Stein 6.10

$\textbf{Exercise 6.10}$ Suppose $\nu_1, \nu_2, \nu_3$ are signed measures and $\mu$ is a positive measure on $(X,\mathcal{M})$. Using the symbols $\perp$ (mutually singular) and $\ll$ (absolutely continuous) prove the following:

(a) If $\nu_1 \perp \mu$ and $\nu_2 \perp \mu $ then $\nu_1 + \nu_2 \perp \mu$
(b) If $\nu_1 \ll \mu$ and $\nu_2 \ll \mu$ then $\nu_1 + \nu_2 \ll \mu$
(c) If $\nu_1 \perp \nu_2$ then $|\nu_1| \perp |\nu_2|$
(d) $\nu \ll |\nu|$
(e) If $\nu \perp \mu $ and $\nu \ll \mu$ then $\nu=0$

$\textit{Proof} : $

(a) Assume that $\nu_1 \perp \mu$ and $\nu_2 \perp \mu $. By the definition given in the book (there are several other equivalent definitions) $\nu_1 \perp \mu$ means that $\nu_1$ and $\mu$ are supported on disjoint subsets. That is, $\exists A,B \in{\mathcal{M}}$ where $A \cap B = \emptyset$ s.t. $$\nu_1(E)=\nu_1(E \cap A) \quad \text{and} \quad \mu(E)=\mu(E \cap B) \quad \forall E \in{\mathcal{M}} $$ $\nu_2 \perp \mu$ means that $\exists C,D \in{\mathcal{M}}$ where $C \cap D = \emptyset$ s.t. $$\nu_2(E)=\nu_2(E \cap C) \quad \text{and} \quad \mu(E)=\mu(E \cap D) \quad \forall E \in{\mathcal{M}} $$ To find disjoint sets of support for $\nu_1+\nu_2$ and $\mu$, we need to pick some combination of $A,B,C,D$ for our disjoint sets. Note that $(A \cup C) \cap (B \cap D)= \emptyset$.
$\mu(E \cap D)=\mu(E)$ means that the $\mu-$measure of any measurable set, intersected with $D$, is the $\mu$-measure of that set. Thus, $$\mu((B \cap D)\cap E) = \mu([E \cap B] \cap D)=\mu[E \cap B]=\mu(E) $$ Also, $$(\nu_1+\nu_2)((A \cup C)\cap E)=\nu_1((A \cup C)\cap E)+\nu_2((A \cup C)\cap E)$$ Since $\nu_1$ and $\nu_2$ are both measures, they are additive on disjoint sets. So, splitting $(A \cup C) \cap E$ into 2 disjoint subsets, we get $$\nu_1((A \cup C) \cap E)=\nu_1((E \cap A) \cup (E \cap (C \cap A^c)))$$ $$=\nu_1((E \cap A)) + \nu_1[E \cap (C \cap A^c)]$$ Since $\nu_1$ is supported on $A$, $$=\nu_1[E \cap (C \cap A^c)] + \nu_1([E \cap (C \cap A^c)] \cap A)= \nu_1(\emptyset)=0$$ Similarly, $$\nu_2((A \cup C) \cap E)=\nu_2((E \cap C) \cup (E \cap (A \cap C^c)))$$ $$=\nu_2(E \cap C) + \nu_2[E \cap (A \cap C^c)]=\nu_2(E)+ \nu_2([E \cap (A \cap C^c)] \cap C)=\nu_2(E)$$ So, we have exhibited two disjoint sets $A \cup C$ and $B \cap D$ s.t. $$(\nu_1+\nu_2)(E) = (\nu_1 + \nu_2)((A \cup C) \cap E) \quad \text{and} \quad \mu(E)=\mu((B \cap D) \cap E) \quad \forall E \in{\mathcal{M}} $$ Therefore, $(\nu_1+\nu_2) \perp \mu$

(b) Assume that $\nu_1 \ll \mu$ and $\nu_2 \ll \mu$ with $E \in{\mathcal{M}}$ and $\mu(E)=0$. Then, $(\nu_1+\nu_2)(E)=\nu_1(E)+\nu_2(E)=0+0=0$. Thus, $\nu_1+\nu_2 \ll \mu$

(c) Assume $\nu_1 \perp \nu_2$. By definition, $\exists A,B \in{\mathcal{M}}$ s.t. $A \cap B = \emptyset$ with $$\nu_1(E) = \nu_1(E \cap A) \quad \text{and} \quad \nu_2(E) = \nu_2(E \cap B) \quad \forall E\in{\mathcal{M}}$$ The variation of a measure is itself a measure and is defined as: $$|\nu_1|(E) \equiv \sup \sum_{j=1}^{\infty}|\nu_1(E_j)|$$ where the supremum is taken over all countable, measurable partitions of $E$. With $\nu_1$ supported on $A$, we get $|\nu_1|(E) \equiv \sup \sum_{j=1}^{\infty}|\nu_1(E_j)|=\sup \sum_{j=1}^{\infty}|\nu_1(E_j \cap A)|=|\nu_1|(E \cap A)$. Similarly, $|\nu_2|(E) \equiv \sup \sum_{j=1}^{\infty}|\nu_2(E_j)|=\sup \sum_{j=1}^{\infty}|\nu_2(E_j \cap B)|=|\nu_2|(E \cap B)$. Thus, $|\nu_1| \perp |\nu_2|$

(d) With the added assumption that $\nu$ is $\sigma$-finite, we get that $$-|\nu| \leq \nu \leq |\nu|$$ So if $|\nu|(E)=0$ this immediately forces $\nu(E) = 0$. \\ Even without this condition, assuming $|\nu|(E) \equiv \sup \sum_{j =1}^{\infty}|\nu(E_j)|=0$ (where $E_j$'s form a measurable, countable partition of $E$) this implies $|\nu(E_j)|=0 \quad \forall j\in{\mathbb{N}}$. In particular, $E \subseteq E$ is a trivial partition of itself. Thus, $\nu(E)=0$. Therefore $|\nu|(E)=0 \Longrightarrow \nu(E)=0$ i.e. $\nu \ll |\nu|$

(e) Assume $\nu \perp \mu$ and $\nu \ll \mu$. So, $\exists A,B \in{\mathcal{M}}$, $A \cap B = \emptyset$ s.t. $$\nu(E)=\nu(E \cap A) \quad \text{and} \quad \mu(E)=\mu(E \cap B) \quad \forall E \in{\mathcal{M}}$$ Since $\mu$ is supported on $B$, consider $$\mu[E \cap A] = \mu([E \cap A] \cap B)=\mu(\emptyset)=0$$ Since $\nu \ll \mu$ this implies $$\nu[E \cap A] = 0 = \nu(E)$$ i.e. $\nu(E)=0 \quad \forall E \in{\mathcal{M}}$ $\quad \blacksquare$