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..

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. 
