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At first, I had thought the average must be zero, since for every positive number there's an equal magnitude negative number to cancel out the positive number's effect on the average, leaving only zero to set the average.

But you can make a similar argument about any number, for example using an arbitrary choice of 9, for every number x units greater than 9, there's another number x units less than 9, which would make 9 the mean. But since I could have chosen any number here instead of 9 that would mean that any and every number is the average of all real numbers.

So, what is the mean of all real numbers?

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    $\begingroup$ $${{{{{{\Large 42}}}}}}$$ $\endgroup$ – dtldarek Jul 9 '13 at 8:15
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As you proved for yourself, there can be be no well defined mean for the real numbers. Another way to think about it would be to think of the mean of a closed interval $[a,b]$ as: $$\mu_{[a,b]}=\int_{a}^{b} \frac{x}{b-a} dx$$ To extend this to all the real numbers, you would have to take $a\to -\infty$, $b\to \infty$. You may be tempted to write the mean as: $$\mu_\mathbb{R}=\lim_{M\to\infty}\int_{-M}^M \frac{x}{2M} dx =0$$, but since no one promises you that $a$ and $b$ go to $\infty$ at the same rate, you must write: $$\mu_\mathbb{R}=\lim_{N\to\infty}\lim_{M\to\infty}\int_{-M}^N \frac{x}{N+M} dx$$ This is undefined, due to the fact that the value depends on your choice of rate (choose $N=2M$ or $N=M$ to see that).

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    $\begingroup$ Your integrals should all be of the function $f(x)=x$. $\endgroup$ – Mariano Suárez-Álvarez Jul 9 '13 at 2:37
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    $\begingroup$ @MarianoSuárez-Alvarez - actually, $x/(b-a)$ :) $\endgroup$ – nbubis Jul 9 '13 at 3:12
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It's undefined.

The mean of a finite collection of numbers $a_1, ... a_n$ is its expected value when regarded as a random variable on the sample space $\{ 1, 2, ... n \}$ given the uniform distribution, where each number occurs with probability $\frac{1}{n}$. To define the mean of a function $f : \mathbb{R} \to \mathbb{R}$ (such as the identity function $f(x) = x$, which is this example), you need to pick a probability distribution on $\mathbb{R}$. It doesn't have an analogue of the uniform distribution, so as it stands the question is underspecified.

A related question, but one that requires somewhat fewer technical details to answer, is "what is the mean of all integers?" The answer is again that this is undefined, and again the reason is because there is no analogue of the uniform distribution on $\mathbb{Z}$.

(The uniform distribution on a finite set is special because it is the maximum entropy distribution, but $\mathbb{Z}$ and $\mathbb{R}$ do not have maximum entropy distributions. However, maximum entropy disributions can still exist with enough extra conditions.)

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    $\begingroup$ Yet, we all know it is zero! :-) $\endgroup$ – Mariano Suárez-Álvarez Jul 9 '13 at 2:27
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    $\begingroup$ I would have thought 42. $\endgroup$ – Andreas Blass Jul 9 '13 at 4:46
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    $\begingroup$ @AndreasBlass Exactly! $\endgroup$ – dtldarek Jul 9 '13 at 8:17

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