Proof of this "Differenting Under the Integral" Generalization

Question

I refer to Theorem 3 at: http://planetmath.org/differentiationundertheintegralsign

I have two questions:

Q1) Why is Theorem 3 considered a generalization of Theorem 2? It seems to me that $f(x,\omega)$ is an absolutely continuous function of $x$ for almost all $w\in\Omega$ is a pretty strong condition, since for Theorem 2 we don't even require for $f$ to be continuous, just Lebesgue-integrable.

Q2) Can anyone provide a proof (or reference to a proof) for Theorem 3? I have been searching but can't find one. (For those who are interested, Theorem 2's proof can be found in e.g. Folland pg 56.)

Thanks for any help.

Answer 1

Only for Q1:

1. Theorem 2 requires $f$ to be integrable for each $x$. Theorem 3 requires $f$ to be integrable for almost all $x$. Both require $f$ to be measureable although this is not stated explicitly in theorem 2.
2. Theorem 2 requires $f$ to be differentiable everywhere with respect to $x$ for almost all $\omega$, while theorem 3 only requires f to be absolutely continuous for almost all $\omega$. It's possible to show that abolute continuous implies differentiability almost everywhere.
3. If the property of theorem 2 holds, i.e. that the derivative with respect to $x$ is bounded by a integrable function independent of $x$ it is easy to see that the integral in theorem 2 is finite (i's bounded by $(b - a) c$, where $c$ is the integral of $\Theta$).