Absolute continuity, shifted set

Question

Let $\mu$ be a finite signed measure on the Borel sets of $\mathbb{R}$, and suppose that $\mu \ll m$ where $m$ is Lebesgue measure. Prove that the function $t\mapsto \mu\{t+x:x\in A\}$ is continuous in $t$ for every Borel set $A$.

An Attempt: Using the sequence definition of continuity and writing $\mu(A)=\int f(x)dm$ as an integral and using the fact that Lebesgue measure is shift invariant I get to where I want to switch limits with integration as well as show that $f(x+t_n)\to f(x+t)$ where $t_n\to t$. This last bit seems especially suspicious since we don't know much about $f$.

Answer 1

The problem is that we may not have $f(x+t_n)\to f(x+t)$ when $t_n\to t$. However, what is true (and sufficient to conclude) is that if $t_n\to t$, then $\int_{\mathbb R}\left|f(x+t_n)-f(x+t)\right|\mathrm d\lambda(x)\to 0$ (to see this, approximate in $\mathbb L^1(\lambda)$ the function $f$ by a continuous function with compact support).

Answer 2

Hint. If $\mu\ll m$, then there exists a function $g\in L^1(\mathbb R)$, such that $d\mu=g\,dm$. Clearly, $$ f(t)=\mu(t+x: x\in A)=\int_Af(x-t)\,dm(x). $$ Then $$ \lvert f(s)-f(t)\rvert\le \int_A\lvert\, g(x-s)- g(x-t)\rvert\,dm(x) \le \int_{\mathbb R}\lvert\, g(x-s)- g(x-t)\rvert\,dm(x) =\|T_{t-s}g- g\|_{L^1} $$ where $(T_ag)(x)=g(x-a)$. It is true that $\lim_{h\to 0}\|T_{h}g- g\|_{L^1}=0$.