How to prove $\nu(A) := \sum_{k\in A}a_k$ is a signed measure on $(\mathbb{N},\mathcal{P}(\mathbb{N}))$

Question

I'm currently reading about signed measures. In doing so, we early on give an example of a signed measure (before the Hanh or Lebesgue-Radon-Nikodym Decomposition theorems).

In particular, we let let $X = \mathbb{N}$ with all subsets being measurable. Then we consider any sequence $(a_k)$ in $\mathbb{R}^*$ which is such that either the sum of its positive terms or the sum of its negative terms is finite. Then for measurable $A$ we let $\nu(A) = \sum_{k\in A}a_k$.

I've managed to convince myself that under these conditions, $\nu$ is well defined. Also, I've shown that $\nu(\emptyset) = 0$ and that $\nu$ assumes at most one of the values $\pm\infty$. I am stuck on the last condition though. That is, showing that for any sequence $(A_i)$ of disjoint sets of $\mathcal{M}$, one has $$\nu(\bigcup_iA_i) = \sum_i\nu(A_i).$$

I've tried splitting into cases like (A) $\cup_iA_i$ is finite; (B) all $A_i$ are finite, etc. but I always get stuck as soon as things become complicated enough, e.g. whenever there is some infinite $A_i$. How should I proceed?

Answer 1

MathMax's comment cleared things up. I post it here for future reference.

If $A_1$ and $A_2$ are disjoint then $$\sum_{k\in A_1\cup A_2}a_k = \sum_{k\in A_1}a_k +\sum_{k\in A_2}a_k.$$ Having showed this, in order to show the general statement, one may use an argument similar to the one typically used to prove that a countable union of countable sets is countable.

Answer 2

$$\nu\left(\bigsqcup_iA_i\right)=\sum_{k\in\bigsqcup_i A_i}a_k\overset{\ast}{=}\sum_i\sum_{k\in A_i}a_k=\sum_i\nu(A_i)$$

The step marked $\ast$ uses the fact that the $A_\bullet$ are disjoint. A $k$ is in $\bigsqcup_i A_i$ iff. it is in exactly one $A_{i_k}$ for some $i_k\in\Bbb N$, and (using the empty sum equals zero convention) that is precisely saying that we can sum over the $A_i$ one-by-one. This partitioning of summation trick is common and useful.

Take a moment to check for yourself that the RHS of $\ast$:

• Sums an $a_k$ for every $k\in\bigsqcup_i A_i$
• Never sums the same $a_k$ twice

Then the equality follows.

Answer 3

In general if $\mu$ and $\lambda$ are positive measures on space $(\Omega,\mathcal A)$ and at least one of them is a finite measure then it can be shown that $\nu$ is a signed measure on $(\Omega,\mathcal A)$ if it is defined like this:$$\nu(A)=\mu(A)-\lambda(A)\text{ for every }A\in\mathcal A\tag1$$This under the convention that $\infty-c=\infty$ and $c-\infty=-\infty$ for any constant $c\in\mathbb R$.

(Can you prove this yourself?)

Now construct $\mu$ and $\lambda$ by stating that $\mu(A)=\sum_{k\in A}\max(0,a_k)$ and $\lambda(A)=\sum_{k\in A}\max(0,-a_k)$.

It is not difficult to see that both are positive measures and the condition on the $a_k$ assures that at least one of them is a finite measure.

Further note that in this situation $(1)$ is a true statement for function $\nu$ as mentioned in your question.