Absolute continuity under the integral Let $f:[0,T]\times \Omega \to \mathbb{R}$ where $\Omega$ is some bounded compact space. Let $t \mapsto f(t,x)$ be absolutely continuous.  Is then
$$t \mapsto \int_\Omega f(t,x)\;dx$$
also absolutely continuous provided the integral exists? 
I think we need $|f(t,x)| \leq g(x)$ for all $t$ where $g$ is integrable. At least. But not sure what else we need!!

When I try to apply the definition of absolutely continuous to the integral, I tried to use the absolute continuity of $f(t)$. But the $\delta$ for $t \mapsto f(t,x)$ depends on $x$, so $\delta=\delta_x$ and that causes problems..
 A: One possible assumption is that we have
$$
\int_\Omega \int_0^T \left|\frac{\partial f}{\partial t} (s,x)\right| \, ds\, dx<\infty. \qquad (\dagger)
$$
For
$$
F(t) := \int_\Omega f(t,x) \, dx \,\,\, \text{ and }\,\,\, G(s) := \int_\Omega \frac{\partial f}{\partial t} (s,x) \, dx,
$$
this will imply
\begin{eqnarray*}
F\left(t\right)-F\left(s\right) & = & \int_{\Omega}f\left(t,x\right)-f\left(s,x\right)\, dx\\
 & = & \int_{\Omega}\int_{s}^{t}\frac{\partial f}{\partial t}\left(r,x\right)\, dr\, dx\\
 & = & \int_{s}^{t}\int_{\Omega}\frac{\partial f}{\partial t}\left(r,x\right)\, dx\, dr\\
 & = & \int_{s}^{t}G\left(r\right)\, dr,
\end{eqnarray*}
where the application of Fubini's theorem is justified by $(\dagger)$.
The assumption $(\dagger)$ also yields
$$
\int_0^T |G(r)| \,dr \leq \int_\Omega \int_0^T \left|\frac{\partial f}{\partial t} (s,x)\right| \, ds\, dx<\infty.
$$
By Lebesgue's differentiation theorem and the (alternative) characterization of absolute continuity ($f$ is absolutely continuous if and only if $f$ is almost everywhere differentiable with $f' \in L^1$ and $f(x) - f(y) = \int_x^y f'(t) \,dt$ for all $x,y$), we see that $F$ is absolutely continuous.
A: Let $\epsilon > 0$ and $(t_i, t_{i+1})_{i=1}^N$ be any non-overlapping set of intervals such that $\sum_j|f(t_j,x) - f(t_{j-1},x)| \leq \epsilon$ whenever $\sum_j |t_{j+1} - t_j| < \delta_x$, where the $\delta_x$ comes from the absolute continuity of $f$.
$\sum_j |F(t_j) - F(t_{j-1})|\leq\sum_j \int_{\Omega} |f(t_j,x) - f(t_{j-1},x)| dx = \int_{\Omega} \sum_j |f(t_j,x) - f(t_{j-1},x)| dx $  Now, $\Omega = \cup_{j=1}^n B(x_j,\delta_{x_j} )$ by compactness.  So for $\sum_j |t_{j+1} - t_j| < \min_j \delta_{x_j}$, we have that $\int_{\Omega} \sum_j |f(t_j,x) - f(t_{j-1},x)| dx \leq \frac{\epsilon}{m(\Omega)} m(\Omega) < \epsilon$ as needed. 
