Measure with respect to complete measure is also complete

[psie]

TL;DR Summary

I haven't read much about the Radon–Nikodym theorem and the necessary background to understand this theorem, but I was wondering about the following basic definition of a measure; if $f$ is a nonnegative, measurable function, and $(X,\mathcal M,\mu)$ is any measure space, we can define $$\nu(A)=\int_A f\,d\mu$$ for $A\in\mathcal M$. That this is a measure on $\mathcal M$ is not so hard to verify, but I wonder under what conditions this measure is complete.

A measure space $(X,\mathcal M,\mu)$ is complete iff $$S\subset N\in\mathcal M\text{ and }\mu(N)=0\implies S\in\mathcal M.$$The meaning of a complete measure is a measure whose domain includes all subsets of null sets.

Suppose now $\mu$ is complete. Under what conditions is $\nu$ also complete? I've heard different claims about this, which now has cast doubt upon my judgement of what is correct and incorrect. I'd be grateful if anyone could clarify this.

 

[pasmith]

Suppose $\nu$ is not complete. Then there exists a set $N$ such that $\nu(N) = 0$ but there exists $S \subset N$ such that $S \notin \mathcal{M}$. If $\mu(N) = 0$ this is a contradiction, so the only way this can happen is if $\nu(N) = 0$ and $\mu(N) > 0$.

 

[psie]

Suppose $\nu$ is not complete. Then there exists a set $N$ such that $\nu(N) = 0$ but there exists $S \subset N$ such that $S \notin \mathcal{M}$. If $\mu(N) = 0$ this is a contradiction, so the only way this can happen is if $\nu(N) = 0$ and $\mu(N) > 0$.

Hmm, I'm not sure I understand. I was thinking; if $\mu$ is complete and $f>0$ $\mu$-a.e., then we will have $\nu(N)=0\implies \mu(N)=0$ (in fact, the other direction holds too). Hence $$S\subset N\in\mathcal M\text{ and }\nu(N)=0\implies S\subset N\in\mathcal M\text{ and }\mu(N)=0\implies S\in\mathcal M.$$Are we saying the same thing?

 

[martinbn]

There has to be some conditiin on the function, may be on the support. Otherwise any set it is zero on will be null for the new measure and all it subsets need to be in the algebra if the new measure is complete.