Regarding the Theorem 3 from here (or pdf ver.).
Let $X$ be an open subset of $\mathbb{R}$, and $\Omega$ be a measure space. Suppose that a function $f\colon X\times\Omega\to \mathbb{R}$ satisfies the following conditions:
- $f(x,\omega)$ is a measurable function of $x$ and $\omega$ jointly, and is integrable over $\omega$, for almost all $x\in X$ held fixed.
- For almost all $\omega\in\Omega$, $f(x,\omega)$ is an absolutely continuous function of $x$. (This guarantees that $\frac{\partial f}{\partial x}(x,\omega)$ exists almost everywhere.)
- $\frac{\partial f}{\partial x}(x,\omega)$ is "locally integrable" --- that is, for all compact intervals $[a,b]\subset X$:
$$ \int_a^b\!\!\int_{\Omega}\left| \frac{\partial f}{\partial x}(x,\omega) \right| d\omega dx<\infty. $$ Then, $\int_\Omega f(x,\omega)d\omega$ is an absolutely continuous function of $x$, and for almost every $x\in X$, its derivatives exists and is given by $$ \frac{d}{dx}\int_{\Omega}f(x,\omega)d\omega = \int_{\Omega}\frac{d}{dx}f(x,\omega)d\omega. $$
Question 1: I do not see how to show $\int_\Omega f(x,\omega)d\omega$ is absolutely continuous in $x$.
Question 2: I also wonder where I can find this result preferably in a book or a paper etc.
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Here is my thought: Notations basically follow the plametmath note by Steven Cheng (the link above).
Let $(\Omega,\mu)$ be the measure space and $(a,b):=X\subset\mathbb{R}$.
For $\mu$-almost all $\omega\in \Omega$, we assumed that $f(\cdot,\omega)$ is an absolutely continuous function in the first variable.
Hence for a.e.-$x$ we have \begin{align} f(x,\omega)&=f(a,\omega)+\int_{a}^x\frac{\partial}{\partial y}f(y,\omega)\,dy \end{align}
Letting $G(x):=\int_{a}^x\int_{\Omega}\frac{\partial}{\partial y}f(y,\omega)\, d\mu(\omega ){d}y$ and $g(x):=\int_{\Omega}\frac{\partial}{\partial x}f(x,\omega)\, d\mu(\omega )$ by virtue of Lebesgue differentiation theorem for the indefinite integral, for a.e. $x$ we have \begin{align} \int_{\Omega}\frac{\partial}{\partial x}f(x,\omega)d\mu(\omega ) &=g(x)= \frac{\partial}{\partial x}G(x) = \frac{\partial}{\partial x}\int_{\Omega}\int_{a}^x \frac{\partial}{\partial y}f(y,\omega)\, dyd\mu(\omega ). \end{align} where the last equality is ok from the assumption $ \int_{\Omega}\int_{a}^x\left|\frac{\partial}{\partial y} f(y,\omega)\right| \,dyd\mu(\omega ) <\infty $. Adding $0=\frac{d}{d t}\int_{\Omega}f(a,\omega)d\mu(\omega )$ to the above yields \begin{align} \int_{\Omega}\frac{\partial}{\partial x}f(x,\omega)d\mu(\omega ) &=\frac{\partial}{\partial x}\left(\int_{\Omega}\left\{ \int_{a}^x \frac{\partial}{\partial y}f(y,\omega)\, dy+f(a,\omega)\right\}d\mu(\omega ) \right)\\ &=\frac{\partial}{\partial x}\int_{\Omega} f(x,\omega)d\mu(\omega ). \end{align} I think the RHS exists for almost every $x$ because LHS does, which is the second claim of Theorem 3. I am not sure if I can immediately say $\int_\Omega f(x,\omega)d\omega$ is AC in $x$, which is the first claim of 3., Theorem 3.