Difference between revisions of "Talk:Extending pre-measures to outer-measures"

Latest revision as of 22:24, 22 May 2016

Proving it extends problem

(These notes are being made before bed) The problem I'm having is showing $\bar{\mu}(A)\le\mu^*(A)$, I have worked out I need to do something involving two infimums. I know that for $A\in\mathcal{R}$ and a $({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{R}$ such that $A\subseteq\bigcup_{n=1}^\infty A_n$ we have $\bar{\mu}(A)\le\sum^\infty_{n=1}\bar{\mu}(A_n)$. However we have:

• $\mu^*(A):=\text{inf}\underbrace{\left\{\left.\sum^\infty_{n=1}\bar{\mu}(A_n)\right\vert (A_n)_{n=1}^\infty\subseteq\mathcal{R}\wedge A\subseteq\bigcup_{n=1}A_n^\infty\right\} }_{\text{exactly the conditions for }\bar{\mu}(A)\le\sum^\infty_{n=1}\bar{\mu}(A_n) }$

But I am struggling to form a statement along the lines of "if we have a set which has members $\le$ every member in $\left\{\left.\sum^\infty_{n=1}\bar{\mu}(A_n)\right\vert (A_n)_{n=1}^\infty\subseteq\mathcal{R}\wedge A\subseteq\bigcup_{n=1}A_n^\infty\right\}$ how can I show the $\text{inf}$ of that set is $\le$ the $\text{inf}$ of $\left\{\left.\sum^\infty_{n=1}\bar{\mu}(A_n)\right\vert (A_n)_{n=1}^\infty\subseteq\mathcal{R}\wedge A\subseteq\bigcup_{n=1}A_n^\infty\right\}$?" I remember doing this once before. I cannot recall what I did. A nudge in the right direction would be useful. Oh wait. I may have just got it. If I use the "epsilon definition" of an infimum which is something like (for $a=\text{inf}(X)$):

• $\forall x\in X[a\le x]$ AND
• $\forall\epsilon>0\exists y\in X[a+\epsilon>y]$

(I'm nearly falling asleep) then I can probably combine this with the epsilon-version of Greater than or equal to Alec (talk) 23:47, 9 April 2016 (UTC)

Outline of proof

• 

  $\mu^*$ is an extension

• 

  $\mu^*$ is an outer measure

Added part 2 of proof Alec (talk) 22:24, 22 May 2016 (UTC)