8
$\begingroup$

Prove that if $\mu_1, \mu_2, \dots$ are measures on a measurable space and $a_1, a_2, \dots \in [0,\infty)$, then $\sum_{n=1}^\infty a_n\mu_n$ is also a measure.

I need some help justifying the third equality in the final line of my proof. My idea was to break this into finite and infinite cases and use facts about absolute convergence, but I'm not sure of the details.

My solution:

First, we let $(X, \mu_n, \mathcal A)$ be a measure space for all $n\in \mathbb N$ and define $\nu_n = a_n\mu_n$. Since $\nu_n(\emptyset) = a_n\mu_n(\emptyset) = a_n\cdot0=0$ and $$\nu_n(\cup_{i=1}^\infty A_i) = a_n\mu_n(\cup_{i=1}^\infty A_i) = a_n\sum_{i=1}^\infty \mu_n(A_i) = \sum_{i=1}^\infty a_n\mu_n(A_i) = \sum_{i=1}^\infty \nu_n(A_i),$$ we know that $\nu_n$ is a measure for all $n\in \mathbb N$. So we are reduced to the case that a countable sum of measures is again a measure.

Let $\mu = \sum_{n=1}^\infty \nu_n$. Since $\nu_n(\emptyset) = 0$ for all $n \in \mathbb N$, then $\mu(\emptyset) = \sum_{n=1}^\infty \nu_n(\emptyset) = 0$. So, we show that $\mu$ is countably additive. That is, if $A_i\in \mathcal A$ for all $i \in \mathbb N$ are pairwise disjoint, we show $\mu(\cup_{i=1}^\infty A_i) = \sum_{i=1}^\infty \mu(A_i)$. Then, \begin{align*} \mu(\cup_{i=1}^\infty A_i) &= \sum_{n=1}^\infty \nu_n(\cup_{i=1}^\infty A_i) =\sum_{n=1}^\infty\sum_{i=1}^\infty \nu_n(A_i) = \sum_{i=1}^\infty\sum_{n=1}^\infty \nu_n(A_i) = \sum_{i=1}^\infty \mu(A_i). \\ \end{align*}

$\endgroup$
5
$\begingroup$

Not absolute convergence, the sum can be infinite. But all terms are non-negative, hence we can change the order of summation. Fitting to the theme, we can appeal to the monotone convergence theorem for the counting measure on $\mathbb{N}$ for the justification.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for the response Daniel. I'm familiar with the MCT, but the question appears much earlier in the book than the MCT. You're saying that since the terms are non-negative, this alone is enough to justify the change in the order of summation? $\endgroup$ – dannum May 22 '14 at 21:07
  • $\begingroup$ Yes, with all terms non-negative, you can rearrange as you please, just as if it were absolutely convergent. Have you a theorem you can appeal to for that, or do you need a proof? $\endgroup$ – Daniel Fischer May 22 '14 at 21:12
  • $\begingroup$ Not that result in particular, but just that if the the double sum over the absolute values converges, then we can switch the order. Is the idea behind this that if the double sum is finite, then we can rearrange as we please, and if the double sum is infinite, then the sum in the reverse order will be infinite as well, and we are still free to rearrange? Thanks for your help. $\endgroup$ – dannum May 22 '14 at 21:20
  • $\begingroup$ Yes. If $$S = \sum_{n=0}^\infty \left(\sum_{k=0}^\infty a_{nk}\right) < +\infty$$ for $a_{nk}\geqslant 0$, then we have a summable family. Any sum of a finite subset of the $a_{nk}$ is bounded by $S$, and by taking enough terms we come arbitrarily close. For the case of an infinite sum, you can use that result, if there were any arrangement with a finite sum, then all arrangements have that same finite sum. $\endgroup$ – Daniel Fischer May 22 '14 at 21:26
  • 1
    $\begingroup$ @Anoldmaninthesea. For the original question, the only important thing is that one can change the order of summation and gets the same result. That follows here since all terms are in $[0,+\infty]$. If there is any arrangement that leads to a finite sum, then we have a summable family and hence all arrangements lead to the same finite sum. If no arrangement leads to a finite sum, the result is $+\infty$ regardless of arrangement. $\endgroup$ – Daniel Fischer Dec 6 '17 at 10:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.