Let $(E,\mathfrak{e},\mu)$ and $(G,\mathfrak{g},\nu)$ be two probability spaces. Let $E\times G$ denote the product space with the product $\sigma-$algebra. Let $L^0$ denote the space of measurable functions from $G$ to $\mathbb{R}$ upto $\nu$ almost sure equality. The space $L^0$ can be equipped with the topology induced by convergence in probability and let $B(L^0)$ denote the Borel $\sigma-$algebra on $L^0,$ that is the $\sigma-$algebra generated by its topology, and consider $(L^0, B(L^0))$ as a measure space.
Let $h \colon E \times G \to \mathbb{R}$ be a measurable function on the product. Then by Fubini theorem, we know that the functions $p(e)=\int_G h(e,g) d\nu$ from $E$ to $\mathbb{R}$ and $r(g)=\int_E h(e,g) d\mu$ from $G$ to $\mathbb{R}$ are measurable.
My question is the following: let $k\colon E \to L^0$ be a measurable function. Or we could see $k$ on the product as follow: $\overline{k} \colon E \times G \to \mathbb{R}$ defined by $\overline{k}(e,g)=k_e(g).$ So if we fix $e$ in $\overline{k}$ then we get $k_e$ a measurable function in $g,$ and these functions change in a measurable way in $e.$ Can we say that $\overline{k}$ is a measurable function on the product?
We could define $s(e)=\int_G k_e(g) d\nu$ and attempt to define $t(g)=\int_E k_e(g) d\mu$. Can we say that $s$ is measurable, and what should we assume more to say that $t$ is well defined and is measurable?
Any kind of hint is appreciated.