$\phi(t) = \int f(t,\omega)d\mu(\omega)$ is everywhere differentiable and moreover, $\phi^/(t)=\int(t,\omega)d\mu(\omega)$

Question

Let $(Ω, A, μ)$ be a measure space and $a < b$ be any two real numbers. Suppose f is a real-valued function on $(a,b)\times Ω$ such that $(i)\forall t \in (a,b)$, the function $ω \mapsto f(t,ω)$ is $A$-measurable and $μ$-integrable and $(ii) \forall ω ∈ Ω$,the function $f(·, ω)$ is everywhere differentiable on $(a,b)$, that is, $h(t,\omega)=\frac{\partial f(t,\omega)}{\partial t}$ exists $\forall t \in (a,b)$ .

Show that if there exists a $μ$-integrable measurable function $g$ on $Ω$ such that $|h(t, ω)| ≤ g(ω) \forall t∈(a, b)$ and $\forall ω ∈ Ω$,then the function $\phi$ defined on $(a,b)$ by $\phi(t) = \int f(t,\omega)d\mu(\omega)$ is everywhere differentiable and moreover, $\phi^/(t)=\int(t,\omega)d\mu(\omega)$

My attempt:

For $t \in (a,b)$, considered sequence $\{t_n\}_{n\ge 1}$ such that $\lim_{n \to \infty} t_n = t$.

Then, $$\lim_{n \to \infty}{\frac{\phi(t_n)-\phi(t)}{t_n - t}}=\lim_{n \to \infty}{\frac{\int f(t_n,\omega)d\mu(\omega)-\int f(t,\omega)d\mu(\omega)}{t_n - t}}$$ $$=\lim_{n \to \infty}\int{\frac{f(t_n,\omega))- f(t,\omega)}{t_n - t}}d\mu(\omega) \text{ ( since for all f(t,.) is $\mu$-integrable )} $$ $$=\lim_{n \to \infty} \int{h(\zeta_n,\omega)}d\mu(\omega) \text{ ( since $h(t,\omega)=\frac{\partial f(t,\omega)}{\partial t}$ exists $\forall t \in (a,b)$ and by MVT )} $$ $$= \int{\lim_{n \to \infty}h(\zeta_n,\omega)}d\mu(\omega) \text { (By DCT)}$$

Now if $\lim_{n \to \infty}h(\zeta_n,\omega) =h(t,\omega) , \forall \omega \in \Omega$, we are done!

But how to conclude that? Like, I don't have continuity of $h$. Does it follow directly or I'm missing something?

Thanks in Advance for help!

Answer 1

The final step is to use the given dominating function. You have for every $n$,

$$\left| \frac{f(t_n,\omega))- f(t,\omega)}{t_n - t} \right| = |h(\zeta_n,\omega)| \le |g| \in L^1$$ Therefore DCT tells us that all relevant limits exist and $$ \lim_{n \to \infty}{\frac{\int f(t_n,\omega)d\mu(\omega)-\int f(t,\omega)d\mu(\omega)}{t_n - t}}\\ = \lim_{n\to \infty} \int \frac{f(t_n,\omega))- f(t,\omega)}{t_n - t} d\mu(\omega)\\ = \int\lim_{n\to \infty} \frac{f(t_n,\omega))- f(t,\omega)}{t_n - t} d\mu(\omega) \\ = \int h(t,\omega) d\mu(\omega)$$

note that no continuity in $h$ was used.

Answer 2

$\frac {f(t_n,\omega)-f(t,\omega)} {t_n-t}\to \frac {\partial f(t,\omega)} {dt}$ and $|\frac {f(t_n,\omega)-f(t,\omega)} {t_n-t}| \leq g(\omega)$ by MVT. So DCT applies.