# Convergence of Lebesgue Integral

This is part of a problem. The first part was to show that if is integrable on $(-\infty,\infty)$, then $$\int f(x)~\text{d}x=\int f(x+t)~\text{d}x$$ which I was able to show.

This is the second part, which I'm having a hard time starting.
Let $g$ be a bounded measurable function. I would like to show that $$\lim_{t\rightarrow 0}\int_{-\infty}^{\infty} \;\left|\;g(x)\left[f(x)-f(x+t)\right]\right.|=0.$$

Any form of help will be appreciated.

Added: Please see if this is ok.

Let $M$ be such that $|g(x)|\leq M$. Sinc $f$ is integrable, there is a continuous function such that $\int|f-h|\lt \frac{\epsilon}{4M}.$ Also $h$ is uniformly continuous, so $\exists ~\delta \gt 0$ such that $|h(x)-h(x+t)|\lt \frac{\epsilon}{2M}.$ Consider, \begin{align*} \int|g(x)[f(x)-f(x+t)|&\leq \int|g(x)[h(x)-h(x+t)]| + \int|g(x)[f(x)-h(x)-f(x+t)+h(x+t)]|\\ &\leq \frac{\epsilon M}{2M} + M\int |f(x)-h(x)|+M\int|f(x+t)-h(x+t)| \\ & \leq \frac{\epsilon}{2} + 2M\int|f(x)-h(x)|\\ & \leq \frac{\epsilon}{2} + \frac{2M\epsilon}{4M}\\ & \lt \epsilon. \end{align*}

-
Hint: Lebesgue Dominated Convergence. –  Jonas Teuwen Oct 25 '11 at 19:22
To expand on @Jonas's hint: prove it first assuming that $f$ is continuous of compact support, using that $\int |g(x)(f(x+t)-f(x))| \,dx \leq \|g\|_{\infty}\cdot2\|f\|_1$. Then use that for every $L^1$-function there is a continuous function $f_n$ such that $\|f-f_n\|_1 \leq \frac{1}{n}$. –  t.b. Oct 25 '11 at 19:30
Why is $h$ uniformly continuous? –  Jonas Teuwen Oct 25 '11 at 20:19
Furthermore, you should be much clearer what quantifiers you use and where and their order. Written like this always somehow suggests to me that the people that do this don't really know which quantifiers they should put there and their order. –  Jonas Teuwen Oct 25 '11 at 20:25

## 1 Answer

First assume that $f$ is continuous and has compact support.

Now pass to a sequence. Let $t_n \to 0$. Then we want to prove that

$$\lim_{n \to \infty} \int |g(x)[f(x) - f(x + t_n)]| \, \textrm{d}x = 0.$$

In this case the result follows from the Lebesgue Dominated Convergence Theorem (LDCT) by using the continuity of $f$. Now an approximation argument is in order, this is very similar as to my answer in this post (actually you can mimic the complete argument since $g$ is bounded anyway).

-
Thanks. I've added to my post. Could you please see if it's ok. –  Nana Oct 25 '11 at 19:59