View source for Dynkin system/Proof that definitions 1 and 2 are equivalent

{{Extra Maths}}<noinclude>
==Claim==
The following definitions are the same:
{| class="wikitable" border="1"
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! Definition 1
! Definition 2
|-
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{{:Dynkin system/Definition 1}}
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{{:Dynkin system/Definition 2}}
|}
==Proof==
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{{Todo|Flesh out the algebra (blue boxes)}}
'''Definition 1 {{M|\implies}} definition 2'''
: Let {{M|\mathcal{D} }} be a subgroup satisfying definition 1, then I claim it satisfies definition 2. Let us check the conditions.
:# {{M|X\in\mathcal{D} }} is satisfied by definition
:# For {{M|A,B\in\mathcal{D} }} with {{M|B\subseteq A}} then {{M|A-B\in\mathcal{D} }}
:#* Note that {{M|1=A-B=(A^c\udot B)^c}} (this is not true in general, it requires {{M|B\subseteq A}}{{Note|Include ven diagram}}
:#: As by hypothesis {{M|\mathcal{D} }} is closed under complements and disjoint unions, we see that {{M|(A^c\udot B)^c\in\mathcal{D} }} thus
:#:* we have {{M|A-B\in\mathcal{D} }}
:# Given {{M|1=(A_n)_{n=1}^\infty\subseteq\mathcal{D} }} being an [[increasing sequence]] of subsets, we have {{M|1=\lim_{n\rightarrow\infty}(A_n)=A}} where {{M|1=A:=\bigcup_{n=1}^\infty A_n}} (See [[Limit of increasing sequence of sets|limit of an increasing sequence of sets]] for more information)
:#: Let {{M|1=(A_n)_{n=1}^\infty\subseteq\mathcal{D} }} be given. 
:#: Define a new sequence of sets, {{M|1=(B_n)_{n=1}^\infty}} by:
:#:* {{M|1=B_1=A_1}}
:#:* {{M|1=B_n=A_n-B_{n-1} }}
:#: This is a pairwise disjoint sequence of sets.
:#: Now by hypothesis {{M|1=\bigudot_{n=1}^\infty B_n\in\mathcal{D} }}
:#* Note that {{MM|1=\bigudot_{n=1}^\infty B_n=\bigcup_{n=1}^\infty A_n}}
:#: So we have {{M|1=\bigcup_{n=1}^\infty A_n\in\mathcal{D} := A}}, thus the limit is in {{M|\mathcal{D} }} - as required.
This completes the first half of the proof.

{{Yellow Note|The second half isn't tricky, the only bit I recommend knowing is {{M|1=A\udot B=(A^c-B)^c}} }}
{{Todo|That second half}}
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==Notes==
<references group="Note"/>
==References==
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{{Theorem Of|Measure Theory}}
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