# Question about a part of the extension theorem proof

My question is how can we justify interchanging the union to get $(*)$. I think I did it see edit 3 or answer.

In the proof(Measures integrals and martingales starts p 39)(step 2): the following set is defined $S_\cup=\{S_1 \sqcup S_2 \ldots \sqcup S_M : M\in \Bbb{N}, S_j \in \mathcal S \}$ where $\mathcal S$ is a semi ring. Now let $(T_k)_{k\in \Bbb{N}} \subset S_\cup$ be such that $T:= \bigsqcup_{k \in \Bbb{N}}T_k$.

By the definition of $S_\cup$ we find $(S_k)_{k\in \Bbb{N}}\subset \mathcal S$ and a sequence of integers $0=n(0)\le n(1) \le n(2) \le \dots$ such that $$T_k=S_{n(k-1)+1} \sqcup \ldots \sqcup S_{n(k)}$$ and $T=U_1 \sqcup \ldots \sqcup U_N$ where $\displaystyle U_l=\bigsqcup_{j\in J_l}S_j \in \mathcal S$ with disjoint index sets $J_1 \sqcup J_2\ldots \sqcup J_N= \Bbb{N}$ that partition $\Bbb{N}$.

$\bigsqcup$ denotes disjoint union,

My attempts to understand this: I am trying to maybe get the $U_i$ from $T$. We know that $T= \bigsqcup_{k\in \Bbb{N}} T_k$ so then we get from the fact that $T_k$ can be written as $S_{n(k-1)+1} \sqcup \ldots \sqcup S_{n(k)}$ that $$T= \bigsqcup_{k\in \Bbb{N}} (S_{n(k-1)+1} \sqcup \ldots \sqcup S_{n(k)})=\bigsqcup_{k\in \Bbb{N}}\bigsqcup_{j=n(k-1)+1}^{n(k)}S_j \ \ \ \ (*)$$ Whats left is to interchange the unions, but that is proving rather tough so, my question is how can we justify interchanging the union to get the above result.

Edit 2: Since $\Bbb{N}$ can be decomposed into $J'_1 \sqcup J'_2\ldots \sqcup J'_N$ (See here) then for each $k$ we have $k \in J'_l$ for only one $l$ hence the union become equal to $$\bigsqcup_{l=1}^N\left(\bigsqcup_{k\in J'_l}\bigsqcup_{j=n(k-1)+1}^{n(k)}S_j\right)$$ then we get $J_l$

Close but not what we wanted.

Edit 3: To get the result wanted we need $$\bigsqcup_{k\in J'_l}\bigsqcup_{j=n(k-1)+1}^{n(k)}S_j=\bigsqcup_{j\in J_l}S_j$$ where $J'_l \neq J_l$. If $J'_l=\{k_1,k_2,\ldots\}$ then $j\in J_l=\{n(k_1-1)+1,..,n(k_1),n(k_2-1)+1),..n(k_2),... \}$ which means that $\bigsqcup_{k\in J'_l}\bigsqcup_{j=n(k-1)+1}^{n(k)}S_j=\bigsqcup_{j\in J_l}S_j$

What's left to show is that $J_1\sqcup J_2\ldots \sqcup J_N=\Bbb{N}$. Let $n\in \Bbb{N}$ then $n(k-1) < n < n(k)$ ,strict since if they are equal the claim is trivial, for some $k \in J'_l$ then $n \in J_l$. Then $J_1\sqcup J_2\ldots \sqcup J_N\subset \Bbb{N}$ the other inclusion is obvious. To show pairwise disjoint-ness let $n \in J_l \cap J_{l'}$ then $n(k-1)+1 \le n\le n(k)$ and $n(k'-1)+1 \le n\le n(k')$ where $k\in J_l$ and $k' \in J_{l'}$. Assume WLOG $k<k'$ they cant be equal since the $J'_l$ are disjoint, then $k' \ge k+1 \implies k'-1 \ge k \implies n(k'-1)+1>n(k)$ but $n \le n(k)$ then $n\le n(k'-1)+1$ which is a contradiction hence they are disjoint.

• Is this stuff hard or am I wording my questions badly? Because I haven't been getting any answers :S – user63697 Jan 10 '14 at 22:31
• Gee! No answer in 18 minutes? What is this world coming to... – Did Jan 10 '14 at 22:59
• No, I mean my questions go into inactivity mode like the previous ones I've asked. For one of them I might have added to much so I divided the question but I got no hints. So I'm just asking maybe the way I'm asking is bad. @did it was 36minutes when you posted this. – user63697 Jan 10 '14 at 23:02
• Yeah, and 18 minutes, as I indicated, when you posted your "I haven't been getting any answers" comment. – Did Jan 10 '14 at 23:35
• @did As you wish, I'm not here to argue. – user63697 Jan 10 '14 at 23:36