4
$\begingroup$

In defining a positive measure $\mu$ over an abstract measure space $(X,\mathcal A)$ isn't saying

for any countable pairwise disjoint collection $\{A_n\}\subset\mathcal A,~\mu(\cup A_n)=\sum\mu(A_n)$ an overstatement? I think it's enough to say for $A,B\in\mathcal A,~\mu(A\cup B)=\mu(A)+\mu(B)?$

Well I can see that if each $\mu(A_n)$ is finite then both the sequence being monotone either convergent to the same limit or diverges to $+\infty.$ However if at least one $\mu(A_k)$ is $+\infty$ then both of the sides equal $\infty$. Whats's wrong with the logic?

I would like to quote from the following lecture note which motivates me to ask the question:

enter image description here

$\endgroup$
10
  • $\begingroup$ Your condition quickly yields finite additivity. That is not enough for many applications. $\endgroup$ Commented Mar 6, 2014 at 2:34
  • $\begingroup$ I think $\mu(\cup A_n)=\sum\mu(A_n)$ follows from finite additivity. $\endgroup$
    – user133432
    Commented Mar 6, 2014 at 2:36
  • 2
    $\begingroup$ @user133432 Unfortunately, you're wrong. $\endgroup$
    – user98602
    Commented Mar 6, 2014 at 2:36
  • $\begingroup$ There are easy examples of set functions which are finitely additive but not countably additive. Google it. $\endgroup$
    – MPW
    Commented Mar 6, 2014 at 2:37
  • 1
    $\begingroup$ If we have a countable set of "atoms" and they all have positive mass, then yes. $\endgroup$ Commented Mar 6, 2014 at 2:38

2 Answers 2

4
$\begingroup$

There is no reason to think that $$\lim_{n\to\infty}\mu\left(\bigcup_{k=1}^n A_k\right) = \mu\left(\bigcup_{k=1}^\infty A_k\right)$$

It's clear that, with finite additivity, you get $$\lim_{n\to\infty}\mu\left(\bigcup_{k=1}^n A_k\right) \leq \mu\left(\bigcup_{k=1}^\infty A_k\right)$$

but you want equality.

A simple example where that would fail is the natural number "density" measure. For $X\subseteq \mathbb N$, define $$\mu(X)=\lim_{n\to\infty} \frac{\left|X\cap [1,n]\right|}{n}$$

This satisfies finite additivity, but not countable additivity.

For example, $A_k=\{k\}$ each have measure zero, but $\mu\left(\cup A_k\right)=1$.

$\endgroup$
15
  • $\begingroup$ If $\lim_{n\to\infty}\mu\left(\bigcup_{1}^n A_k\right)$ not necessarily equals $\mu\left(\bigcup_{k=1}^\infty A_k\right)$ then what is the interpretation for $\mu\left(\bigcup_{1}^\infty A_k\right)?$ $\endgroup$
    – user133432
    Commented Mar 6, 2014 at 2:45
  • $\begingroup$ That's a confusing question. Can you be clearer? $\mu$ is a function defined on (some) subsets of your original set, so $\mu\left(\bigcup\dots\right)$ is defined if it is defined. Look at the example which is finite but not countably additive. There is some use of such things, but they don't let us define integration, for example. $\endgroup$ Commented Mar 6, 2014 at 2:47
  • $\begingroup$ But how would I interpret $\cup A_n?$ I don't have any idea for convergence of unions. $\endgroup$
    – user133432
    Commented Mar 6, 2014 at 2:50
  • 2
    $\begingroup$ $\cup_{1}^\infty A_k$ just means the set of all points that are in at least one of the $A_k$. Doesn't require any notion of convergence - we can define the union of any collection of sets, no matter the size of the collection. $\endgroup$ Commented Mar 6, 2014 at 2:54
  • $\begingroup$ With a finitely additive measure, the value of $\mu\left(\bigcup_{k=1}^{\infty}A_{k}\right)$ cannot be deduced from the values of $\mu(A_{k})$. That is the point! However, if you have lower semicontinuity ($\mu\left(\bigcup_{k=1}^{\infty} A_{k}\right)=\lim_{j\to\infty} \mu\left(\bigcup_{k=1}^{j}A_{k}\right)$), together with finite additivity, then countable additivity follows. $\endgroup$
    – Unwisdom
    Commented Mar 6, 2014 at 2:56
1
$\begingroup$

What you are talking about is called a finitely additive measure. There are finitely additive measures that are not proper measures (i.e., not countably additive). See here for a counterexample.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .