# Measure on the power set of the natural number

The problem is the following.

Let $$X = \mathbb{N}$$, and $$A = P(X)$$(A is the power set of the natural number). Fix some sequence $$(a_n)_{n=1}^{\infty} \subset [0, \infty)$$ such that $$\sum^\infty_{n=1} a_n < \infty$$. Define

$$\mu(A) = \sum_{n \in A} a_n$$ Show that $$\mu$$ is a measure.

The condition I am having trouble with is the countable additivity.

So I would have to show that for disjoint collection $$(A_n)_{n = 1}^\infty \subset P(X)$$, we have

$$\mu(\cup A_n) = \sum_{n \in \cup A_n} a_n = \sum_{i = 1}^\infty \mu(A_i) = \sum_{i = 1}^\infty \sum_{n \in A_i} a_n$$

Intuitively, this seems obvious as $$A_i$$ forms a partition, so summing over the partition and then adding them again should yield the same result. But I don't know how to write this intuition out mathematically, and I also know that intuition can be wrong with things like infinite sums. So anyone could help me on writing the proof out, I would be really grateful.

Thank you!

• @SL_MathGuy, seems like a completely different question. Am I missing something?
– Phil
Commented Jan 9, 2020 at 21:38
• How have you defined $\sum_{n\in A}a_n$? Is it a limit of some partial sums? A supremum of finite sums? Commented Jan 10, 2020 at 0:28
• @MiloBrandt, It wasn't specified in class, but I guess I would have to go with supremum of finite sums.
– Phil
Commented Jan 10, 2020 at 0:44
• Does this answer your question? Counting measure proof Commented Jan 10, 2020 at 9:08

There's a nice way to show this if one has defined, for non-negative $$a_n$$ $$\sum_{n\in S}a_n = \sup\left\{\sum_{n\in S'}a_n:S'\text{ is a finite subset of }S\right\}.$$ Of course, it is not too hard to show that this is true from any other definition of a sum, since any quantity less than the supremum is eventually exceeded by a partial sum - therefore by every partial sum thereafter - which, together with the fact that the supremum is an upper bound, implies convergence.

After this step, we only need the following properties of the supremum:

1. The supremum is an upper bound. If $$x\in X$$ then $$x\leq \sup X$$.

2. The supremum is the least upper bound. If $$x \leq c$$ for every $$x\in X$$ then $$\sup X \leq c$$.

3. Constants can be taken outside a supremum. For any set $$X$$ and any $$c$$ we have $$\sup(X+c)=(\sup X)+c$$.

Notably, this means that our proof will proceed without any $$\varepsilon$$'s, which is neat - and, in fact, I promise that there is no interesting step of the proof: we are just going to shuffle symbols around to show that this statement is true. Our argument essentially amounts to the fact that a set is a finite subset of $$\bigcup A_i$$ exactly if it intersects a finite number of sets $$A_i$$ and intersects each $$A_i$$ at only finitely many points.

We want to show that if $$I$$ is some indexing set and $$A_i$$ is a set of pairwise disjoint sets and for each $$n\in A_i$$ there is some $$a_n \in [0,\infty]$$, we have $$\sum_{n\in \bigcup A_i}a_n = \sum_{i\in I}\sum_{n\in A_i}a_n.$$ We can do this by showing two inequalities.

Direction 1: "Finite sums of a union are finite sums of finite sums of elements"

First, let $$a_{1}+a_2+\ldots+a_k$$ be some sum of finitely many elements of $$\bigcup A_i$$. Let $$I'$$ be the set of indices $$i\in I$$ such that $$A_i$$ contains at least one of the $$a_n$$. We then observe that $$a_1+a_2+\ldots+a_k\leq \sum_{i\in I'}\sum_{n\in A_i}a_n \leq \sum_{i\in I}\sum_{n\in A_i}a_n$$ because the terms $$\sum_{n\in A_i}a_n$$ are individually upper bounds to sums of finitely many elements from $$A_i$$ - and the sum $$a_1+\ldots+a_k$$ is necessarily a sum of sums of finitely many elements from each $$A_i$$. For instance, if $$a_1$$ and $$a_2$$ came from $$A_1$$ and $$a_3$$ and $$a_4$$ came from $$A_2$$, this would say that $$a_1+a_2+a_3+a_4$$ is bounded above by $$\sum_{n\in A_1}a_n + \sum_{n\in A_2}a_n$$ because it may be grouped as $$(a_1+a_2)+(a_3+a_4)$$ and each group is subject to a bound via the supremum.

