User:Alec/Modules/Measure theory

Lecture notes summary

Week 1

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

This is so hard to read