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 $$ \lim_{t\longrightarrow 0}\int_{\mathbb{R}} |f(x+t) - f(x)|\;d\mu(x) = 0.$$

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    $\begingroup$ This measure must be inner and outer regular (see, for instance, theorems 1.16 and 1.18 of Folland's Real Analysis book). Therefore, the space of continuous and compactly supported functions will be dense in $L^{1}(\mu)$, and you can apply the following argument. $\endgroup$ – Vinícius Novelli Nov 19 '17 at 7:27
  • $\begingroup$ @ViníciusNovelli I don't see how that argument applies. It uses the fact that $||f-\phi||_1<\epsilon/3$ implies $||\tau_tf-\tau_t\phi||_1<\epsilon/3$ (where $\tau_t$ denotes translation). $\endgroup$ – David C. Ullrich Nov 19 '17 at 16:43
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    $\begingroup$ I don't think this is true. The counterexample I have in mind is a little intricate, but it seems to me you can jigger up an example where $f\in L^1(\mu)$ but there exist $t_n\to0$ such that $\int|f(x-t_n)|\,d\mu(x)=\infty$. $\endgroup$ – David C. Ullrich Nov 19 '17 at 16:45
  • $\begingroup$ I had the same thought as Novelli, but my teacher told me the the problem is indeed wrong, can anyone help me with a counter example? $\endgroup$ – Sean Nov 19 '17 at 16:54

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 $f\in L^1(\mu)$.).

  • $\begingroup$ What is the intuition behind your example? Can you help me with that? $\endgroup$ – Sean Nov 20 '17 at 23:15
  • $\begingroup$ Well first II noticed that the argument suggested in a comment was wrong because the $L^1(\mu)$ norm was not translation-invariant. Then I said hmm, in fact a translate of an integrable function need not even be integrable. So I tried to construct an example with non-integrable translates by $t_n\to0$... $\endgroup$ – David C. Ullrich Nov 20 '17 at 23:29

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..

  • $\begingroup$ Of course now we have to show there exists a continous measure that's not absolutely continuous, which is not entirely obvious at first. A non-constant non-decreasing continuous $F$ with $F'=0$ almost everywhere? Clearly impossible [not]... $\endgroup$ – David C. Ullrich Nov 19 '17 at 17:38
  • $\begingroup$ We do indeed. Historical query: Who first referred to Cantor's function as the "Devil's staircase"? $\endgroup$ – John Dawkins Nov 19 '17 at 17:54

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