Let $\mu$ be a positive measure and $\nu_1, \nu_2$ arbitrary measures, all defined on the same measurable space $(S,\Sigma)$. We say that two arbitrary measure $\mu, \nu$ are mutually singular (notation $\nu \perp \mu$) if there exist disjoint sets $E$ and $F$ such that $\nu(A)=\nu(A \cap E)$ and $\mu(A) = \mu(A \cap F)$ for all $A \in \Sigma$. We say that $\nu$ is absolutely continuous w.r.t. $\mu$ (notation $\nu \ll \mu$), if $\nu(E) = 0$ for every $E \in \Sigma$ with $\mu(E)=0$.
Now, I want to prove that
- If $\nu_1 \perp \mu$ and $\nu_2 \perp \mu$, then $\nu_1 + \nu_2 \perp \mu$.
- If $\nu_1 \ll \mu$ and $\mu_2 \perp \mu$, then $\nu_1 \perp \nu_2$.
To start with 1.:
$\exists E,F, G,H \in \Sigma$ such that
\begin{align}
\nu_1(A) = \nu_1(A \cap E)\ \text{ and }\ \mu(A) = \mu(A \cap F)\ \text{ for all $A \in \Sigma$}.\\
\nu_2(B) = \nu_2(B \cap G)\ \text{ and }\ \mu(B) = \mu(B \cap H)\ \text{ for all $B \in \Sigma$}.\\
\end{align}
So, for which sets $C,I,J \in \Sigma$ do I have to show that $(\nu_1 + \nu_2)(C) = (\nu_1 + \nu_2)(C \cap I)$ and $\mu(C) = \mu(C \cap J)\ \text{ for all $C \in \Sigma$}$?
Secondly, I do not have any suggestions for 2. Do you have any suggestions?
Best Answer
Suppose $$ \nu_1 \perp \mu,\qquad \nu_2 \perp \mu $$ with given decompositions $$ A_1 \cup B_1 = X,\qquad A_2 \cup B_2 = X. $$ We wish to find a decomposition $A \cup B$ such that $$ (\nu_1 + \nu_2)(B) = 0,\qquad \mu(A) = 0. $$ Given what we know, there is not much room for imagination since if $(\nu_1 + \nu_2)(B) = \nu_1(B) + \nu_2(B) = 0$ we must (in general) assume that $B \subseteq B_1 \cap B_2$. Likewise, if $\mu(A) = 0$ we must have $A \subseteq A_1 \cup A_2$. Since $$ (A_1 \cup A_2) \cup (B_1 \cap B_2) = X, $$ and $$ (A_1 \cup A_2) \cap (B_1 \cap B_2) = \emptyset, $$ we have found a decomposition by taking $A = A_1 \cup A_2$ and $B = B_1 \cap B_2$. Therefore $\nu_1 + \nu_2 \perp \mu$.