Jordan Measures, Open sets, Closed sets and Semi-closed sets I cannot understand:
$$\bar{\mu} ( \{Q \cap (0,1) \} ) = 1$$ 
and (cannot understand this one particularly)
$$\underline{\mu} ( \{Q \cap (0,1) \} ) = 0$$ 
where $Q$ is rational numbers, why? I know that the measure for closed set $\mu ([0,1]) = 1$ so I am puzzled with the open set solution. Is $\underline{\mu} ( (0,1) ) = 0$ also? How is the measure with open sets in general? 
So far the main question, history contains some helper questions but I think this is the lion part of it what I cannot understand. More about Jordan measure here.
Related


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*Jordan measure with semi-closed sets here

*Jordan measure and uncountable sets here

The inner measure of $S:=\mathbb Q \cap [0,1]$ is, by definition, the "largest" simple subset contained in S (, with largest defined as the sup of the measures of simple sets contained in $S$). But there is no non-trivial , i.e., non-empty, simple set contained in $S'$, since , by density of both rationals and irrationals in $\mathbb R$, any simple set $S':=[a,a+e)$ contained in $S$ (i.e., with $0<a<a+e<1$
,  will necessarily hit some irrational, i.e., no non-empty simple subset $S'$ of $S$ can be contained in $S$, so the only simple set contained in $S$ is the trivial , empty set. And the empty set is defined to have measure $0$.
For the outer measure, you want to find the "smallest" set $T$ containing $S:=\mathbb Q \cap [0,1]$. As pointed above, by density of $\mathbb Q$ in $\mathbb R$ , no strict subset of $[0,1]$ can contain $S$. We then only have the option of having sets of the type $S'':=[0,1+e)$ covering $S$; we can rewrite $S'':=[0, 1+\frac{1}{n})$, and $m^*(S'')=1
+ \frac{1}{n}$. The infimum of the measures over all $S''$ is then $ inf$ {  1+$\frac{1}{n}$ : n in $\mathbb N$ }, which is 1.