In any case, taking the supremum of the left hand side yields $$\sum_{n\in \bigcup A_n}a_n \leq \sum_{i\in I}\sum_{n\in A_i}a_n.$$

Direction 2: "Finite sums of finite sums of elements are finite sums of a union"

For this direction, we start by considering an arbitrary finite subset $$I'\subseteq I$$. For each $$i\in I'$$, let us consider a finite subset $$A'_i$$ of $$A_i$$. Observe that $$\bigcup_{i\in I'} A'_i$$ is a finite subset of $$\bigcup_{i\in I} A_i$$, therefore that $$\sum_{i\in I'}\sum_{n\in A'_i}a_n\leq \sum_{n\in \bigcup_I A_i}a_n$$ Note that we can fix some $$i_0\in I'$$ and take the supremum of this inequality over all finite subsets of $$A_{i_0}$$ to get $$\sum_{n\in A_{i_0}}a_n + \sum_{i\in I'\setminus \{i_0\}}\sum_{n\in A'_i}a_n\leq \sum_{n\in \bigcup_I A_i}a_n$$ where essentially the only difference is that a sum over $$A'_{i_0}$$ has been changed to a sum over $$A_{i_0}$$ by taking the supremum over all possible $$A'_{i_0}$$.

Since $$I'$$ is finite, we can proceed inductively, taking the supremum over all finite subsets $$A'_i$$ of each $$A_i$$ one at a time to get $$\sum_{i\in I'}\sum_{n\in A_i}a_n\leq \sum_{n\in \bigcup_I A_i}a_n.$$ Then we can take the supremum of this inequality over all $$I'$$ to get $$\sum_{i\in I}\sum_{n\in A_i}a_n \leq \sum_{n\in \bigcup_I A_i}a_n.$$

Having shown both inequalities, we may then conclude that $$\sum_{i\in I}\sum_{n\in A_i}a_n = \sum_{n\in \bigcup_I A_i}a_n.$$

• Thank you so much!
– Phil
Commented Jan 14, 2020 at 16:32

To show that $$\sum_{n \in \cup A_i} a_n = \sum_{i = 1}^\infty \sum_{n \in A_i} a_n$$ you can show both

$$\sum_{n \in \cup A_i} a_n \le \sum_{i = 1}^\infty \sum_{n \in A_i} a_n$$

and

$$\sum_{n \in \cup A_i} a_n \ge \sum_{i = 1}^\infty \sum_{n \in A_i} a_n$$

For both of these you can use proof by contradiction. For example $$\sum_{n \in \cup A_i} a_n > \sum_{i = 1}^\infty \sum_{n \in A_i} a_n$$ means that for an $$n_0$$ the partial sum $$\sum_{n \in \cup A_i, n<{n_0}} a_n$$ is greater than $$\sum_{i = 1}^\infty \sum_{n \in A_i} a_n$$. Since the first part is a finite sum and $$a_n \ge 0$$ for all $$n$$ you can arrive at a contradiction. I hope you can take it from here... (the second inequality is a bit trickier but you can still use the same reasoning).

PS I used $$\sum_{n \in \cup A_i} a_n$$ instead of $$\sum_{n \in \cup A_n} a_n$$ and $$\sum_{i = 1}^\infty \sum_{n \in A_i} a_n$$ instead of $$\sum_{i = 1}^\infty \sum_{n \in A_i} A_i$$. I took the liberty to assume that this is what you meant.

• Thank you so much!
– Phil
Commented Jan 14, 2020 at 16:32