# Disjoint Sets Seemingly Not Disjoint in Description of Lebesgue Measure

I am currently working through Knapp's Basic Real Analysis. I am currently working in Chapter 5 on "Lebesgue Measure and Abstract Measure Theory." The book states the following, giving Lebesgue Measure as an example of a measure.

"$\bf{Lebesgue \ Measure}$ $m$ on the ring $R$ of elementary sets of $\mathbb{R}$. If $E$ is a finite disjoint union of bounded intervals, we let $m(E)$ be the sum of the lengths of the intervals...Consider the case that $J=I_{1} \cup...\cup I_{r}$ disjointly with $I_{k}$ extending from $a_{k}$ to $b_{k}$, with or without endpoints. Then we can arrange the intervals in order so that $b_{k}=a_{k+1}$ for $k=1,...,r-1$."

I believe that I am misunderstanding something. If the union of the intervals is disjoint, how can it be that $b_{k}=a_{k+1}$? Then, wouldn't $I_{k} \cap I_{k+1}\neq \emptyset$? This seems counterintuitive, and any help would be much appreciated.

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$(0,1]$ is disjoint from $(1,2]$. – Andrés E. Caicedo Jan 24 '13 at 22:54
@Andres: That should be an answer. – Rahul Jan 24 '13 at 23:23

There are several ways we can have disjoint intervals with coinciding end-points. For example, we could have an interval $(a,b)$ or $(a,b]$ next to an interval $(b,c)$. If points are admitted as intervals, $[a,a]=\{a\}$, then we could have situations such as $(a,b)$ next to $\{b\}$ next to $(b,c)$, etc.