# $\lim\limits_{t\rightarrow\infty}\int\limits_{E}\phi(x+t)dx=0$

Suppose $$E\subset\mathbb{R}$$ has finite Lebesgue measure and $$\varphi\in L^1(\mathbb{R})$$. Show that $$\lim\limits_{t\rightarrow\infty}\int\limits_{E}\varphi(x+t)dx=0$$.

I guess first I have to show $$\lim\limits_{t\rightarrow\infty}\varphi(x+t)=0$$ for $$x\in\mathbb{R}$$ and use the Lebesgue Dominated Convergence Theorem but I am not sure.

• Maybe change of variables $x+t=\tau$ and use the integral convergence at infinity. – thing Dec 25 '19 at 1:36
• Can you prove it for an interval first? – copper.hat Dec 25 '19 at 1:43

Given $$\epsilon>0$$, choose a $$\psi\in C_{00}$$ such that $$\|\varphi-\psi\|_{L^{1}(\mathbb{R})}<\epsilon$$, then \begin{align*} \int_{E}|\varphi(x+t)|dx&\leq\int_{E}|\varphi(x+t)-\psi(x+t)|dx+\int_{E}|\psi(x+t)|dx\\ &\leq\|\varphi-\psi\|_{L^{1}(\mathbb{R})}+\int_{E}|\psi(x+t)|dx. \end{align*} Now $$|\psi(x+t)|\chi_{E}\leq\|\psi\|_{L^{\infty}}\chi_{E}\in L^{1}(\mathbb{R})$$, apply Lebesgue Dominated Convergence Theorem to make the term $$\displaystyle\int_{E}|\psi(x+t)|dx$$ arbitrarily small as $$t\rightarrow\infty$$.

• Does $C_{00}$ denote the set of continuous functions that vanish outside a bouned set? – user682705 Dec 25 '19 at 2:21
• Yes, or you can say compact supported continuous functions. – user284331 Dec 25 '19 at 2:22

Note that $$\int \phi(x+t)1_E(x) d x = \int \phi(x)1_E(x-t) d x$$

If $$E$$ is a bounded set then $$\lim_{t \to \infty} 1_E(x-t) = 0$$ and so $$\int \phi(x)1_E(x-t) d x = 0$$.

Since $$\phi$$ is integrable, for any $$\epsilon>0$$, there is some $$\delta>0$$ such that if $$mA < \delta$$ then $$\int_A|\phi| < \epsilon$$ (absolute continuity).

Suppose $$\epsilon>0$$. Choose the $$\delta>0$$ as above.

If $$E$$ has bounded measure, then we can write $$E = B \cup C$$, where $$B$$ is bounded and $$mC < \delta$$.

Then $$\int \phi(x+t)1_E(x) d x = \int \phi(x+t)1_C(x) d x + \int \phi(x)1_B(x-t) d x< \epsilon + \int \phi(x)1_B(x-t) d x$$ and there is some $$T$$ such that if $$t>T$$ we have $$\int \phi(x)1_B(x-t) d x < \epsilon$$. Hence $$\int \phi(x+t)1_E(x) d x \to 0$$.