It is given $f:[0,\infty)\to\mathbb{R}$ is a continuous function and $\lim_{n\to \infty}\int_{0}^{1} f(x+n)dx=2$ then prove that $\lim_{n\to \infty}\int_{0}^{1} f(nx)dx=2$.

I can prove that $\lim_{x\to \infty} f(x)=2$ using the mean value theorem.

But then i used M.V.T. in second integral and getting $\lim_{n\to \infty}\int_{0}^{n} \frac{f(u)}{n}du=f(c)$ for some $c\in (0,n)$, but i cant conclude that $c\to\infty$ and $f(c)\to 2$.

I have thought of using Lebesgue dominated convergence theorem and I have proven it but is there any other way to prove this with only elementary real analysis?

  • 2
    $\begingroup$ I doubt that you can prove $\lim_{x \to \infty} f(x) = 2$. Consider for example $f(x) = 2 + \sin(2 \pi x)$. $\endgroup$
    – Martin R
    Commented Jun 15 at 13:05
  • $\begingroup$ okay i didnt thought about this, using m.v.t. i think that was $c\in (n,n+1)\forall n\in\mathbb{N}$ such that $f(c)=2$. $\endgroup$
    – Ricci Ten
    Commented Jun 15 at 13:33

1 Answer 1


Change of variables gives $\int_0^1 f(x+n)dx=\int_n^{n+1}f(x)dx$, and $\int_0^1 f(nx)dx=\frac{1}{n}\int_0^n f(x)dx$.

So if we define the sequence $a_n=\int_{n-1}^n f(x)dx$ then we have $\int_0^1 f(nx)dx=\frac{a_1+...+a_n}{n}$. And it is a very standard result about limits that if $a_n\to L$ then its sequence of arithmetic means tends to $L$ as well.

  • 3
    $\begingroup$ Indeed, that is the Stolz–Cesàro theorem $\endgroup$
    – Martin R
    Commented Jun 15 at 12:39
  • $\begingroup$ @MartinR The Stolz–Cesàro theorem is more general. I think that this particular case is due to Cesàro. $\endgroup$
    – jjagmath
    Commented Jul 5 at 18:56

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