$f \in L^1(\mathbb{R})$ and $\lim_{|x| \to +\infty} f(x) = 0$ implies $\lim_{|x| \to +\infty} \int_{\mathbb{R}} f(x-y) d\mu(y) = 0$

Question

Let $f \in L^1(\mathbb{R})$ be a Lebesgue integrable function on $\mathbb{R}$. Assume that $$ \lim_{|x| \to +\infty} f(x) = 0. $$ Let $\mu$ be a non-singular (i.e., if $\mu = \mu_a + \mu_s$, where $\mu_a$ is absolutely continuous part, and $\mu_s$ is singular part, then $\mu_a \neq 0$) Borel probability measure on $\mathbb{R}$.

Is it true that $$ \lim_{|x| \to +\infty} \int_{\mathbb{R}} f(x-y) d\mu(y) = 0? $$

Answer 1

I think this is a counterexample.

Consider the absolutely continuous probability measure $$ \mu(E) = \int_E g(y) \ d\lambda(y), $$ with Radon-Nikodym derivative $$ g(y)=\sum_{n=1}^\infty \frac{1}{2^{n+1}\sqrt{y-n}}\mathbb 1_{(n, n+1)}.$$

[Here, $\lambda$ denotes the Lebesgue measure.]

Consider the Lebesgue-integrable function

$$ f(x)=\frac{1}{\sqrt{-x}}\mathbb 1_{(-1, 0)},$$

which clearly tends to zero as $|x| \to \infty$.

Now observe that for any $n \in \mathbb N$,

$$ f(n-y)=\frac{1}{\sqrt{y-n}} \mathbb 1_{(n, n+1)}$$

so

$$ \int_{\mathbb R}f(n-y) \ d\mu(y)=\int_{\mathbb R}f(n-y)g(y) \ d\lambda (y)= \int_{(n, n+1)}\frac{1}{2^{n+1}(y-n)}=+\infty.$$

Since this holds for arbitrary $n \in \mathbb N$, it cannot be the case that $\int_{|x| \to \infty} f(x-y) \ d\mu(y) = 0$.