In a measurable partition of an interval, the sum of the measures of the subsets in the partition equals the length of the interval

Question

I am working through A User-Friendly Introduction to Lebesgue Measure and Integration, by Gail S. Nelson. On page 67 Nelson defines a measurable partition of the interval $[a,b]$ to be finite collection $ \{E_j \}_{j=1}^{n}$ of subsets of $[a,b]$ such that: (1) $E_j$ is measurable for each $j$, (2) $\bigcup_{j=1}^{n}E_j=[a,b]$, and (3) $m(E_i \cap E_j)=0$ whenever $i\not=j$. On page 68 she then remarks that given any measurable partition of $[a,b]$, we can convert it into a pairwise disjoint measurable partition like so: set $F_1=E_1$, and $F_j=E_j \setminus(\bigcup_{i=1}^{j-1}E_i)$ for $j=2,3,...n$. I can see why these sets are pairwise disjoint, and I can see how each one is measurable, but I'm unable to prove that (1) their union is $[a,b]$ and (2) that $m(E_j)=m(F_j)$ for each $j=2,3,...n$. I suspect I'm overlooking something simple. Any help is greatly appreciated!

Answer 1

Let's start with $(1)$. Since $F_j \subset E_j \subset [a,b]$, we only have to see that $[a,b] \subset \bigcup_j F_j$. Take $x \in [a,b]$ and let $s = \min\{n : x \in E_j\}$. This is well defined because there exists some $j$ for which $x \in E_j$ since the $E_j$s partition $[a,b]$. By construction, $x \in E_s$ but necessarily, $x \not \in E_t$ when $t < s$. Hence,

$$ x \in E_s \setminus \bigcup_{t < s}E_t = E_s \setminus \bigcup_{t=1}^{s-1}E_t = F_s. $$
This proves that, in effect, $[a,b]$ is the disjoint union of $F_1, \dots, F_n$.

Finally, note that for each $j \in [n]$,

$$ F_j = E_j \setminus \bigcup_{i=1}^{j-1}E_i = E_j \setminus \bigcup_{i=1}^{j-1}E_j \cap E_i $$

and so since everything here is measurable and of finite measure,

$$ |F_j| = |E_j| - \left|\bigcup_{i < j}E_j \cap E_i\right|. \tag{$\star$} $$

However, by hypothesis we have that

$$ \left|\bigcup_{i < j}E_j \cap E_i\right| \leq \sum_{i < j}|E_j \cap E_i| = 0, $$

and so plugging that in $(\star)$ proves that, in effect,

$$ |F_j| = |E_j|. $$

The intuition here is that $F_i$ is $E_i$ except removing some of the intersections with other $E_j$, but since all of these are of zero measure by hypothesis, $F_i$ and $E_i$ have the same measure: the sets we have 'cropped' are (measure wise) negligible.