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Let $(X,\mathcal{H})$ be a measurable space. A measure on $(X,\mathcal{H})$ is a function $\mu : \mathcal{H} \rightarrow [0, \infty]$ such that

  1. $\mu(\emptyset)=0.$

  2. For any sequence of mutually disjoint sets $\{A_i\}_{i=1}^{\infty}$ in $\mathcal{H}$, then $\displaystyle \mu(\displaystyle\cup_{i=1}^{\infty}A_i)= \sum_{i=1}^{\infty}\mu(A_i) .$

Question 1: I would like to understand the true meaning of axiom two since the left hand side is independent of order and the right hand side depends on the order $A_1, A_2, \cdots$ or better still what is the correct way to define this.

Question 2: Is my definition of a measure correct?

Thanks.

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    The RHS is a series of non-negative (extended) real numbers. So its value is independent from the order of summation. – 2017-02-20

2 Answers 2

1

It's actually just as typical to write $$\mu \left(\bigcup_{n \in \mathbb N} A_n \right)=\sum_{n \in \mathbb N} \mu (A_n).$$

The real meaning behind this is the notion of

summing over sets: Let $S$ be a set, and let $f:S \to [0,\infty]$ be a function, then we can formally define

$$\sum_{s \in S} := \sup\{f(s_1)+...+f(s_n) : n \in \mathbb N\}$$

(or alternatively, that $\{s_1,...,s_n\}$ is a countable subset of $s$, so we are taking a supremum over all partial sums.

This is known as countable additivity, but the index set is countable countable in the definition, to see why, perhaps see the definition of a $\sigma$-algebra.

Also note that this would be a poorly defined "sum" if we allowed negative terms, since conditionally convergent sums depend on the way we choose subsets. Here is a quick proof that it does not matter which notion you prefer:

Let $f$ be as before. It is clear that $$\sum_{n \in \mathbb N}f(n) \geq \sum_{n=1}^{N} f(n)$$

for all $N \in \mathbb N$, so we have that $$\sum_{n \in \mathbb N}f(n) \geq \sum_{n=1}^{\infty} f(n).$$

On the other hand, let $\{n_1,..,n_k\}$ be a finite set. Choose $N$ to be larger than any integer in this set. Then

$$f(n_1)+\dots f(n_k) \leq \sum_{n=1}^{N} f(n) \leq \sum_{n=1}^{\infty} f(n)$$

for all finite subsets, implying the reverse inequality.

2

Answer 1. As all term in the series of right is no-negative, the order of the sum don't change the series value.

Answer 2. It's correct. However you can define a measure in something smaller than a measurable space, (e.g. $\mathcal{H}$ being a Algebra instead a $\sigma$-algebra) and the second property would say: $$\lbrace A_i\rbrace_{i=1}^{\infty}\subset\mathcal{H}\land\bigcup_{i=1}^{\infty}A_i\in\mathcal{H}\implies\mu\left(\bigcup_{i=1}^{\infty}A_i\right)=\sum_{i=1}^{\infty}\mu(A_i)$$