Consider the following sequence: $$x_n:= \frac{1}{n} \int _0 ^n f(x)dx$$ Where $f:\mathbb{R} \rightarrow \mathbb{R}$ is a continuous function vanishing at infinity (i.e. such that $\lim_{|x| \rightarrow +\infty} f(x)=0$). Then I want to prove that $x_n \rightarrow 0$ for $n \rightarrow +\infty$.
The first thing I thought was that the improper integral $\int_0 ^{+\infty} f(x)dx$ was always finite, but this is clearly not true since $$f(x):=\begin{cases} \frac{1}{|x|} && x \in \mathbb{R} \setminus [-1,1] \\1 && x \in [-1,1]\end{cases} \Rightarrow \int_0 ^{+\infty} f(x)dx=1+\int_1^{+\infty} \frac{1}{x}dx =+\infty$$ Nevertheless since the primitive of $1/x$ is $\log x$ on $[1, +\infty)$ the sequence becomes $$x_n=\frac{\log n}{n} \longrightarrow 0 \qquad \text{for } n \longrightarrow +\infty$$ Because $\log x=o(x)$. So fundamentally it seems to me that the proof reduces to demonstrate that $F(x)=\int _0 ^x f(t)dt=o(x)$.
Is this the way? How can be proven?
Let $\epsilon > 0$. Split the integral into two parts: $\int_0^nf(x)\,dx = \int_0^N f(x)\,dx + \int_N^n f(x)\,dx$. First choose $N$ so large that $x>N$ implies $|f(x)|<\epsilon/2$. Next choose $n_0$ so large that $\int_0^N |f(x)|\,dx < n_0\epsilon/2$. For $n > \max\{N,n_0\}$, what can you say about $\frac{1}{n}\int_0^n f(x)\,dx$?
I leave it to you to prove the existence of such $N, n_0$.