If $\mu$ is a semifinite mesure and $\mu(E) = \infty$, then for any $C > 0$ there exists $F \subset E$ with $C < \mu(F) < \infty$

Asked Apr 05 '22 at 06:52

Active Jun 19 '25 at 13:23

Viewed 156 times

This is Problem 1.14 from Folland's Real Analysis.

I am aware of a few solutions to this problem (for example, If a measure is semifinite, then there are sets of arbitrarily large but finite measure). However I am having trouble with one step. It is clear that we need to show that $$ s = \sup\{\mu(F) : F \in \mathcal{M}, F \subset E, \mu(F) < \infty\} = \infty.$$

In all of the proofs the approach is to assume by way of contradiction that $s < \infty$. Then they claim that there must exist a sequence of sets $F_n \subset E$ with $\mu(F_n)< \infty$ and $\mu(F_n)→ s$. But why? What ensures the existence of such a sequence of sets? What if my set of measures is finite so I do not have such a nice continuum?

Similarly, another solution states that if $s = \alpha$ and $\alpha<\infty$, then $\alpha-1/n<\alpha$, so for every $n$ there exists $F_n\subset E$ with $\alpha-1/n<\mu(F_n)<\infty$. What guarantees the existence of such a set?

Would anyone be able to walk me through this proof in more detail that is in the post? Thanks!

edited Jun 19 '25 at 13:23

Dean Miller

7,015
1
7
34

asked Apr 05 '22 at 06:52

CBBAM

7,149

4

$F_n$'s exist by definition of supremum of a set of real numbers. This has nothing to do with measure theory. – Kavi Rama Murthy Apr 05 '22 at 07:32
@KaviRamaMurthy Thanks! Is there a name for that property? – CBBAM Apr 05 '22 at 17:21

If $\mu$ is a semifinite mesure and $\mu(E) = \infty$, then for any $C > 0$ there exists $F \subset E$ with $C < \mu(F) < \infty$

0 Answers0