A sequence of sets increasing to a limit $A$

measure-theory

Terminology and definition question:

When we say

A sequence of sets $A_1,A_2,…$ belong to a field $\mathscr{F}$ and increases to a limit $A$.

Does this equivalently mean

$A_1\subset A_2\subset …$ and $\cup_{n=1}^\infty A_n=A$

What I am confused is that when someone says "a sequence of sets increase to a limit A", do we assume it is an increase sequence of sets converging to a set in the limit? Is it possible to state a sequence of sets increase to a limit A without presupposing the sequence is an increasing sequence?

Reference:
$\textit{Probability and Measure Theory}$ (Robert B. Ash and Catherine A. Doleans-Dade), Harcourt/Academic Press, 1999.

Best Answer

Your interpretation is correct.

There is a concept of limit of a sequence $(A_n)_{n\in\mathbb N}$ of sets, which is as follows: a set $A$ is the limit of such a sequence if $A$ satisfies both conditions:

  1. $A=\{x\in\mathscr F\mid x\in A_n\text{ for infinitely many $n$'s}\}$:
  2. $A=\{x\in\mathscr F\mid(\exists N\in\mathbb N)(\forall n\in\mathbb N):n\geqslant N\implies x\in A_n\}$.

But if the sequence $(A_n)_{n\in\mathbb N}$ is increasing and if you take $A=\bigcup_{n\in\mathbb N}$, then both of those conditions hold and therefore you have $A=\lim_{n\to\infty}A_n$ indeed.

Note that if the if the sequence $(A_n)_{n\in\mathbb N}$ is decreasing and if you take $A=\bigcap_{n\in\mathbb N}$, then both conditions hold too and therefore you have $A=\lim_{n\to\infty}A_n$ ind this case too.

There are cases in which the limit exists but the $A_n$'s are not nested. Take, for instance, $\mathscr F=\mathbb Q$ and, for each $n\in\mathbb N$, $A_n=\{0,n\}$. Then $\lim_{n\to\infty}A_n=\{0\}$.

Related Question