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!


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.