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Let $\Omega = {\omega_1, \omega_2, ...}$ be some countable set. Let $\mathfrak{F} = 2^{\Omega}$. Consider a sequence {$p_n$} in [0,1] s.t. $\sum_{n=1}^{\infty} p_n = 1$.

Define P: $\mathfrak{F} \to [0,1]$ as follows:

$\forall A \in \mathfrak{F}, P(A) = \sum_{\omega_{n} \in A} p_n$.

Then P is a probability measure. To show, we need to show it satisfies Kolmogorov's axioms.

Pf:

1 $P(\Omega) = 1$

2 $P(A) \geq 0 \ \forall A \in \mathfrak{F}$

3 For a pairwise disjoint collection of sets {$A_k$}$_{k=1}^{\infty}$, $P(\bigcup _{k=1}^{\infty} A_k) = \sum_{k=1}^{\infty} P(A_k)$

To establish $\sigma$-additivity, we must show that for a pairwise disjoint collection of sets {$A_k$}$_{k=1}^{\infty}$, $\sum_{\omega_{n} \in \bigcup _{k=1}^{\infty} A_k} p_n = \sum_{k=1}^{\infty} \sum_{\omega_{n} \in A_k} p_n$.

LHS = $\sum_{\omega_{n} \in \bigcup _{k=1}^{\infty} A_k} p_n$

$= \sum_{\omega_{n} \in \bigcup _{k=1}^{\infty} A_k} p_n (1_{A_1}(\omega_n) + 1_{A_2}(\omega_n) + ...)$

where $1_{A_k}(\omega_n) = 1$ if $\omega_n \in A_k$ and $0$ o/w. That is, $1_{A_1}(\omega_n) + 1_{A_2}(\omega_n) + ... = 1$ iff $\omega_{n} \in \bigcup _{k=1}^{\infty} A_k$

$= \sum_{\omega_{n} \in \bigcup _{k=1}^{\infty} A_k} p_n (\sum_{k=1}^{\infty} 1_{A_k}(n))$

$= \sum_{\omega_{n} \in \bigcup _{k=1}^{\infty} A_k} \sum_{k=1}^{\infty} p_n 1_{A_k}(n)$

$= \sum_{k=1}^{\infty} \sum_{\omega_{n} \in \bigcup _{k=1}^{\infty} A_k} p_n 1_{A_k}(n)$ (*)

$= \sum_{k=1}^{\infty} \sum_{\omega_{n} \in A_k} p_n$ = RHS

QED

(*) What is the justification for this step?

My prof says that $\Sigma_{n=1}^{\infty} \Sigma_{m=1}^{\infty} a_{mn} = \Sigma_{m=1}^{\infty} \Sigma_{n=1}^{\infty} a_{mn}$ holds if $a_{mn} \geq 0$. Is this true? This seems to suggest otherwise. I have a feeling that this step is instead justified by $\sum_{n=1}^{\infty} p_n = 1 < \infty$ rather than $p_n \geq 0$, but then again $\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} p_n = \sum_{m=1}^{\infty} 1 = \infty$.

I find it hard to believe that just because the summand is nonnegative, we can switch order of summation since I recall there were certain conditions to say $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y) dx dy = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y) dy dx$...something like uniform continuity? Definitely not just f being $\geq 0$.

BCLC
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  • http://en.wikipedia.org/wiki/Absolute_convergence justfies what your professor says (she/he is right), indeed positivity of a measurable function always justifies interchange of orders of summation. – Adam Hughes Jul 11 '14 at 04:23
  • @AdamHughes You sure? I don't see anything like that here. http://www.math.ubc.ca/~feldman/m321/twosum.pdf

    What about for integrals?

     – BCLC Jul 11 '14 at 04:31
  • Yes, that's the first theorem on the page, for positive sums you can interchange the order of summation. Integrals are limits of sums, so the same holds for them. – Adam Hughes Jul 11 '14 at 04:35
  • @AdamHughes 1 Do you mean that the hypothesis is equivalent to saying $a_{jk} \geq 0$? 2 So swapping integration order is justified for $f \geq 0$ ? – BCLC Jul 11 '14 at 04:47
  • summable and positive, yes. there's a proof out there somewhere (I forget at the moment where) that the positive sum converging is equivalent to being able to interchange sums however you like. – Adam Hughes Jul 11 '14 at 04:50
  • @AdamHughes Wait, you're saying yes to both questions? – BCLC Jul 11 '14 at 05:11
  • Assuming you mean that arbitrary order of summation is justified, then I am saying this is true when your link (and mine) say it is: namely when the positive sum converges for any order of summation. – Adam Hughes Jul 11 '14 at 05:13
  • @AdamHughes Apparently convergence is not needed: Tonelli's theorem for non-negative functions – BCLC Sep 21 '15 at 09:38
  • yes, I meant for the limit to exist. – Adam Hughes Sep 21 '15 at 16:32

1 Answers1

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If $a_{mn} \geq 0$, then

$$\sum_m \sum_n a_{mn} = \sum_n \sum_m a_{mn}$$

by Tonelli's theorem for non-negative functions

Convergence is not necessary.

BCLC
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