Difference between revisions of "The (pre-)measure of a set is no more than the sum of the (pre-)measures of the elements of a covering for that set"

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{{Stub page|I could probably write the statement a bit better}}
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{{Stub page|review=true|msg=I could probably write the statement a bit better<br/>'''Page review:'''<br/>The proof needs to be checked before this page may be considered no longer a stub}}
 
==[[The (pre-)measure of a set is no more than the sum of the (pre-)measures of the elements of a covering for that set/Statement|Statement]]==
 
==[[The (pre-)measure of a set is no more than the sum of the (pre-)measures of the elements of a covering for that set/Statement|Statement]]==
 
{{:The (pre-)measure of a set is no more than the sum of the (pre-)measures of the elements of a covering for that set/Statement}}
 
{{:The (pre-)measure of a set is no more than the sum of the (pre-)measures of the elements of a covering for that set/Statement}}
 
==Proof==
 
==Proof==
{{Requires proof|See outline}}
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Suppose that {{MSeq|A_i|i|1|n|in=\mathcal{R} }} is a finite [[sequence]], in this case we shall consider the ''[[countably infinite]]'' [[sequence]]:
'''Outline:'''
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* {{MSeq|A_k'|k|1|\infty|in=\mathcal{R} }} given by {{M|1=A_k':=A_k}} when {{M|k\le n}} and {{M|1=A_k'=\emptyset}} otherwise.
# Construct a new sequence, {{M|(B_i)}} where the elements are [[pairwise disjoint]]
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As such we need only prove the statement for infinite sequences (as we implicitly associate each finite sequence with the corresponding infinite sequence by the above construction) ({{WLOG}})
# Define {{M|B}} as the union of all {{M|(B_i)}}
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# As {{M|A\subseteq B}} we can use the [[monotonic|monotonicity]] of a measure/pre-measure.
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'''Proof:'''
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* Let {{M|A\in\mathcal{R} }} be given.
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** Let {{MSeq|A_n|in=\mathcal{R} }} be given such that {{M|1=A\subseteq\bigcup_{n=1}^\infty A_n}}
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*** Define a new sequence: {{MSeq|B_n}} where {{M|1=B_n:=A\cap A_n}}.
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**** Notice that {{M|1=\bigcup_{n=1}^\infty B_n=A}}<ref group="Note">Should I bother to prove this?</ref>
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**** By hypothesis, {{M|A\in\mathcal{R} }}, and so is each {{M|A_n}}. We may invoke [[a (pre-)measure is sigma-subadditive]] which states:
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***** Given {{MSeq|A_n|in=\mathcal{R} }} (with {{M|1=\bigcup_{n=1}^\infty A_n\in\mathcal{R} }} - which isn't guaranteed to be true for [[pre-measure|pre-measures]]) then:
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****** {{M|1=\mu\left(\bigcup^\infty_{n=1}A_n\right)\le\sum^\infty_{n=1}\mu(A_n)}}
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**** Applying this to the sequence {{MSeq|B_n}} we see:
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***** {{M|1=\mu(A)=\mu\left(\bigcup^\infty_{n=1}B_n\right)\le\sum^\infty_{n=1}\mu(B_n)=\sum^\infty_{n=1}\mu(A\cap A_n)}}
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*** Now we have: {{M|1=\mu(A)\le\sum^\infty_{n=1}\mu(A\cap A_n)}}
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*** By [[monotonicity of (pre-)measures]]<ref group="Note">Monotonicity is also proved on the [[pre-measure]] page</ref> we know that:
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**** {{M|1=\forall A,B\in\mathcal{R}[A\subseteq B\implies\mu(A)\le\mu(B)]}}
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*** By [[the intersection of sets is a subset of each set]] we know that {{M|A\cap A_n\subseteq A_n}} (and, less importantly, {{M|A\cap A_n\subseteq A}})
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*** Combining these we see that {{M|\mu(A\cap A_n)\le\mu(A_n)}}
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*** Thus {{M|1=\sum^\infty_{n=1}\mu(A\cap A_n)\le\sum^\infty_{n=1}\mu(A_n)}}<ref group="Note">Adding a proof wouldn't hurt, although it is REALLY obvious, a construction ought to be given though. For any readers following this note because they cannot see the step, note:
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* {{M|\mu(A\cap A_i)\le\mu(A_i)}}, then {{M|1=\mu(A\cap A_1)+\mu(A\cap A_2)+\ldots \le \mu(A_1)+\mu(A\cap A_2)+\ldots\le\mu(A_1)+\mu(A_2)+\mu(A\cap A_3)+\ldots}} and so forth.
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* Remember also that {{M|1=\sum^\infty_{n=1}a_n=\mathop{\text{Lim} }_{n\rightarrow\infty}\left(\sum_{i=1}^n a_i\right)=\mathop{\text{Lim} }_{n\rightarrow\infty}\left(a_1+a_2+\ldots+a_n\right)}}</ref>
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*** Combine this with {{M|1=\mu(A)\le\sum^\infty_{n=1}\mu(A\cap A_n)}} from before and we see:
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**** {{M|1=\mu(A)\le\sum^\infty_{n=1}\mu(A\cap A_n)\le\sum_{n=1}^\infty\mu(A_n)}}
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*** By the [[transitive relation|transitive property]] of a [[partial order]], we conclude:
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*** {{M|1=\mu(A)\le\sum^\infty_{n=1}\mu(A_n)}}
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** Since {{MSeq|A_n|in=\mathcal{R} }} was an arbitrary sequence with {{M|1=A\subseteq\bigcup_{n=1}^\infty A_n}} we have shown this for all such sequences
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* Since {{M|A\in\mathcal{R} }} was arbitrary, we have shown this for all {{M|A\in\mathcal{R} }}
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This completes the proof.
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==Notes==
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<references group="Note"/>
 
