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!

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

2 Answers 2


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 $$\sum_{i\in I}\sum_{n\in A_i}a_n = \sum_{n\in \bigcup_I A_i}a_n.$$

  • $\begingroup$ Thank you so much! $\endgroup$
    – 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

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


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

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.

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

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