Question on a pre-measure defined by Folland's real analysis book

My simple question might suggest I'm missing something quite basic. I, therefore, apologize in advance.

In Prop 1.15, Folland defines a pre-measure on h-intervals (left open right closed intervals).

1.15 Proposition. Let $F: \mathbb{R} \rightarrow \mathbb{R}$ be increasing and right continuous. If $\left(a_j, b_j\right]$ $(j=1, \ldots, n)$ are disjoint $h$-intervals, let

$$\mu_0\left(\bigcup_1^n\left(a_j, b_j\right]\right)=\sum_1^n\left[F\left(b_j\right)-F\left(a_j\right)\right],$$

and let $\mu_0(\emptyset)=0$. Then $\mu_0$ is a premeasure on the algebra $\mathcal{A}$.


My problem is the following. Folland also includes in the h-intervals Algebra sets of the following form $(a,\infty)$
How are these defined under his definition of $\mu_0$?

Measure Theory
Dannyboy990 Dannyboy990
Report

Answer

Answers can be viewed only if
  1. The questioner was satisfied and accepted the answer, or
  2. The answer was disputed, but the judge evaluated it as 100% correct.
View the answer

1 Attachment

Erdos Erdos
  • Erdos Erdos
    0

    Please leave a comment if you need any clarifications.

    • Dannyboy990 Dannyboy990
      0

      Why should it be closed under countabe union? The pre-measure is defined on an Algebra which is closed under finite unions.

    • Erdos Erdos
      0

      If you read the bottom of page 30 in Folland, in defining a premature the goal is to extend measure from algebras to \sigma-algebras. So by definition a pre-measure is defined on the smallest \sigma-algebra that contain the given algebra. So you need the sets to be closed under countable union, and not just finite union. I attached page 30 of Folland.

The answer is accepted.
Join Matchmaticians Affiliate Marketing Program to earn up to 50% commission on every question your affiliated users ask or answer.