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

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?

• If $A_1$ and $A_2$ are disjoint, you have $\sum\limits_{k \in A_1\cup A_2} a_k = \sum\limits_{k \in A_1} a_k + \sum\limits_{k \in A_2} a_k$. If this is cleat to you, this should help solve your problem ; if you don't see why, maybe try showing that first (which is a series question more that a measure question). Nov 22, 2022 at 17:29
• As long as at least one of the values $-\infty$ or $\infty$ Are not attained, things are kosher and your set function $\mathcal{P}(\mathbb{Z})$ is indeed a signed measure Nov 22, 2022 at 18:49
• @MathMax I've shown that now. Not sure how to use that for the general case though. Nov 22, 2022 at 20:34
• @MathMax I think I got it, actually! I used an argument similar to the one typically used to show that a countable union of countable sets is countable. You should post your comment as an answer so that I can mark it as the one most helpful in finding my solution! :-) Nov 22, 2022 at 21:20

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

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.

• Suppose the sum of all positive $a_k$ is finite. In $*$ could it not be the case that $\sum_{k\in A_{i_0}}a_k = -\infty$ for some $i_0$ (implying that the RHS is $-\infty$) but that, on the LHS, all $a_k$ for $k\in A_{i_0}$ are cancelled out by $a_j$ for some $j\in\cup_iA_i\setminus A_{i_0}$ (and that the LHS turns out to be finite)? Nov 22, 2022 at 18:22
• If the positive part is finite and $\sum_{k\in A_{i_0}}a_k=-\infty$, it is impossible for this $-\infty$ to be cancelled out! The finite positive cannot beat the infinite negative (in measure theory we must also be careful with doing algebra with extended reals) Nov 22, 2022 at 18:25

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.

• This seems helpful, so definitely a vote up. Though MathMax's comment above helped my resolve my question by anpther method. Nov 23, 2022 at 11:36

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.