Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

in Paul R. Halmos - Measure Theory: for a sequence ${x_n}$ the supremum of the sequence is \, and the infimum of $x_n$ is the intersection
of $x_n$ for n from 1 to infinity, how to understand that?
by the way is it a right book for undergrad study for measure theory?
Thanks!

share|improve this question

closed as not a real question by Andres Caicedo, Micah, kahen, ncmathsadist, Chris Eagle Feb 27 '13 at 2:05

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

8  
I don't understand "the supremum of the sequence is \,". –  Gerry Myerson Feb 26 '13 at 11:34

1 Answer 1

up vote 2 down vote accepted

If $(B_n)$ is a sequence of subsets of $X$, then the pointwise supremum of the indicator functions $(1_{B_n})$ coincides with the indicator function of $\bigcup_n B_n$, so it makes sense to call $\bigcup_n B_n$ the supremum of the sequence $(B_n)$. It is also the order theoretic supremum under the relation $\subseteq$ on the powerset of $X$.

The approach of Halmos in doing everything on $\sigma$-rings instead of $\sigma$-algebras is a bit outdated, but the book is not bad.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.