Measure of the irrational numbers? I have read that the measure of the irrational numbers on an interval $[a,b] = b-a$. This both makes sense and doesn't make sense to me. If you consider that the union of the irrationals with the rationals are the reals, then if the rationals have measure 0, then the irrationals must have the same measure as the reals. (right?)
But if you consider that the irrationals are totally disconnected, then it is a collection of disconnected points. If it's a collection of disconnected points, like the rationals, how can it have a non-zero measure?
Another fact I'm not sold on, which may be the source of the problem, is that there isn't a bijection between the rationals and the irrationals. If both of these sets are totally disconnected by each other and both dense, that is, "Between any $ \forall a,b \in \mathbb{R} $, such that $a < b$, there is exists an irrational $i$ and a rational $q$ satisfying $a < i < b$, and $a < q < b$."
I know of the diagonalization argument but I've just never been completely sold on this fact. For the irrationals to be uncountable and the rationals to be countable, in my head it would make more sense if there exists an $\epsilon > 0$ such that around any irrational number there exists only other irrational numbers. I know that if it exists you could just sum enough of those epsilons together to get the length of it greater than some $\frac{p}{q}$ thus finding a contradiction... but I'm still not sold on it.
Can someone help me finally understand these concepts?
 A: The argument in your first paragraph is correct.
For the second paragraph, why do you think that being disconnected is of any relevance? The rational numbers are of zero measure because they are countably many of them. The set of irrationals is not countable, therefore it can (and indeed does) have a non-zero measure.
On your third paragraph: It is true that between any two rationals there's an irrational, and between any two irrational there's a rational. However, there is not only one irrational between two rationals, and not only one rational between two irrationals. Rather there are countably many rationals between two irrationals, but uncountably many irrationals between two rationals.
In other words, it's not that you have an alternating sequence of rational and irrational numbers (which is what you seem to have in mind). In particular, you cannot define ”the next larger rational” nor ”the next larger irrational.”
In some sense, although the rationals are already dense, the irrationals are in a sense even more dense.
Maybe it helps if instead of the rationals, you consider the numbers with finite decimal expansion, which also is a countable dense subset. Here, it might be more intuitive that, you have many more possibilities for numbers if you can arbitrarily continue your number string infinitely than if you are forced to stop after finitely many digits. Yet again, between any two finite-expansion numbers, you always can find an infinite-expansion number (just add infinitely many non-zero digits to the smaller one, after first padding it to the length of the larger with zeroes if necessary), and between any two infinite-expansion numbers you can find a finite-expansion number (just cut the larger one at an appropriate point).
A: The set of rational number is numerable thus $Q=(u_n)_{n\in N}$. Each rational is a Borelian set of measure 0, thus $\mu(u_n)=0$, $\mu(Q)=lim_{n>0}\mu(U_n)$ where $U_n=\{u_0,...,u_n\}$. Thus $\mu(U_n)=0$ and henceforth $\mu(Q)=0$. Since $R$ is the disjoint union of $Q$ and $R-Q$, and $\mu(Q)=0$, then $\mu(R-Q)=+\infty$.  
You also have $[a,b]\cap Q\subset Q$ thus $\mu([a,b]\cap Q)=0$.
$[a,b]$ is the disjoint union of $[a,b]\cap Q$ and $[a,b]-Q$ thus $\mu([a,b])=\mu([a,b]\cap Q)+\mu([a,b]-Q)=b-a$.
