2.2

Since continuous functions with compact support are dense in $L^1(\mathbb{R}^d)$, we know that $\forall \epsilon > 0$ there exists a $g$, continuous with compact support such that $|| f - g ||_{L^1} < \frac{\epsilon}{3}$. It follows that since: $$f(\delta x) - f(x) = f(\delta x) - g(\delta x) + g(\delta x) -g(x) + g(x) -f(x)$$ We have, by the triangle inequality: $$||f(\delta x) - f(x) ||_{L^1} \leq ||g(\delta x) - g(x) ||_{L^1} + ||f(\delta x) - g(\delta x) ||_{L^1} + ||f(x) - g(x) ||_{L^1}$$ ...and thus, since $|| f - g ||_{L^1} < \frac{\epsilon}{3}$, we have: $$||f(\delta x) - f(x) ||_{L^1} \leq ||g(\delta x) - g(x) ||_{L^1} + ||f(\delta x) - g(\delta x) ||_{L^1} + \frac{\epsilon}{3}$$ Glance ahead to exercise 2.16 and note the dilation property of $L^1$ functions: $$\int_{\mathbb{R}^d} h(x) \hspace{0.1cm}dx = |\delta|^d \int_{\mathbb{R}^d} h(\delta x) \hspace{0.1cm}dx$$ It's clear from this property that: $$||f(\delta x) - g(\delta x) ||_{L^1} = \frac{1}{|\delta|^d} ||f - g||_{L^1} \leq \frac{\epsilon}{3|\delta|^d} \leq \frac{\epsilon}{3}$$ Now, given that $g(x)$ is continuous, with compact support, we have that $g(x)$ and $g(\delta x)$ are uniformly continuous. Thus, there exists a bound $M$ for $g$. Next, see that since the sets $E$ and $\delta E$ are compact, the sets $E \Delta \delta E$, and $E \cap \delta E$ must be compact as well. It follows directly that: $$\int_{E \cup \delta E} |g(x) - g(\delta x)| \hspace{0.1cm}dx = \ldots$$ $$\ldots = \int_{E \cap \delta E} |g(x) - g(\delta x)| \hspace{0.1cm}dx + \int_{E \Delta \delta E} |g(x) - g(\delta x)| \hspace{0.1cm}dx \leq \ldots$$ $$\ldots \leq \int_{E \cap \delta E} |g(x) - g(\delta x)| \hspace{0.1cm}dx + 2M m(E \Delta \delta E)$$

Now observe that for every $\delta \neq 1 \hspace{0.25cm} \exists \alpha$ such that $\delta x = x + \frac{x}{\alpha}$. Since $E \cap \delta E$ is compact, we can select the diameter, $r = \max_{x,y \in E \cap \delta E} |d(x,y)|$. Certainly now, $x$, and $x + \frac{x}{\alpha} \in B_{|r| / |\alpha|}(x)$. Thus, for any $\xi > 0$ we can have for a $\delta$ sufficiently close to 1, (i.e. $\alpha$ sufficiently large.) we have $$|x - \delta x| \leq \frac{|r|}{|\alpha|} < \xi$$ Thus, for any $\epsilon > 0$, there exists a $\delta$ sufficiently close to $1$ such that: $$\int_{E \cap \delta E} |g(x) - g(\delta x)| \hspace{0.1cm}dx < \frac{\epsilon}{6}$$ and, since $m(E \Delta 2E) < \infty$ (by compactness) and $E \Delta \delta E \searrow \emptyset$ we have from corollary 3.3 (in chapter 1) that for some $\delta$ sufficiently close to $1$, $m(E \Delta \delta E) < \frac{\epsilon}{12}$. Thus, choose $\delta$ close enough to $1$ such that both of these things happen, and we'll have: $$||g(x) - g(\delta x)||_{L^1} = \int_{E \cup \delta E} |g(x) - g(\delta x)| \hspace{0.1cm}dx < \frac{\epsilon}{6} + 2M\frac{\epsilon}{12M} = \frac{\epsilon}{3}$$ Finally, we have for $\delta$ sufficiently close to 1: $$||f(\delta x) - f(x) ||_{L^1} \leq ||g(\delta x) - g(x) ||_{L^1} + ||f(\delta x) - g(\delta x) ||_{L^1} + ||f(x) - g(x) ||_{L^1} \leq \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$$