absolutely continuous ($\sigma$-) finite measures

Question

I got two questions:

1. If I have two measures $\nu$ and $\mu$ with $\nu << \mu$ and we know that $\nu$ is (i) finite (e.g. a probability measure) or (ii) $\sigma$-finite. Can we draw any conclusion about $\mu$ in terms of it being also finite or $\sigma$ finite?

2. My other question is the following: Given a probability measure $Q$ on $(\mathcal{X}, \mathcal{A})$ and a dominating measure $\mu$, then we have the Radon-Nikodym derivative $q=dQ/d\mu$. In a book I am reading I found that $Q(\{x\in \mathcal{X}: q(x)>0\})=1$. Why does this hold? Is it necessary that $\mu$ is $\sigma$ finite here?

Answer 1

1. No we cannot conclude from $\nu << \mu$ that $\mu$ is finite/$\sigma$-finite from $\nu$ finite. For example, let $\mu$ be Lebesgue measure on $\mathbb{R}$, and for an $\mu-$integrable $f$ with $f>0$, $\int_{\mathbb{R}}f d\mu = 1$, let $\nu(A) = \int_A f d\mu$. The $\mu$ is not finite, but $\nu$ is, and $\nu << \mu$.

The only examples I'm aware of of non-$\sigma$-finite measures are on the pathological side, but it is evident that we can come up with silly counter examples here too. For example, we might let $\mu$ be counting measure on $\mathbb{R}$, and let $\nu(A)=0$ for all $A$ (trivial measure). Then $\nu << \mu$, but $\mu$ is not $\sigma-$finite.

2. Let $A = \{ q(x) > 0\}$, then $Q(A) = \int_A q(x)d\mu$. But we must also have since $Q$ is a probability measure that, $1=Q(\mathcal{X})=Q(A)+ Q(A^c)$.

Notice that we must have that $Q(A^c) = \int_{A^c} q(x)d\mu = 0$. $A^c = \{q(x)\le 0\}$. Let $B_n = \{ q(x)\le -1/n\}$. Then we must have $\mu(B_n)=0$, since if $\mu(B_n)>0$, $$Q(B_n)= \int_{B_n}q(x)d\mu(x) \le-\mu(B_n)/n <0,$$ contradicting the positivity of $Q$. Since $B= \{q(x)<0\}=\cup_{n=1}^\infty B_n$, $0\le \mu(B) \le \sum_{n=1}^\infty \mu(B_n) = 0$. Therefore $Q(B)=0$, and since $A^c/B=\{q=0\}$, $Q(A^c)= Q(B)+ Q(A^c/B)= 0+ \int_{\{q=0\}}q(x)d\mu(x)=0$.

Therefore $Q(A)=1$.