Let $\mu$ be the Lebesgue measure on $\mathbb R^n$. For any $f\in L^1(\mu)$, prove that $$\lim_{a\to 1}\int_{\mathbb R^n} |f(x)-a^nf(ax)|d\mu=0.$$
My attempt:
Since \begin{align} \lim_{a\to 1}\int_{\mathbb R^n} |f(x)-a^nf(ax)|d\mu&=\lim_{a\to 1}\int_{\mathbb R^n}\left((f(x)\chi_{f(x)\ge a^nf(ax)}-a^nf(ax)\chi_{f(x)\ge a^nf(ax)}\right. \\ &\quad \left.+a^nf(ax)\chi_{f(x)< a^nf(ax)}-f(x)\chi_{f(x)<a^nf(ax)})\right)d\mu\\ \end{align} By change of variables $y=ax$, we have: \begin{align} \int_{\mathbb R^n}a^nf(ax)(\chi_{f(x)< a^nf(ax)}-\chi_{f(x)\ge a^nf(ax)})d\mu=\int_{\mathbb R^n}f(y)(\chi_{f(\frac ya)< a^nf(y)}-\chi_{f(\frac{y}{a})\ge a^nf(y)})d\mu\\ \end{align} Then by the dominated convergence theorem we can swap the limit and integral and we are done.
Is the proof correct? Thanks.
Hints: there exists a continuous function $g$ with compact support such that $\int |f-g| <\epsilon$. Write $g(x)-a^{n}g(ax)$ as $(1-a^{n})g(x)+a^{n}(g(x)-g(ax))$ and use Uniform continuity for $\int |g(x)-g(ax)|dx$.
Let me know if you want a detailed proof.