Talk:Extending pre-measures to outer-measures

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

I have completed an outline of one part of the proof, that it is indeed an extension. It's really easy once one is armed with passing to the infimum and another theorem, both of which are used (in lighter grey/blue on the original) on the image Alec (talk) 22:35, 19 May 2016 (UTC)