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I am trying to solve the following homework problem, where the notation $\nu \perp \mu$ means that $\nu$ and $\mu$ are mutually singular:

Suppose $\{\nu_j\}$ is a sequence of positive measures. If $\nu_j \perp \mu$ for all $j$, then $\sum_1^\infty \nu_j \perp \mu$; and if $\nu_j \ll \mu$ for all $j$, then $\sum_1^\infty \nu_j \ll \mu$.

I have seen a similar question asked for finite measures, but here we only restrict $\nu_j \ge 0$ for all $j$.

My proof relies on two key points:

  • $\sum_1^\infty \nu_j$ is a measure;
  • $\left(\sum_1^\infty \nu_j\right)(E) \le \sum_1^\infty \nu_j(E)$ for all appropriate $E$.

However, I am having difficulty justifying the second point. All of the references I would use hinge on $\nu_j$ being finite. Some guidance is appreciated.

Emily
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    I presume the definition of $\sum_j v_j$ is $(\sum_j v_j)(E)= \sum_j (v_j(E))$? – copper.hat Mar 25 '15 at 21:56
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    If $E$ is a set, what is your definition of $(\sum_1^\infty \nu_j)(E)$? – Umberto P. Mar 25 '15 at 21:56
  • @copper.hat Since $\sum_1^\infty \nu_j$ is a measure, $\left(\sum_1^\infty \nu_j\right)(E)$ denotes that measure of the set $E$. If it is simpler, let $\lambda = \sum_1^\infty\nu_j$; I need to explore $\lambda(E)$. – Emily Mar 25 '15 at 21:58
  • In other words, is $\left(\sum_1^\infty \nu_j\right)(E) = \sum_1^\infty \nu_j(E)$ by definition and I am just being daft? Unfortunately, I cannot find this notation defined in my textbook. – Emily Mar 25 '15 at 22:00
  • I am asking how you define $(\sum_j v_j)$. If it is defined as the sum of measures of a set then it is indeed a measure. – copper.hat Mar 25 '15 at 22:00
  • I would imagine that it is defined that way. Since it is positive, it is easy to check that it is a measure. – copper.hat Mar 25 '15 at 22:01
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    See http://math.stackexchange.com/questions/805857/countable-sum-of-measures-is-a-measure. The definition of $(\sum_n \nu_n)(E)=\sum_n \nu_n(E)$ – Moya Mar 25 '15 at 22:01
  • @copper.hat I guess it must be. I will have to explore where finiteness is used in other proofs, as my approach seems to be identical but does not require that property. – Emily Mar 25 '15 at 22:02
  • You don't need finiteness. I mean, it may a pretty useless measure if it is $\infty$ everywhere (except for $\emptyset$) but it is still a measure. – copper.hat Mar 25 '15 at 22:05
  • @copper.hat In such a case, I wish I had back my restless 45 minutes of overthinking this problem last night. – Emily Mar 25 '15 at 22:07
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    Mathematics is a dangerous and merciless siren. I have done far worse for far longer. – copper.hat Mar 25 '15 at 22:08
  • @copper.hat That was beautiful. It made me laugh. And cry. And then cry some more. – Moya Mar 25 '15 at 22:12
  • @Moya: I wish I was joking ;-). – copper.hat Mar 25 '15 at 22:13

1 Answers1

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To provide an answer to the question: the sum of positive measures is a measure (not difficult to show), and by convention we write $\left(\sum \nu_j \right)(E) = \sum \nu_j (E)$, where the sum can be finite or infinite, as each term is positive.

Emily
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