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

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!

## 1 Answer

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.