Simple partitioning of sequence in Proposition 1.15 in Folland's text

[psie]

TL;DR Summary

I'm stuck at a simple idea in Proposition 1.15 in Folland's real analysis text, where he makes an assumption that I'm trying to grasp rigorously.

Below is Proposition 1.15 in Folland at the beginning of the section of Borel measures on $\mathbb R$ (he is trying to construct a measure from $F$). Here the algebra $\mathcal{A}$ is the finite disjoint union of h-intervals, where h-interval is a set of the form $(a,b]$, $(a,\infty)$ or $\varnothing$ for $-\infty\leq a<b<\infty$.

I assume the reader knows what a premeasure is (more or less a set function on an algebra that maps $\varnothing$ to $0$ and is countably additive for any sequence of disjoint sets in the algebra). First Folland shows that $\mu_0$ is well-defined. Then he goes on to show countable additivity, and in doing so, he makes a statement that I understand intuitively but that I don't understand how to write down rigorously.

Given a sequence of disjoint h-intervals $\{I_j\}_1^\infty$ such that $\bigcup_1^\infty I_j\in\mathcal{A}$, he makes the following claim.

Since $\bigcup_1^\infty I_j$ is a finite disjoint union of h-intervals, the sequence $\{I_j\}_1^\infty$ can be partitioned into finitely many subsequences such that the union of the intervals in each subsequence is a single h-interval.

Again, intuitively this makes sense, but I'm looking for an explanation with symbols (e.g. where one proves that the union of the intervals in each subsequence is in fact a single h-interval). How would you do this?

 

[pasmith]

Again, intuitively this makes sense, but I'm looking for an explanation with symbols (e.g. where one proves that the union of the intervals in each subsequence is in fact a single h-interval).

Have you shown that $\bigcup_{j=1}^\infty I_j$ is indeed a finite disjoint union of h-intervals $\bigsqcup_{n=1}^N H_n$? Without that you cannot proceed, since it is this which allows you to construct the subsequences.

For each $H_n$, define a subsequence consisting of those $I_j$ which intersect it. Each $I_j$ intersects exactly one $H_n$ or they would not be disjoint, so each $I_j$ appears in exactly one of those subsequences.

It is then straightforward that the union of the intervals in a subsequence is the corresponding $H_n$.

 

[psie]

It is then straightforward that the union of the intervals in a subsequence is the corresponding $H_n$.

Thank you. This is exactly the bit I'm struggling with. Here's my attempt:

There are two cases two consider regarding the sequence $\{I_j\}_1^\infty$:

1. For $j>K$, the sequence equals $\varnothing$ and perhaps also before that. Then the statement of partitioning the sequence is trivially satisfied.
2. If there isn't such $K$, then we could simply consider $\{I_j:j\in\mathbb N\}\setminus\{\varnothing\}$ since $\varnothing$ doesn't contribute to the union. Let us proceed with this case.

We have $$\bigcup_{j=1}^\infty I_j = \bigcup_{i=1}^k J_i,$$ where $J_i$ are disjoint h-intervals. We can assume $J_i$ are maximal in the sense that for $m\neq n$, the sets $J_m,J_n$ don't share endpoints and that none of them are the empty set; if e.g. $J_m=(0,1]$ and $J_n=(1,2]$ we could simply join them (likewise for the empty set). If we define the sets $A_n = \{j\in\mathbb N : I_j\subset J_n\}$ for $n=1,\ldots,k$, then $A_n$'s are disjoint and $\bigcup_{j\in A_n} I_j=J_n$.

Showing that the $A_n$'s are disjoint follows from the $J_n$'s being disjoint, and showing $\bigcup_{j\in A_n} I_j\subset J_n$ is also straightforward from the definition of $A_n$. But how do you show $\bigcup_{j\in A_n} I_j\supset J_n$? Also, I'm a bit unsure where I'm using the assumption of maximality. I guess it is simply not possible to state $\bigcup_{j\in A_n} I_j=J_n$ without it.

EDIT: For $\bigcup_{j\in A_n} I_j\supset J_n$, I guess one could simply argue like this. If $x\in J_n$ and $x\notin \bigcup_{j\in A_n} I_j$, then $\bigcup_{j=1}^\infty I_j\supset \bigcup_{i=1}^k J_i$ would be false, a contradiction.

 

[pasmith]

Yes, the idea is that the $H_n$ are disjoint nonempty intervals and the union of any two or more distinct $H_n$ is disconnected. It follows by connectedness of intervals that if $I_j$ intersects more than one $H_n$ then the union of those $H_n$ is connected, which is a contradiction. Hence for every $j \geq 1$ either $I_j = \varnothing$ or else there exists a unique $1 \leq n \leq N$ such that $I_j \subset H_n$.

By definition, if $I_j$ is empty then it does not intersect any $H_n$ so it will not be included in any subsequence. "Partitioning" the sequence into subsequences does imply that every $I_j$ must be placed in some subsequence, so either the assumption is that none of the $I_j$ are empty, or else we can add a subsequence consisting of exactly those $I_j$ which are empty. Adding one to a finite number yields a finite number.

Let $(I_{j_k})_{k \geq 1}$ be the subsequence of intervals which intersect the interval $H_n$.

• By the above, each $I_{j_k} \subset H_n$ so $\bigcup_{k=1}^\infty I_{j_k} \subset H_n$ is immediate.
• If $x \in H_n \subset \bigcup_{j=1}^\infty I_j$ then $x \in I_j$ for some $j$. However, by construction the subsequence $(I_{j_k})_{k \geq 1}$ contains all (and only) those $I_j$ which intersect $H_n$, so that $x \in I_{j_k}$ for some $k$. Hence $H_n \subset \bigcup_{k=1}^\infty I_{j_k}$.