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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.

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  • $\begingroup$ $\nu$ is just the restriction of $\mu$ in the $\sigma$-algebra $(\Omega, \mathcal F_E, \mathbb P)$ trace of $(\Omega, \mathcal F, \mathbb P).$ $\endgroup$ – Zbigniew Mar 12 '15 at 12:10
  • $\begingroup$ @Zbigniew what do you mean by "trace of"? $\endgroup$ – Mike Brown Mar 12 '15 at 20:40
  • $\begingroup$ @Zbigniew The trace of $\mathcal{F}$ in $\mathcal{F}_{E}$ is defined as $\mathcal{F}_{E}:=\{E\cap A: A\in\mathcal{F}\}$. It is the intersection not the union. $\endgroup$ – Mike Brown Mar 13 '15 at 19:47
  • $\begingroup$ Excuse me for this mistake. In this case why did you not take $\nu(X)=\mu(X)$ if $X\in \mathcal F$ and $\nu(E)$ equal to every real in $[0,1]$? $\endgroup$ – Zbigniew Mar 14 '15 at 6:56
  • $\begingroup$ @Zbigniew. Maybe I have a stupid question:how can you make sure $\nu$ is countable additive measure on $\mathcal{F}_{E}$? $\endgroup$ – Mike Brown Mar 14 '15 at 16:02
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I will also assume that the measure is a probability measure - This doesn't really matter. Let $a = \text{sup} \{\mu(K) : K \subseteq E, K \in \mathcal{F}\}$. Let $K \subseteq E, K \in \mathcal{F}$ be such that $\mu(K) = a$. Note that $\mathcal{F}_E = \{(A \cap E) \bigcup (B \cap E^c) : A, B \in \mathcal{F}\}$. Put $\nu((A \cap E) \bigcup (B \cap E^c)) = $ $\mu(A \cap K) + \mu(B \cap K^c)$. Now check.

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  • $\begingroup$ why such $K$ must exist? $\endgroup$ – user115608 Oct 24 '15 at 16:15

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