I have two questions Question 1 and Question 2 see below.
This is an extension of(Problem 32.9 from the Billiglsey Probability and measure.)
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Let $\mu$ and $ \nu$ be measures on $(\mathscr \Omega , \mathscr F )$ and suppose that $\mathscr F^o$ is a σ-field contained in $\mathscr F$.
Then the restrictions $\mu^ο$ and $ \nu^ο$ of $\mu$ and $\nu$ to $\mathscr F^o$ are measures on $(\mathscr \Omega , \mathscr F^o )$.
Assume that $ \nu^ο$ is σ-finite (which implies that $ \nu$ is also σ-finite). Let $ \nu_{ac}$, $ \nu_{s}$, $ \nu_{ac}^o$, and $ \nu_{s}^o$ be, respectively, the absolutely continuous and singular parts of $ \nu$ and $ \nu^ο$ with respect to $\mu$ and $\mu^ο$.
Part A : Show $ \nu_{ac}^o (E) \ge \nu_{ac} (E) $ and $ \nu_{s}^o (E) \le \nu_{s} (E) $ for $E \in \mathscr F^o$.
I think I managed to solve Part A , since for every $E \in \mathscr F^o$
Case 1 : IF $\nu_{s} (E)=0$ , then $\nu_{s} (E) \ge \nu_{s}^o (E) $
Case 2 : IF $\nu_{s} (E)>0$ , then $\mu^o (E) =\mu^o (E)=0 $ , then $ \nu_{ac}^o (E) = \nu_{ac} (E) =0 $ and $\nu_{s} (E) = \nu (E) - \nu_{ac} (E) = \nu (E) = \nu^o (E) = \nu^o (E) = \nu^o (E) - \nu_{ac}^o (E) = \nu_{s}^o (E) $.
So , $ \nu_{s} (E) \ge \nu_{s}^o (E)$
Question 1 Does this mean $ \nu_{ac}^o (E) \ge \nu_{ac} (E) $ and $ \nu_{s}^o (E) \le \nu_{s} (E) $ for $E \in \mathscr F^o$ ???
My Question 2 is in Part B below:
Part B : I need to give an example to show that strict inequality may hold.
As a hint I should let $\mu$ and $ \nu$ be probability measures measures on a two-point space and then take $\mathscr F^o = {(\emptyset , \Omega )} $
(Problem 32.9 from the Billiglsey Probability and measure.) a short solution for this part which is not complete is the below
For the case where $\mathscr F^o = { (\emptyset , \Omega) } $ , in which $ \nu^o$ is absolutely continouse with respect to $ \mu^o$ ( provided that $ \mu (\Omega) \ge 0 )$ and hence $\nu_{s}^o$ vanishes.
I do not understand this solution, and why it vanishes.