Is the average of all $x$ such that $0 \le x \le 1$ $ \forall $ $x \in \mathbb{R}$ equal to $\frac{1}{2}$? 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..
 A: 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 $a<b$, but its length $b-a$ might be a good notion of it size for our purposes. By extending this notion of lenght to more complicated sets of real numbers (but not to all sets of real number!) we arrive at the notion of Lebesgue 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.    
