let $ f $ be integrable function in any interval such [0,M]. assume $ \lim_{x\to\infty}f\left(x\right)=L $ for some $ L\in \mathbb{R} $
and prove that
$ \lim_{n\to\infty}\intop_{0}^{1}f\left(n\cdot x\right)dx=L $.
I've managed to prove that
$ \intop_{0}^{1}f\left(n\cdot x\right)dx=\frac{1}{n}\intop_{0}^{n}f\left(x\right)dx $ .
Im not sure how to continue. Thanks in advance
For $\epsilon >0$, take $N >0$ such that $\vert f(x) - L \vert < \epsilon$ for $x >N$
Then for $n >N$ $$\begin{aligned}\left\vert \frac{1}{n}\intop_{0}^{n}f(x)dx - L \right\vert &= \left\vert\frac{1}{n} \intop_0^N f(x) \ dx + \frac{1}{n} \intop_N^n \left(f(x)-L\right) \ dx + \frac{n-N}{n} L - L\right\vert \\ &\le \frac{1}{n} \intop_0^N \vert f(x) \vert \ dx + \epsilon \frac{n - N}{n} + \frac{N}{n} \vert L \vert \end{aligned}$$
As $\lim\limits_{n \to \infty} \frac{1}{n} = 0$, and $\int_0^N \vert f(x) \vert \ dx$ is finite by hypothesis, you can bound the RHS of the inequality by $3 \epsilon$ for $n$ large enough which provides the desired conclusion.