Let $(Ω, A, μ)$ be a measure space and $a < b$ be any two real numbers. Suppose $f$ is a real-valued function on $(a, b) × Ω$ such that (i) $\forall t ∈ (a, b)$, the function $ω \mapsto f(t,ω)$ is A-measurable and (ii) $\forall ω ∈ Ω$, the function $t \mapsto f(t,ω)$ is continuous on $(a,b)$.
Show that if $ \exists μ-integrable$ function $g$ on $Ω$ such that $|f(t,ω)| \le g(ω) , \forall(t,ω)$, then $φ(t) = \int f(t, ω)dμ(ω)$ defines a real-valued continuous function on $(a,b)$.
My attempt:
$\forall ω ∈ Ω$,the function $t \mapsto f(t,ω)$ is continuous on $(a,b)$ and hence measurable on $\beta(a,b)$. Also, $\forall t ∈ (a, b)$, the function $ω \mapsto f(t,ω)$ is A-measurable.
I don't know how to conclude from the above two lines that $f : (a,b) \times \Omega \to \Bbb R$ is measurable w.r.t the product measure.
If, $f : (a,b) \times \Omega \to \Bbb R$ is indeed measurable w.r.t the product measure and $ \exists μ-integrable$ function $g$ on $Ω$ such that $|f(t,ω)| \le g(ω) , \forall(t,ω)$ , then it seems that Dominated Convergence Theorem should suffice.
It will be extremely helpful if details can be provided. Thanks in advance for help!
You don't have to prove joint measurability. Let $t_n \to t$. To show that $\int f(t_n,\omega) d\mu(\omega) \to \int f(t,\omega) d\mu (\omega)$ you can apply DCT.