Let $f\in L_1(-\infty,\infty)$ be a Lebesgue-summable function on $\mathbb{R}$. I read that the function$$\varphi(x)=\int_{[\xi_0,\xi]}f(x+t)d\mu_t$$is absolutely continuous on any real closed interval. I knew that if $g\in L_1[a,b]$ then $\int_{[a,x]}g(t)d\mu_t$ is absolutely continuous as a function of $x$, but I cannot prove the absolute continuity of $\varphi$ to myself.
Moreover, it is said that it is Lebesgue-integrable on $\mathbb{R}$.
How can we prove those two statements?
As to the Lebesgue-integrability, I see that it would follow, because of Fubini's theorem, from the measurability of $\int_{\mathbb{R}}|f(x+t)|d\mu_x$ as a function of $t$, but I have no idea of how to prove it, if it is true.
I $\infty$-ly thank you all!!!
Regarding the integrability of $\varphi$, rather than showing the measurability of
$$t \mapsto \int_\mathbb{R} \lvert f(x+t)\rvert\,d\mu_x,$$
use the mesurability of
$$(x,t) \mapsto f(x+t)$$
which is for example shown here. Then we can compute
\begin{align} \int_\mathbb{R} \lvert \varphi(x)\rvert\,d\mu_x &= \int_\mathbb{R} \left\lvert \int_{[\xi_0,\xi]} f(x+t)\,d\mu_t\right\rvert\,d\mu_x\\ &\leqslant \int_\mathbb{R} \int_{[\xi_0,\xi]} \lvert f(x+t)\rvert\,d\mu_t\,d\mu_x\\ &= \int_{[\xi_0,\xi]}\int_{\mathbb{R}} \lvert f(x+t)\rvert\,d\mu_x\,d\mu_t\\ &= \int_{[\xi_0,\xi]} \lVert f\rVert_{L^1(\mathbb{R})}\,d\mu_t\\ &= (\xi-\xi_0)\lVert f\rVert_{L^1(\mathbb{R})}\\ &< +\infty \end{align}
to see that $\varphi \in L^1(\mathbb{R})$.
For the absolute continuity of $\varphi$, write (for $x < y$ to avoid dealing with signs)
\begin{align} \varphi(y) - \varphi(x) &= \int_{[\xi_0,\xi]} f(y+t)\,d\mu_t - \int_{[\xi_0,\xi]} f(x+t)\,d\mu_t\\ &= \int_{y+\xi_0}^{y+\xi} f(u)\,d\mu_u - \int_{x+\xi_0}^{x+\xi} f(u)\,d\mu_u\\ &= \int_{x+\xi}^{y+\xi} f(u)\,d\mu_u - \int_{x+\xi_0}^{y+\xi_0} f(u)\,d\mu_u\\ &= \int_{x}^y f(\xi+t)\,d\mu_t - \int_x^y f(\xi_0+t)\,d\mu_t\\ &= \int_x^y f(\xi+t) - f(\xi_0+t)\,d\mu_t. \end{align}
Thus we see that $\varphi$ is the integral of the integrable function
$$t \mapsto f(\xi + t) - f(\xi_0+t),$$
and hence $\varphi$ is absolutely continuous.