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

  1. If $\nu_1 \perp \mu$ and $\nu_2 \perp \mu$, then $\nu_1 + \nu_2 \perp \mu$.
  2. 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?

  • $\begingroup$ When we adopt Schilling's definition of mutual singular measures: two measure $\mu, \nu$ are called mutually singular if there is a set $N \in \Sigma$ such that $\nu(N) = 0 = \mu(N^c)$, statement 2. follows quite easily. Statement 1. stays unclear for me. $\endgroup$
    – iJup
    Dec 29, 2016 at 12:30

1 Answer 1


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 $\nu_1 + \nu_2 \perp \mu$.

  • $\begingroup$ Do $A_1,B_1$; $A_2,B_2$ and $A,B$ have to be disjoint? $\endgroup$
    – iJup
    Dec 30, 2016 at 9:43
  • $\begingroup$ You use that $A \cap B = \emptyset$. So do $A_1,B_1$ and $A_2,B_2$ have to be disjoint? $\endgroup$
    – iJup
    Dec 30, 2016 at 10:00
  • $\begingroup$ I'm using what you call the Schilling definition, writing $A = N, B = N^c$. $\endgroup$ Dec 30, 2016 at 10:57
  • $\begingroup$ where could i read more about this Schilling definition? $\endgroup$
    – Brofessor
    Jun 19, 2019 at 7:29

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