Let $(M, \mathcal{A}, \mu)$ a complete separable probability space. Recall that complete means that any subset of a measurable set with zero measure is measure (and has zero measure) and separable means that there exists a countable family $E\subset \mathcal{A}$ such that for any $\varepsilon >0$ and any $B \in \mathcal{A}$ the exists $A_k \in E$ such that $\mu(B\triangle A_k)<\varepsilon$.
Let $Y$ compact metric space.
We define the family $\mathcal{F}$ so that $\phi \in \mathcal{F}$ sss
- $\phi$ is measurable ($Y$ with borel sigma algebra)
- for all $x\in M$, $\ $ $\phi(x,.): Y \rightarrow \mathbb{R}$ is continuous i.e. $\phi(x,.)\in (C^0(Y),\Vert . \Vert_0 )$ (supremum norm)
- $x \mapsto \Vert \phi(x,.) \Vert_0 \in L^1(\mu)$
equip $\mathcal{F}$ of a norm: $\Vert \phi \Vert_1=\int \Vert \phi(x,.) \Vert_0 d\mu$
Then $(\mathcal{F}, \Vert . \Vert_1 )$ is a separable Banach space.
I tried that $(\mathcal{F}, \Vert . \Vert_1 )$ is a Banach space
I have difficulty with the separability: I show my progress
using the fact that $C^0(Y)$ is separable, and the assumption that the probability space is separable we obtain: $\lbrace f_i\rbrace $ countable and dense in $C^0(Y)$ and $\Gamma= \lbrace \sum _{k=1}^{r} c_k\chi_{A_k} : A_k \in E, c_k\in \mathbb{Q} \rbrace $ countable and dense in $L^1(\mu)$.
Now note that for $\phi \in \mathcal{F}$, $\phi(x,.)$ can be approximated by $\lbrace f_i\rbrace $ and $\phi(.,y) \in L^1(\mu)$
since for fixed $y$ $$\int \vert\phi(x,y)\vert d\mu(x)\leq \Vert \phi\Vert_1$$ then $\phi(.,y)$ can be approximated by $\Gamma$ but how to unite these facts...
I appreciate the patience to read my query and hope you can help
What you want to do is think of your normed space as $L^1(M; C^0(Y))$, the Banach space of integrable maps taking values in the Banach space $C^0(Y)$. More generally let $(V,\|\cdot \|_V$ be any separable Banach space; then $L^1(M;V)$ is separable. To see this take a countable dense set $N=\{f_i\}_{i\in \mathbb{N}}\subset V$ and consider the collection $ L$ of linear combinations $$ \sum_{i=0}^m f_i\chi_{E_i}$$ where $f_i\in N$, $E_i\in E$ and $m\in \mathbb{N}$ which belong to $L^1(M;V)$, i.e. $$\int_M\|\sum_{i=0}^m f_i\chi_{E_i}\|_Vd\mu<\infty.$$ Let us show this (countable) collection is dense in $L^1(M;V)$.
Firstly, take $\phi\in L^1(M;V)$. For natural numbers $m,i$ let $A_{i,m}'= \phi^{-1}(B(f_i,2^{-m}))\subset M$. Here $B(f,r)=\{g\in V: \|f-g\|_V<r\}$. This set is measurable (in your particular case, the measurability of this set follows from your assumption that $x\mapsto \|\phi(x,\cdot)\|_0$ is integrable and the Pettis measurability theorem ) and we modify it to obtain a partition of $M$, by setting $$ A_n^m:=A_{n,m}'\setminus \bigcup_{i<n}A_{i,m}.$$ Now $$\| \sum_{i} f_i\chi_{A_i^m}-\phi\|_V\le 2^{-m}$$ for every (almost every) $x\in M$ by construction, and thus the sequence $\displaystyle g_m= \sum_{i} f_i\chi_{A_i^m}$ converges to $\phi$: $$\int_M\|g_m-\phi\|_V d\mu\le 2^{-m}.$$
For every $i,m$ we may choose a set $B_i^m$ from the countable collection $E$ so that $$\mu(B_i^m\triangle A_i^m)<2^{-i-m}\|f_i\|_V^{-1}.$$ Hence $$ \int_M \|g_m-\sum_{i} f_i\chi_{B_i^m}\|_Vd \mu \le \sum_i \|f_i\|_V\int_M|\chi_{A_i^m}-\chi_{B_i^m}|d\mu<\sum_i \|f_i\|_V(\|f_i\|_V^{-1} 2^{-i-m})=2^m,$$ since $$\int_M|\chi_{A_i^m}-\chi_{B_i^m}|d\mu= \mu(B_i^m\triangle A_i^m).$$ Therefore we may find a sequence of linear combinations from $L$, converging to $\phi$ in norm.