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 [ilmath]\bar{\mu}(A)\le\mu^*(A)[/ilmath], I have worked out I need to do something involving two infimums. I know that for [ilmath]A\in\mathcal{R} [/ilmath] and a [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] such that [ilmath]A\subseteq\bigcup_{n=1}^\infty A_n[/ilmath] we have [ilmath]\bar{\mu}(A)\le\sum^\infty_{n=1}\bar{\mu}(A_n)[/ilmath]. However we have:

• [ilmath]\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) }[/ilmath]

But I am struggling to form a statement along the lines of "if we have a set which has members [ilmath]\le[/ilmath] every member in [ilmath]\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\}[/ilmath] how can I show the [ilmath]\text{inf} [/ilmath] of that set is [ilmath]\le[/ilmath] the [ilmath]\text{inf} [/ilmath] of [ilmath]\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\}[/ilmath]?" 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 [ilmath]a=\text{inf}(X)[/ilmath]):

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

(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)