Jordan outer content of rationals in [0, 1] How to prove Jordan outer content of rationals in $[0,1]$ is 1 just by definition of Jordan outer content? I mean how to prove this without using "Jordan content of a set is equal to that of its closure"? 
 A: HINT: Show that if $S$ is a finite union of left-closed, right-open intervals, and $S\supseteq[0,1]\cap\Bbb Q$, then $S\supseteq[a,b)$ for some $a,b\in\Bbb R$ such that $a\le 0$ and $1<b$; that gives you one inequality. For the other, note that if $1<b$, then $[0,1]\cap\Bbb Q\subseteq[0,b)$.
A: Your goal is to prove that m(S)=1; every elementary set within [0,1] must belong to S because otherwise they will contain some rational number q which is contradictory (because q is supposed to be in S). But if m(S)<1, S must have excluded some elementary set in [0,1] that has Jordan measure greater than 0 (otherwise m(S)>=1). On the other hand, [0,1] clearly contains S (S is infimum of all that contains all rationals in [0,1]). THAT gives you m(S)=1.
A: I suppose, what other people haven't completely (though the hints are sufficient) proved is the fact that if $(I_k)_{k=1}^n$ is a finite sequence of intervals (left closed, right open) covering all rationals in $[0,1]$, then it must cover the whole interval $[0,1]$. A detailed proof is given below:
Let $E=[0,1]\cap \mathbb{Q}$ and $m^*_J(E)$ denote the Jordan outer content of $E$.
Let $I_k=[a_k,b_k), a_k\neq b_k$.
Without loss of generality, assume that $(I_k)_{k=1}^n$ is ordered in the increasing order of $\{a_k\}$. i.e, $a_k\leq a_{k+1}$ for $1\leq k<n$. We claim that $(I_k)_{k=1}^n$ covers the whole interval $[0,1]$. 
i.e. $[0,1]\subset\cup_{k=1}^n I_k$. Suppose not. 
Then there exists a point $x_0\in [0,1]$ such that $x_0\notin I_k$ for every $k=1,2,\dots,n$. Clearly $x_0$ is not a rational number. 
Now let, 
$$a_{k_0}=\min\{a_k\;|\;x_0< a_k,1\leq k\leq n\}$$
If $x_0=a_{k_0}$, then $x_0\in [a_{k_0},b_{k_0})$ contradicting our hypothesis (or trivially because minimum of finitely many numbers strictly greater than $x_0$ can not be $x_0$). Observe that, for $k\leq k_0$, $I_k\subset (-\infty,x_0]$ and for $k>k_0$, $I_k\subset [a_{k_0},\infty)$.
Hence as $x_0\neq a_{k_0}$, there exists a (infinitely many in fact) rational number in $(x_0,a_{k_0})$ which is not in any $I_k$, again contradicting the hypothesis that $\cup_{k=1}^nI_k$ contains all rationals in $[0,1]$. 
This proves that $(I_k)_{k=1}^n$ covers whole of $[0,1]$ and hence by subadditivity property of the Lebesgue outer measure (or Jordan outer content),$$1=m^*_J([0,1])\leq \sum_{k=1}^n l(I_k)$$
To prove equality, note that the interval $[0,1+\epsilon)$ contains all rationals in $[0,1]$ for every $\epsilon>0$ and hence by the definition of Jordan outer content, we have, $m^*_J(E)\leq 1$.
