Let $(\Omega,\mathcal{A},\mu)$ be a measure space and $a<b$ be any two real numbers. Suppose $f$ is a real-valued function on $(a,b)\times\Omega$ such that
(i)for each $t\in(a, b)$, the function $\omega \mapsto f(t,\omega)$ is $\mathcal{A}$-measurable and
(ii) for each $\omega\in\Omega$, the function $t\mapsto f(t, \omega)$ is continuous on $(a, b)$. Show that if there is a $\mu$-integrable function $g$ on $\Omega$ such that$|f(t, \omega)|\leq g(\omega) \forall (t, \omega)$, then $\phi(t)=\int{f(t,\omega) d\mu(\omega)} $ defines a real-valued continuous function on $(a, b)$.
I tried to show that $\phi$ is Lipschitz. I faced some problems as $(a,b)$ is open. My next attempt was to approximate $f(t,\omega)$ by simple valued functions $\{f_n\}_{{n\in\mathbf{N}}} $ for a fixed $\omega$, which will increase to $f$. This should lead us to a sequence of a continuous functions $\{\phi_n\}_{n\in\mathbf{N}}$ which should increase to $\phi$ (I think we have to use DCT here). But here again uniform continuity is a problem. Please help with a detailed solution.
This is an immediate consequence of the Lebesgue's Dominated Convergence Theorem. If $t_n\to t$, the functions $f(t_n,\,\cdot\,)$ converge pointwise to $f(t,\,\cdot\,)$ and are dominated by $g$. Therefore $$ \phi(t_n) = \int_\Omega f(t_n,\omega)\,d\mu(\omega) \to \int_\Omega f(t,\omega)\,d\mu(\omega) = \phi(t) . $$
The function $\phi$ is not necessarily Lipschitz. Take for instance $\Omega=\{0\}$, $\mu=\delta_0$, $a=0$, $b=1$, $f(t,0)=\sqrt t$ and $g(0)=1$.