Difference between revisions of "User:Alec/Modules/Measure theory"

Latest revision as of 05:20, 19 October 2016

Lecture notes summary

Week 1

• Goal: [ilmath]\mu:\mathcal{P}(\mathbb{R})\rightarrow\[0,\infty]:=\mathbb{R}_{\ge 0}\cup\{+\infty\}[/ilmath]
  • Additivity: [ilmath]A\cap B=\emptyset\implies\mu(A\cup B)=\mu(A)+\mu(B)[/ilmath]
• Task: Find [ilmath]\mu[/ilmath] such that [ilmath]\mu(A)[/ilmath] is defined for all [ilmath]A\in\mathcal{P}(\mathbb{R})[/ilmath] and [ilmath]\mu[/ilmath] is [ilmath]\sigma[/ilmath]-additive
• Problem: if demanding in addition that:
  • [ilmath]\mu(x+A)=\mu(A)[/ilmath] [ilmath]\forall A\in\mathcal{P}(\mathbb{R})\ \forall x\in\mathbb{R} [/ilmath] it is not possible.
  • Claim: [ilmath]\not\exists \mu:\mathcal{P}(\mathbb{R})\rightarrow[0,\infty][/ilmath] such that:
    1. [ilmath]\mu(x+A)=\mu(A)[/ilmath] [ilmath]\forall A\in\mathcal{P}(\mathbb{R}),\ \forall x\in\mathbb{R} [/ilmath]
    2. [ilmath]\mu(\bigcup_{i=1}^\infty A_i)=\sum^\infty_{i=1}\mu(A_i)[/ilmath] if the [ilmath]A_i[/ilmath] are pairwise disjoint
    3. [ilmath]\mu((a,b)) = b-a[/ilmath] for every interval [ilmath](a,b)[/ilmath]
  • Notice if [ilmath]B\subseteq A[/ilmath] then [ilmath]\mu(A)=\mu(B)+\mu(A-B)+\sum_{i=3}^\infty \mu(\emptyset)\ge\mu(B)[/ilmath]
• Vitali's set
  • [ilmath]\exists V\subseteq [0,1][/ilmath] such that all [ilmath]V+r[/ilmath] for [ilmath]r\in\mathbb{Q} [/ilmath] are mutually disjoint and
    • [ilmath]\bigcup_{r\in\mathbb{Q} }(V+r)=\mathbb{R}[/ilmath]
  • Then for [ilmath]\mu[/ilmath] satisfying 1-3 above:
    • Consider a sequence [ilmath] ({ r_i })_{ i = 1 }^{ \infty } [/ilmath] of all rational numbers in [ilmath](-1,1)[/ilmath], then:
      • [ilmath](0,1)\subseteq\bigcup_{i=1}^\infty (r_i+V)\subseteq (-1,2)[/ilmath]
        • (*): [ilmath]\bigcup_{r\in\mathbb{Q} }(V+r)=\mathbb{R}[/ilmath] and [ilmath]\bigcup_{r\in\mathbb{Q}-(-1,1)}(V+r)\cap(0,1)=\emptyset[/ilmath]
        • [ilmath]v_i\in(-1,1)[/ilmath] and [ilmath]V\subset[0,1][/ilmath]
  • Hence:
    1. [ilmath]3\ge\mu(\bigcup_{i=1}^\infty(R_i+v))=\sum_{i=1}^\infty \mu(r_i+V)=\sum^\infty_{i=1}\mu(V)[/ilmath] [ilmath]\implies \mu(V)=0[/ilmath]
    2. [ilmath]1\le\sum \mu(r_i+V)=\sum^\infty_{i=1}\mu(V)=\sum 0=0[/ilmath]
• Proof of existence of Vitali's set:

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