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Here's a little question I saw in a book recently, which I can see but can't set out my own formal proof and it's annoying me.

Say $\lim_{x\to\infty} g(x) = a$

and $g$ is continuous,

so now prove that

$\lim_{x\to\infty} \frac{1}{x} \int_0^x g(y) \mathrm{d}y = a$ .

I can see that if we define $\int_0^x g(y)\mathrm{d}y = G(x) - G(0)$

then $\frac{1}{x} \int_0^x g(y) \mathrm{d}y = \frac{G(x) - G(0)}{x}$

which clearly looks a lot like the limit of a differential, but I'm not sure how to handle the limit?!

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Is $g$ continuous? – Davide Giraudo Oct 10 '12 at 9:19
Yes, we can assume $g$ is continuous. – Donald Peters Oct 10 '12 at 9:29
Can you include it in the OP? Thanks! – Davide Giraudo Oct 10 '12 at 9:32

Perhaps you assume $g$ is continuous, at least some kind of integrability has to be assumed.


  1. Since $\lim_{x\to\infty} g(x) =a$ exists, we can for each $\varepsilon>0$ find $x_0$ such that $$|g(x)-a|<\varepsilon$$ for all $x\geq x_0$.

  2. Split the integral $$\int_0^x=\int_0^{x_0}+\int_{x_0}^x$$

  3. What can we say about $$\frac{1}{x}\int_0^{x_0} ?$$

  4. What can we say about $$\frac{1}{x}\int_{x_0}^x ?$$

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So for 3, we know that the integral will be greater than $a\epsilon$ and for 4 the opposite? – Donald Peters Oct 10 '12 at 9:37
3: No, we cannot say that. However, that integral does not depend on $x$, what can we say when $x\to\infty$? – AD. Oct 10 '12 at 9:44
4: Try to estimate the integral using 1, before letting $x\to\infty$ – AD. Oct 10 '12 at 9:46
So maybe I have it: for each $\epsilon > 0$ I can find $x_0$ s.t.: $a-\epsilon \le g(x) \le a$ for all $x \ge x_0$. So.... $(a-\epsilon)x = \int_0^x (a-\epsilon)\mathrm{d}y \le \int_0^x g(y) \mathrm{d}y \le \int_0^x a \mathrm{d}y$ so: $\frac{(a-\epsilon)x}{x} \le \frac{1}{x}\int_0^x g(y)\mathrm{d}y \le a$ And so the limit is $a$. – Donald Peters Oct 10 '12 at 9:46
Right, (typo $g(x)\leq a+\varepsilon$). – AD. Oct 10 '12 at 9:48

We don't need $g(x)$ to be continous, we only need $g(x)$ to be integrable.

Take any $\varepsilon > 0$. Since $\lim_{x\to +\infty} g(x) = a$, for any $x>M_\varepsilon$ we have $|g(x)-a|<\varepsilon$. Take

$$I_\varepsilon = \int_{0}^{M_\varepsilon}\left|g(x)-a\right|\,dx$$


$$n_\varepsilon = \left\lceil\frac{I_\varepsilon}{\epsilon}\right\rceil.$$

For any $x>0$ we have:


so, for any $x> \max(n_\varepsilon, M_\varepsilon)$ we have:

$$\left|-a+\frac{1}{x}\int_{0}^{x}g(x)dx\right|\leq \frac{I_\varepsilon}{x}+\frac{1}{x}\int_{M_\varepsilon}^{x}|g(x)-a|\,dx \leq 2\varepsilon.$$

Since $\varepsilon$ can be taken arbitrarily small, the last inequality gives:

$$\lim_{x\to +\infty}\frac{1}{x}\int_{0}^{x}g(x)\,dx = a,$$


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