Show $\lim\limits_k\int_{[0,a]}f(kx) = ar$ where $f:[0,\infty) \to \mathbb{R}$, bounded, Lebesgue measurable, and $\lim\limits_{x\to\infty} f(x) = r$.
$$ \int_{[0,a]}f(kx) = \int \chi_{[0,a]}(x)f(kx) = \int \chi_{[0,ka]}(kx)f(kx) = \frac{1}{k}\int\chi_{[0,ka]}(x)f(x) \\ = a\frac{1}{ak}\int\chi_{[0,ka]}(x)f(x) $$
I'm stuck at showing $$ \frac{1}{ak}\int\chi_{[0,ka]}(x)f(x) \to r. $$
Denote $f_k(x)=f(kx)$ then we have $$\lim_{k\to\infty}f_k(x)=r\;\text{ a.e.}$$ so by the dominated convergence theorem we have
$$\lim_{k\to\infty}\int_0^a f_k(x)dx=\int_0^a rdx=ra$$