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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?2012-10-10
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    Yes, we can assume $g$ is continuous.2012-10-10
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    Can you include it in the OP? Thanks!2012-10-10

2 Answers 2

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

Hints:

  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?2012-10-10
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    3: No, we cannot say that. However, that integral does not depend on $x$, what can we say when $x\to\infty$?2012-10-10
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    4: Try to estimate the integral using 1, before letting $x\to\infty$2012-10-10
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    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 $a$.2012-10-10
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    Right, (typo $g(x)\leq a+\varepsilon$).2012-10-10
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    You have got it. :)2012-10-10
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    Great. Thanks for your help.2012-10-10
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    Happy to help, just note that $$a-\varepsilon\leq g(x)\leq a+\varepsilon$$ and not $$a-\varepsilon\leq g(x)\leq a....$$2012-10-10
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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

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

and

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

For any $x>0$ we have:

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

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:

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

QED.

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    @BillDubuque: honestly? Because I am a bit disgruntled by the amount of trivial questions on MSE these days. It's becoming Yahoo Answers v2. To tell a beginner "you should probably study on your own, before desperately ask for help" is not really polite, I agree, but that does not mean it is the wrong thing to do. *You have no idea? Well, make one, then ask.*2018-11-01
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    That's true, but when the noise/signal ratio raises over the 80% also a general-level site ceases to be useful.2018-11-01
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    Judging "useful" will of course be very subjective. It seems to me that many students are learning much from the site (at least in the tags I'm active in).2018-11-01