$\int |f(x+t) - f(x)| \,dx \to 0$ for integrable function $f$

Question

If $f$ is integrable, then prove that $$\lim_{t \to 0} \int \left| f(x+t) - f(x) \right| \,dx = 0~.$$

Source: Probability and Measure (Billingsley), Problem 17.4

The given hint says to use the result that if $f$ is integrable, then for any $\varepsilon > 0$, there exists a step function $g = \sum_{i = 1}^k x_i \mathbb{I}_{A_i}$ where $A_i$ are bounded intervals, such that $\int |f - g| \,dx < \varepsilon$.

But I am a little confused whether I'm going in the right direction. Any help will be appreciated. Thanks!

You are going in the right direction indeed. But I recommend to choose a function $g \in C_0^0(\mathbb{R})$ instead of a step function - but I do not know whether you have already learned that result. — Hyperbolic PDE friend, Mar 08 '23 at 13:35
@Meowdog You mean that I should take $g$ to be a continuous function? Yes, I have learnt that result. But couldn't proceed. — JRC, Mar 08 '23 at 13:45
I mean a continuous function with compact support. So for the first integral, you can do a transfrom $y=x+t$. For the second one use continuity+compact support+dominated convergence. For the last one simply use the approximation. So all of them should get small — Hyperbolic PDE friend, Mar 08 '23 at 13:47
Thanks for your reply. It will be really appreciable if you can write a brief answer. Thank you. — JRC, Mar 08 '23 at 13:53
Does this answer your question? Continuity of $L^1$ functions with respect to translation — coudy, Mar 08 '23 at 14:11

Hyperbolic PDE friend · Answer 1 · 2023-03-09T08:57:10.867

I assume that you mean $$ \int_{\mathbb{R}^n} \lvert f(x+t) - f(x) \rvert~\mathrm{d}x, $$ i.e. an integral over the entire space. This can always be assumed, if we continue $f$ by $0$. Let $\varepsilon > 0$ be arbitrary and choose $g \in C_0^0(\mathbb{R}^n)$ such that $$ \int_{\mathbb{R}^n} \lvert f(x) - g(x) \rvert~\mathrm{d}x < \frac{\varepsilon}{3}. $$ As you stated, $$ \int_{\mathbb{R}^n} \lvert f(x+t) - f(x) \rvert~\mathrm{d}x \leq \int_{\mathbb{R}^n} \lvert f(x+t) - g(x+t) \rvert~\mathrm{d}x + \int_{\mathbb{R}^n} \lvert g(x+t)-g(x) \rvert~\mathrm{d}t + \int_{\mathbb{R}^n} \lvert f(x)-g(x) \rvert~\mathrm{d}t $$ The last integral, by definition of $h$ is smaller than $\frac{\varepsilon}{3}$. For the first integral introduce a transformation $y := x+t$. Then $$ \int_{\mathbb{R}^n} \lvert f(x+t) - g(x+t) \rvert~\mathrm{d}x = \int_{\mathbb{R}^n} \lvert f(y) - g(y) \rvert~\mathrm{d}y < \frac{\varepsilon}{3}. $$ So we still have to get $$ \int_{\mathbb{R}^n} \lvert g(x+t)-g(x) \rvert~\mathrm{d}t = \int_{\mathrm{supp}\left(g(\cdot +t)\right) \cup\mathrm{supp}\left(g(\cdot)\right)} \lvert g(x+t) - g(x) \rvert~\mathrm{d}x $$ small. Note that by continuity, $\lim_{t \rightarrow 0} \lvert g(x+t) - g(x) \rvert = 0$ for all $x \in \mathbb{R}^n$. Now we want to prove that $$ \mathrm{supp}\left(g(\cdot +t)\right) \cup\mathrm{supp}\left(g(\cdot)\right) $$ is bounded for small $\lvert t \rvert$. For this let $\lvert t \rvert \leq \delta$ for arbitrarily small $\delta > 0$ and choose $y \in \mathrm{supp}(g(\cdot + t ) )$. It follows $y+t \in \mathrm{supp}(g(\cdot))$. Since $\mathrm{supp}(g(\cdot))$ is compact, there is some $M>0$ that does not depend on $y \in \mathrm{supp}(g(\cdot))$ and $t$ such that $\lvert y +t \rvert < M$. Now: $$ \lvert y \rvert - \lvert t \rvert \leq \lvert y + t \rvert \leq M $$ And hence $\lvert y \rvert \leq M + \lvert t \rvert \leq M + \delta$. So we also get boundedness of $\mathrm{supp}(g(\cdot + t ))$ for small $t$ by some $R>0$. It follows: $$ \int_{\mathrm{supp}\left(g(\cdot +t)\right) \cup\mathrm{supp}\left(g(\cdot)\right)} \lvert g(x+t) - g(x) \rvert~\mathrm{d}x \leq 2R \sup_{x \in \mathbb{R}^n} \lvert g(x) \rvert < \infty $$ So by dominated convergence, for small $t>0$ we can also achieve $$ \int_{\mathbb{R}^n} \lvert g(x+t)-g(x) \rvert~\mathrm{d}t < \frac{\varepsilon}{3}. $$ So in total, for $\lvert t \rvert$ sufficiently small $$ \int_{\mathbb{R}^n} \lvert f(x+t) - f(x) \rvert~\mathrm{d}x \leq \int_{\mathbb{R}^n} \lvert f(x+t) - g(x+t) \rvert~\mathrm{d}x + \int_{\mathbb{R}^n} \lvert g(x+t)-g(x) \rvert~\mathrm{d}t + \int_{\mathbb{R}^n} \lvert f(x)-g(x) \rvert~\mathrm{d}t \leq 3 \frac{\varepsilon}{3} = \varepsilon. $$

while DCT is fine, I think it is much more transparent to use uniform continuity (afterall that’s why we’re working with continuous compactly supported functions). — peek-a-boo, Mar 09 '23 at 09:14

$\int |f(x+t) - f(x)| \,dx \to 0$ for integrable function $f$

1 Answers1