Question on "Measure and Cardinality" - Briggs ans Schaffter - There is a part of a proof in the article "Measure and Cardinality" by Briggs and Schaffter that I don't understand.
Theorem : If $E$ is a measurable set of reals of cardinality less than $c$, then $E$ has measure 0.
Proof : We show that if $E$ has positive measure then the cardinality of $E$ is $c$. Accordingly, suppose $mE > 0$ and let $F$ be a closed subset of $E$ of positive measure. Then there are two disjoint, closed intervals $I_0, I_1$ of length less than mF which intersect F in sets of positive measure... (I'm ok with the rest of the proof).
Intuitively, it makes sense that such intervals $I_0, I_1$ exist. I'm pretty sure it's not something sophisticated but I'm not familiar enough with measure theory to see why they necessarily exist. 
 A: One way to see this is to tile $\Bbb R$ with the closed intervals $C_j:=[jmF/2, (j+1)mF/2]$, $j\in\Bbb Z$.  Since
$$
F=\bigcup_j F\cap C_j \text{  and, for all } j,\ m(F\cap C_j)\le mF/2,
$$
there must be two distinct intervals $C_j$ and $C_k$ ($j<k$) such that $m(F\cap C_j)$ and $m(F\cap C_k)$ are positive.  Then, if these intervals are adjacent, lower the upper endpoint of $C_j$ by any amount less than $ m(F\cap C_j)$.
A: Since $\bigcup_n[-n,n]=\mathbb{R}$, the intersection of $F$ and a closed bounded interval must have positive measure. Sowe can assume without loss of generality that $E$ is the subset of a bounded interval $[a,b]$. 
Take the intervals $[a,a+(b-a)/3]$, $[a+(b-a)/3,a+2(b-a)/3]$, and $[a+2(b-a)/3,b]$. If the first and the last intersect $F$ on a set of positive measure, we are done. If not, we have two adjacent intervals wih union being a closed interval of length $2/3(b-a)$ that contains all of $F$ except for a null set. We can divide this interval again, and proceed the same way. Eventually, the process must stop with two nonadjacent subinterval that both intersect $F$ on a set of positive measure. If not, we have $\mu(F)<(2/3)^n (b-a)$ for all $n$, which contradicts $\mu(F)>0$.   
If one of the resulting subintervals happens to have length $\mu(F)$, we can aplly the same method to find smaller subintervals that intersect $F$ on a set of positve measure.
