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Is the average of all $x$ such that $0 \le x \le 1$ $ \forall $ $x \in \mathbb{R}$ equal to $\frac{1}{2}$ ?

It may seem this way but how can we show that indeed it is $\frac{1}{2}$. We know that the number of elements in $0$ to $1$ of real numbers is $\aleph_1$. therefore we have uncountable over uncountable..? Is this true or not? thanks..

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    You cannot really average the elements, but you can take expectations with respect to the [uniform distribution](http://en.wikipedia.org/wiki/Uniform_distribution_(continuous)), and this expectation is indeed $1/2$. You obtain it as the integral $\int_0^1 x~dx=1/2$.2012-03-28
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    If you think from a probability perspective, i.e. $x$ is a uniform distribution over (0,1), then it makes sense.2012-03-28
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    So this means that it is impossible to find the average of any uncountable number of elements? and what could be the distinction between the average point-of-view and the uniform distribution?2012-03-28
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    "We know that the number ... is $\aleph_1$." No, this is (CH). We know that it is $2^{\aleph_0}$.2012-03-28
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    It means that you have to be careful with what you mean by "average". The definition that works for a finite set of points does not work for an infinite set.2012-03-28
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    @ martini: isn't it $2^{\aleph_0} = \aleph_1 ?$2012-03-28
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    @Keneth Adrian: Whether that is true cannot be decide by the usual axioms of set theory. Take a look at the [continuum hypothesis](http://en.wikipedia.org/wiki/Continuum_hypothesis).2012-03-28

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If $F$ is a finite set and $f:F\to\mathbb{R}$ is a function, we can take its average as $1/N \sum_{x\in F}f(x)$, where $N$ is the number of elements of $F$. The average of a finite set $F$ of real numbers is simply the average of the identity on $F$.

If we want to extend this to a nonnegative function $f:X\to\mathbb{R}$ with $X$ an infinite set, we encounter two problems. First, since the cardinality of $X$ is not a real number, it is not clear what dividing by that number means. Second, if $X$ is uncountable and $f(x)\neq 0$ for uncountably many $x$, the sum $\sum_{x\in X}f(x)$ is necessarily infinite. If $f$ is not nonnegative, we can in addition get problems with convergence of the sum.

In order to solve the first problem, we observe that we can sometimes use a different notion of size for the underlying set $N$. For eaxample, a closed interval $[a,b]$ is necessarily uncountable if $aLebesgue measure.

To deal with the second problem, we can ask ourselves what a good alternative to adding up might be. If $f$ is constant with value $c$ on the interval $[a,b]$ and zero everywhere else, we would want that "replacement of sum"$/$"size of $[a,b]$"$=$ "replacement of sum"$/(b-a)=c$. So "replacement of sum"$=c(b-a)$. Now the sum of the values of $f+g$ is in the finite case the sum of the values of $f$ plus the sum of the values of $g$. Also, the sum of the values of $\alpha f$ equals the sum of the values of $f$ times $\alpha$ when $\alpha$ is a real number. So we can sum up functions that are linear combinations of "blocks". Extending this by a limiting operation to a wider class of functions gives us the integral as a replacement of summation.