# Extension of measure on sigma-algebra

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1. Mar 12, 2015

### Mike.B

Suppose $\mu:\mathcal{F}\rightarrow[0,\infty)$ be a countable additive measure on a $\sigma$-algebra $\mathcal{F}$ over a set $\Omega$. Take any $E\subseteq \Omega$. Let $\mathcal{F}_{E}:=\sigma(\mathcal{F}\cup\{E\})$. Then, PROVE there is a countable additive measure $\nu:\mathcal{F}_{E}\rightarrow [0,\infty)$ such that $\nu(A)=\mu(A)$ for any $A\in\mathcal{F}$. I already know the measure extension theorem. But it is based on algebra and the extension is only about $\sigma(\mathcal{F})$. Can someone give me hint? Or how can I make use of measure extension theorem.

Last edited: Mar 12, 2015
2. Mar 14, 2015

### andrewkirk

I think it may be possible to show that $\mathscr{F}_E=\sigma(\mathscr{F}\cup\{E\})=\mathscr{F}\cup\mathscr{A}_\cup\cup\mathscr{A}_\cap\cup\mathscr{A}_\smallsetminus$
where
$\mathscr{A}_\cup\equiv\{S\cup E\vert S\in \mathscr{F}\}$
$\mathscr{A}_\cap\equiv\{S\cap E\vert S\in \mathscr{F}\}$
$\mathscr{A}_\smallsetminus\equiv\{S\smallsetminus E\vert S\in \mathscr{F}\}$

I can't see a way to use the Caratheodory Extension Theorem here, because the point of that is in going from finite to countably infinite unions, and that is not the obstacle, if the above is correct.

But what if we were to define measure $\nu$ on $\mathscr{F}_E$ by
$\nu(U)\equiv \sup\{\mu(V)\vert V\in\mathscr{F}\wedge V\subseteq U\}$

This clearly matches $\mu$ on $\mathscr{F}$. I think it may be possible to show that it matches on
$\mathscr{A}_\cup,\mathscr{A}_\cap,\mathscr{A}_\smallsetminus$, taking them one at a time.

It then remains to show that $\nu$ is a countably additive measure, which again I think should be possible.

3. Mar 15, 2015

### andrewkirk

We need one more component of $\mathscr{F}_E$, which is $\mathscr{A}_{\cup\smallsetminus}\equiv\{S\cup (\Omega\smallsetminus E)\vert S\in \mathscr{F}\}$. It is pretty straightforward to show that $\mathscr{F}_E=\sigma(\mathscr{F}\cup\{E\})=\mathscr{F}\cup\mathscr{A}_\cup\cup\mathscr{A}_\cap\cup\mathscr{A}_\smallsetminus\cup \mathscr{A}_{\cup\smallsetminus}$ and the union is disjoint.