(x, \cdot) \in L^p_{\text{loc}} (Y)$ for a.e. $x \in X$. The map $G: X \to \mathbb R, x \mapsto \|f(x, \cdot)\|_{L^p_{\text{loc}}}$ is measurable.
There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?
Let $f_n:X \times Y \to \mathbb R, (x, y) \mapsto f(x, y) 1_{B(y_n, 1)}(y)$. Then $f_n$ is the restriction of $f$ to the measurable subset $X \times B(y_n, 1)$. Then $f_n$ is measurable. Let $$ F_n: X \to L^p (Y), x \mapsto f_n(x, \cdot),\\ G_n: X \to \mathbb R, x \mapsto \|f_n(x, \cdot)\|_{L^p}. $$
Then $G =\sup_n G_n$ everywhere. It suffices to prove that $F_n$ is measurable.
Lemma Let $f:X \times Y \to \mathbb R$ be measurable such that $f(x, \cdot) \in L^p (Y)$ for a.e. $x \in X$. Then $F: X \to L^p (Y), x \mapsto f(x, \cdot)$ is measurable.
By Lemma, $F_n$ is measurable. On the other hand, the norm $\|\cdot\|_{L^p}$ is continuous and thus (Borel) measurable. Then $G_n = \|\cdot\|_{L^p} \circ F_n$ is measurable. This completes the proof.
