Translation is continuous in $L^1$ for finite Borel measures

Question

I have the following question which I couldn't find a way to solve, let $\mu$ be a positive, finite Borel measure on $\mathbb{R}$. Assume that $x\longmapsto \mu(-\infty, x]$ is continuous. Is that true that if $f\in L^1(\mu)$ then $$ \lim_{t\longrightarrow 0}\int_{\mathbb{R}} |f(x+t) - f(x)|\;d\mu(x) = 0.$$

Answer 1

It's not true.

Let $$\phi(x)=x^{-1/2}\chi_{(0,1)}(x)$$and $$\phi_n(x)=\phi(x-1/n).$$

Let $$f(x)=|x|^{-1/2}$$and $$I_n=\int f(x)\phi_n(x)\,dx.$$

Since $I_n<\infty$ (for $n=1,2,\dots$) there exist $a_n>0$ with $$\sum a_n<\infty$$and $$\sum_1^\infty a_nI_n<\infty.$$Define $$\psi=\sum_1^\infty a_n\phi_n$$and $$d\mu=\psi dx$$(that is, $\mu(E)=\int_E\psi(x)\,dx$.) Then $\mu$ is a finite Borel measure (since $\psi\in L^1(\mathbb R)$), and dominated convergence shows that $\mu$ satisfies the conintuity hypothesis. And $\sum a_nI_n<\infty$ shows that $f\in L^1(\mu)$, although $$\int f(x-1/n)\,d\mu(x)=\infty$$for every $n$ (so that $\int|f(x)-f(x-1/n)|\,d\mu(x)=\infty$, since $f\in L^1(\mu)$.).

Answer 2

Another approach, based on a recent question ($\mu(A+x)$ is continuous implies $\mu \ll m$) in S.E: If the statement were true then, in particular, $t\mapsto\mu(B+t)$ would be continuous for each Borel set $B$, which is only true if $\mu$ is absolutely continuous with respect to Lebesgue measure..