Below is a question, which I asked before, from Stein's Real Analysis. I've provided a partial solution, which I think it's pretty along the lines of what needs to be done, however, I have no finished the solution. If anyone has a hint for the part I'm stuck at, I would very much appreciate it. :)
Question: Prove that if $f$ is integrable on $\mathbb{R}^d$ and $\delta > 0$, then $f(\delta x)$ converges to $f(x)$ in $L^1$-norm as $\delta \to 1$.
My Attempt:
Let $\epsilon > 0$ be given and suppose $\delta > 1$ (if not, just take $1/\delta$ in place of $\delta$). Since $f$ is integrable on $\mathbb{R}^d$, there exists a function $g$ continuous on $\mathbb{R}^d$ supported on a compact set $k$, with $\|f(x)-g(x)\| < \epsilon/3$. Applying the triangle inequality, one finds that $$ \|f(\delta x) - f(x)\| \leq \|f(\delta x) - g(\delta x)\| + \|f(x) - g(x)\| + \|g(\delta x) - g(x)\|. $$ With the assumptions on $f$ and $g$, together with the observation that $$ \|(f-g)(\delta x)\| = \int_{\mathbb{R}^d} |(f-g)(\delta x)| dx = \frac{1}{\delta^d} \int_{\mathbb{R}^d} |(f-g)(x)| dx = \frac{1}{\delta^d} \|(f-g)(x)\| $$ one can estimate $\|(f(\delta x) - f(x)\|$ by $$ \|f(\delta x) - f(x)\| < \frac{2 \epsilon}{3} + \|g(\delta x) - g(x)\|. $$ To complete the proof, it remains only to show that $$ \|g(\delta x) - g(x)\| \to 0 \quad \text{ as } \quad \delta \to 1. $$
Observe that the function $g(\delta x)$ is supported on the set $\delta K$, a compact set; hence, the difference function $g(\delta x) - g(x)$ is supported on the compact set $\delta K \cup K$. Since $g(\delta x)$ is uniformly continuous on $\delta K$ and since $g(x)$ is uniformly continuous on $K$, there exist positive constants $L$ and $M$, with $|g(\delta x)| \leq L$ And $|g(x)| \leq M$; hence writing, $$ \delta K \cup K = (\delta K \Delta K) \cup (\delta K \cap K), $$ a disjoint union, where $\delta K \Delta K = (\delta K \setminus K) \cup (K \setminus \delta K)$, and applying the triangle inequality, it follows that $$ \|g(\delta x) - g(x)\| \leq \int_{\delta K \Delta K} |g(\delta x)| dx + \int_{\delta K \Delta K} |g(x)| dx + \int_{\delta K \cap K} |g(\delta x) - g(x)| dx. $$ Using the definition of the Lebesgue integral, one then finds that $$ \|g(\delta x) - g(x)\| \leq 2(L+M) m(\delta K \Delta K) + \int_{\delta K \cap K} |g(\delta x) - g(x)| dx. $$ Now, if ${c_n}$ is any sequences of positive numbers such that $\delta = c_1 \geq c_2 \geq \dots$ and $c_n \geq 1$ for all $n$, decreasing to $1$, then the corresponding sequences of compact set $K_n = c_n K \Delta K$ decreases to the empty set, that is, $$ \bigcap_{n=1}^{\infty} K_n = \emptyset. $$ Moreover, since $m(K_1)$ has finite measure, by the continuity of Lebesgue measure, it follows that $m(K_n) \to 0$ as $c_n \to 1$. Hence, there exists $N \geq 1$ with $m(\delta K \Delta K) < \epsilon/12(L+M)$ whenever $n \geq n$ and $\delta \in (c_n, 1]$. Thus, $$ \|f(\delta x)- f(x)\| < \frac{2\epsilon}{3} + \frac{\epsilon}{6} + \int_{\delta K \cap K} |g(\delta x) -g(x)|dx $$ whenever $\delta$ is sufficiently close to $1$.
Note: The issue I'm having is dealing with the final integral $\int_{\delta K \cap K} |g(\delta x)-g(x)| dx$. Any hints would be appreciated. : )
Like I said in my comment, the best way to solve this problem is probably using the dominated convergence theorem. But I'll address your actual question which was to show $$\int_K|g(\delta x)-g(x)|dx\to0$$ as $\delta\to1$ where $K$ is compact and $g\in L^1$. What you should observe is that by density of $C$ in $L^1$, for every $\epsilon>0$, there is a $g_\epsilon\in C\cap L^1$ such that $\|g-g_\epsilon\|<\epsilon$. Thus $$\int_K|g(\delta x)-g(x)|dx\leq\int_K|g(\delta x)-g_\epsilon(\delta x)|dx+\int_K|g_\epsilon(\delta x)-g_\epsilon(x)|dx+\int_K|g_\epsilon(x)-g(x)|dx\\ \leq 2\|g-g_\epsilon\|+\int_K|g_\epsilon(\delta x)-g_\epsilon(x)|dx.$$ Taking $\delta\to1$ makes the second term zero since $g$ is continuous*, and taking $\epsilon\to0$ makes the first term vanish by assumption.
*note that making this line precise is where you require compactness of $K$.