==References==
 
==References==
 
<references/>
 
<references/>
 
{{Measure theory navbox|plain}}
 
{{Measure theory navbox|plain}}
 
{{Theorem Of|Measure Theory}}
 
{{Theorem Of|Measure Theory}}

Latest revision as of 21:03, 31 July 2016

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I could probably write the statement a bit better
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Statement

Suppose that [ilmath]\mu[/ilmath] is either a measure (or a pre-measure) on the [ilmath]\sigma[/ilmath]-ring (or ring), [ilmath]\mathcal{R} [/ilmath] then[1]:

  • for all [ilmath]A\in\mathcal{R} [/ilmath] and for all countably infinite or finite sequences [ilmath](A_i)\subseteq\mathcal{R} [/ilmath] we have:
    • [ilmath]A\subseteq\bigcup_i A_i\implies\mu(A)\le\sum_{i}\mu(A_i)[/ilmath]

Note: this is slightly different to sigma-subadditivity (or subadditivity) which states that [ilmath]\mu\left(\bigcup_i A_i\right)\le\sum_i\mu(A_i)[/ilmath] (for a pre-measure, we would require [ilmath]\bigcup_i A_i\in\mathcal{R} [/ilmath] which isn't guaranteed for countably infinite sequences)

Proof

Suppose that [ilmath] ({ A_i })_{ i = 1 }^{ n }\subseteq \mathcal{R} [/ilmath] is a finite sequence, in this case we shall consider the countably infinite sequence:

  • [ilmath] ({ A_k' })_{ k = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] given by [ilmath]A_k':=A_k[/ilmath] when [ilmath]k\le n[/ilmath] and [ilmath]A_k'=\emptyset[/ilmath] otherwise.

