Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Define $$ \alpha(E)=\sup\{\mu(F)-\nu(F)\,|\,F\subset E~\text{and}~\nu(F)<\infty\}.$$ where $\mu$ and $\nu$ are measures on some $\sigma$-algebra.

How do I show that $\alpha$ is countably additive?

I've already got one direction. That is $\sum \alpha(E_n)\geq \alpha(\cup E_n).$ I am however struggling with the other direction: showing that $\sum\alpha(E_n)\leq \alpha(\cup E_n)$.

thanks.

share|improve this question
    
I guess the sets $E_n$ must be disjoint. Then, for every positive $\varepsilon$, consider $F_n\subset E_n$ such that $\mu(F_n)-\nu(F_n)\geqslant\alpha(E_n)-\varepsilon/2^n$ and their union $F$. –  Did Jan 20 '12 at 13:24
2  
@Didier: Good start, but of course, it might be that $\nu(F)=\infty$. This objection is not hard to work around, but I thought it worth mentioning. –  Harald Hanche-Olsen Jan 20 '12 at 13:38
1  
@Harald: Of course, comment $\ne$ answer. –  Did Jan 20 '12 at 13:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.