# $\varphi(x)=\int_{[\xi_0,\xi]}f(x+t)d\mu_t$ absolutely continuous and summable on $\mathbb{R}$

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!!!

• $\mu$ is the Lebesgue measure on $\mathbb{R}$? – Daniel Fischer Nov 19 '14 at 13:50
• Thank you for the comment! Yes, it is. – Self-teaching worker Nov 19 '14 at 13:51

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.

• I heartily thank you! As always, your asnwers are deep and precious. Only one thing isn't clear to me in the proof of $(x,y)\mapsto f(x\pm t)$: why all measurable sets of $\mathbb{R}\times[\xi_0,\xi]$ are in the form $G_\delta \cup N$ with $N$ of null measure? If you consider it more proper, I could open a new thread... Thank you again!!!!! – Self-teaching worker Nov 19 '14 at 20:09
• They are $G_\delta \setminus N$ (or, equivalently, $F_\sigma \cup N$). The Lebesgue measure is regular, so for every measurable set $M$ we have $\lambda(M) = \inf \{ \lambda(U) : U\text{ open and } M\subset U\}$. Take a sequence $(U_n)$ of open sets with $M\subset U_n$ and $\lambda(U_n) < \lambda(M) + 2^{-n}$. Then $A =\bigcap U_n$ is a $G_\delta$ with $M\subset A$, and $\lambda(A) \leqslant \lambda(M) + 2^{-n}$ for all $n$, hence $\lambda(A) = \lambda(M)$, so $A\setminus M$ is a null set [and $M = A\setminus (A\setminus M)$]. – Daniel Fischer Nov 19 '14 at 20:23
• Ouch, I spent the afternoon trying to use the sets $U_n$ such that $\lambda(A\triangle U_n)< \varepsilon_n$ defining the Lebesgue measurable set. I heartily thank you again!!!!! – Self-teaching worker Nov 19 '14 at 21:03
• If I'm not bothering you too much: I see that the same holds for any topological subspace of $\mathbb{R}^n$ endowed with Lebesgue "ordinary" measure $\lambda$ (constructed by using the volume of $n$-parallelepipeds): am I right? Thank you so much again!!! – Self-teaching worker Nov 19 '14 at 21:51
• What "the same"? The $G_\delta \setminus N$? Yes, that holds for all Lebesgue-measurable subspaces of $\mathbb{R}^n$. [In non-Lebesgue-measurable subspaces, we have the question what we call measurable subsets, if Lebesgue-measurable subsets of $\mathbb{R}^n$ which happen to be contained in our subspace, the same holds.] – Daniel Fischer Nov 19 '14 at 22:08