0

In my book for a measurable space $(X,\mathcal{A})$, a signed measure on it is a function $\mu: \mathcal{A}\rightarrow (-\infty,\infty]$ such that

$(1)$ $\mu(\varnothing) = 0$; $(2)$ If $A_1,\cdots, A_k, \cdots$ are pairwise disjoint sets, then we have $$ \mu(\bigcup_{i = 1}^\infty A_i) = \sum_{i = 1}^\infty \mu(A_i), $$ and if $\mu(\bigcup_{i = 1}^\infty A_i)$ is finite then the right side must absolutely converge to a number.

My question is, if $\mu(\bigcup_{i = 1}^\infty A_i) = \infty$, how can we check whether the right side is infinity? I took real analysis a while ago and forgot a lot about series. Is it possible that $\sum_i^\infty a_i = \infty$ but after reordering it terms, it sums to a number or is even divergent (no value)?

Thanks!

Coco
  • 613
  • If $\sum_{i=1}^\infty a_i = +\infty$ we can assume without loss of generality that $a_i\geqslant 0$ for all $i$ since $\infty-x=\infty$ for any $x\in\mathbb R$ and $\infty-\infty$ is not defined. – Math1000 Nov 09 '23 at 08:21

1 Answers1

0

In general, it is certainly possible to have $\sum_{i=0}^\infty a_i=\infty$, and yet a reordering of the terms gives a convergent series - consider the alternating harmonic series $\sum_{n=1}^\infty \frac{(-1)^n}{n}$, which converges, but can be reordered to diverge to either $\infty$ or $-\infty$ (or to oscillate between any number of values).

However, this definition asserts that that does not happen in the case of a signed measure, since reordering the terms does not change the left hand side of

$$\mu(\bigcup_{i = 1}^\infty A_i) = \sum_{i = 1}^\infty \mu(A_i).$$

In fact, the statement that the sum on the right must converge absolutely to a number did not have to be included in the definition, since it is an immediate consequence of the fact that the value of sum is independent of ordering (see here or here).

Note also that since $\mu$ does not ever take a value of $-\infty$, the sum of all the negative terms on the right hand side must be finite.

M W
  • 9,866