As such we need only prove the statement for infinite sequences (as we implicitly associate each finite sequence with the corresponding infinite sequence by the above construction) (Template:WLOG)

Proof:

  • Let [ilmath]A\in\mathcal{R} [/ilmath] be given.
    • Let [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] be given such that [ilmath]A\subseteq\bigcup_{n=1}^\infty A_n[/ilmath]
      • Define a new sequence: [ilmath] ({ B_n })_{ n = 1 }^{ \infty } [/ilmath] where [ilmath]B_n:=A\cap A_n[/ilmath].
        • Notice that [ilmath]\bigcup_{n=1}^\infty B_n=A[/ilmath][Note 1]
        • By hypothesis, [ilmath]A\in\mathcal{R} [/ilmath], and so is each [ilmath]A_n[/ilmath]. We may invoke a (pre-)measure is sigma-subadditive which states:
          • Given [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] (with [ilmath]\bigcup_{n=1}^\infty A_n\in\mathcal{R}[/ilmath] - which isn't guaranteed to be true for pre-measures) then:
            • [ilmath]\mu\left(\bigcup^\infty_{n=1}A_n\right)\le\sum^\infty_{n=1}\mu(A_n)[/ilmath]
        • Applying this to the sequence [ilmath] ({ B_n })_{ n = 1 }^{ \infty } [/ilmath] we see:
          • [ilmath]\mu(A)=\mu\left(\bigcup^\infty_{n=1}B_n\right)\le\sum^\infty_{n=1}\mu(B_n)=\sum^\infty_{n=1}\mu(A\cap A_n)[/ilmath]
      • Now we have: [ilmath]\mu(A)\le\sum^\infty_{n=1}\mu(A\cap A_n)[/ilmath]
      • By monotonicity of (pre-)measures[Note 2] we know that:
        • [ilmath]\forall A,B\in\mathcal{R}[A\subseteq B\implies\mu(A)\le\mu(B)][/ilmath]
      • By the intersection of sets is a subset of each set we know that [ilmath]A\cap A_n\subseteq A_n[/ilmath] (and, less importantly, [ilmath]A\cap A_n\subseteq A[/ilmath])
      • Combining these we see that [ilmath]\mu(A\cap A_n)\le\mu(A_n)[/ilmath]
      • Thus [ilmath]\sum^\infty_{n=1}\mu(A\cap A_n)\le\sum^\infty_{n=1}\mu(A_n)[/ilmath][Note 3]
      • Combine this with [ilmath]\mu(A)\le\sum^\infty_{n=1}\mu(A\cap A_n)[/ilmath] from before and we see:
        • [ilmath]\mu(A)\le\sum^\infty_{n=1}\mu(A\cap A_n)\le\sum_{n=1}^\infty\mu(A_n)[/ilmath]
      • By the transitive property of a partial order, we conclude:
      • [ilmath]\mu(A)\le\sum^\infty_{n=1}\mu(A_n)[/ilmath]
    • Since [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] was an arbitrary sequence with [ilmath]A\subseteq\bigcup_{n=1}^\infty A_n[/ilmath] we have shown this for all such sequences
  • Since [ilmath]A\in\mathcal{R} [/ilmath] was arbitrary, we have shown this for all [ilmath]A\in\mathcal{R} [/ilmath]

This completes the proof.

Notes

  1. Should I bother to prove this?
  2. Monotonicity is also proved on the pre-measure page
  3. Adding a proof wouldn't hurt, although it is REALLY obvious, a construction ought to be given though. For any readers following this note because they cannot see the step, note:
    • [ilmath]\mu(A\cap A_i)\le\mu(A_i)[/ilmath], then [ilmath]\mu(A\cap A_1)+\mu(A\cap A_2)+\ldots \le \mu(A_1)+\mu(A\cap A_2)+\ldots\le\mu(A_1)+\mu(A_2)+\mu(A\cap A_3)+\ldots[/ilmath] and so forth.
    • Remember also that [ilmath]\sum^\infty_{n=1}a_n=\mathop{\text{Lim} }_{n\rightarrow\infty}\left(\sum_{i=1}^n a_i\right)=\mathop{\text{Lim} }_{n\rightarrow\infty}\left(a_1+a_2+\ldots+a_n\right)[/ilmath]

References

  1. Measure Theory - Paul R. Halmos