Fall 2001 Problem 1

Exercise 12.2 (Integrals are continuous in mean)   Show that if $f\in L^1(\mathbb{R})$ then $\int_{-\infty}^\infty \vert f(x+h)-f(x)\vert dx \to 0$ as $h\to 0$.

Proof. Let $h_n\to 0$. Set $f_n(x)=f(x+h_n)$. Select $F\subseteq \mathbb{R}$ such that $\mu(F)<\infty$ and

$$\int_{\mathbb{R}\setminus F} 2\vert f\vert < \epsilon.$$ (12.7)

From this arises the estimate

$$\int_{\mathbb{R}} \vert f_n-f\vert = \int_{\mathbb{R}\setminus F}\vert f_n-f\vert + \int_F\vert f_n-f\vert$$ (12.8)

$$\leq \int_{\mathbb{R}\setminus F}2\vert f\vert + \int_F\vert f_n-f\vert$$ (12.9)

$$\leq \epsilon + \int_F\vert f_n-f\vert.$$ (12.10)

Now select $\delta>0$ so that if $\mu(B)<\delta$, then

$$\int_B 2\vert f\vert < \epsilon.$$ (12.11)

By Egorov's theorem there exists $E\subseteq F$ such that $\mu(F\setminus E)<\delta$ and $f_n\to f$ uniformly on $E$. Then

$$\int_F\vert f_n - f\vert = \int_{F\setminus E} \vert f_n-f\vert + \int_E \vert f_n-f\vert$$ (12.12)

$$\leq \int_{F\setminus E} 2\vert f\vert + \mu(E)\Vert f_n-f\Vert _E$$ (12.13)

$$\leq \epsilon + \mu(E)\Vert f_n-f\Vert _E.$$ (12.14)

Sending $n\to\infty$ and $\epsilon\to 0$ shows

$$\lim_{n\to\infty} \int_{\mathbb{R}}\vert f_n - f\vert = 0.$$ (12.15)

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