set which has outer measure as infinity can be expressed as disjoint union of countable measurable sets

Question

I am studying the 'Real Analysis' by Royden, 4th edition.

In page $41$, Royden quote the following:

If $m^*(E) = \infty$ and measurable , then $E$ can be expressed as the disjoint union of countable collection $\{ E_k \}$ of measurable sets, each of which has finite outer measure.

How to prove the above statement?

The following link asked the same question. http://math.stackexchange.com/questions/430292/if-me-infty-then-e-bigcup-k-1-inftye-k-e-k-measurable-and-m — Idonknow, Mar 03 '17 at 15:58

score 3 · Accepted Answer · answered Mar 01 '17 at 15:08

3

This statement is essentially equivalent to the $ \sigma $-finiteness of the measure space. Simply take it to be the disjoint union

$$ E = \bigcup_{n= -\infty}^{\infty} E \cap [n, n+1) $$

answered Mar 01 '17 at 15:08

Ege Erdil

17,747

Why we need $m^*(E) = \infty$? – Idonknow Mar 01 '17 at 15:21
@Idonknow You don't, but if the outer measure of $ E $ is finite; then the disjoint union can be taken to consist of just the single set $ E $. – Ege Erdil Mar 01 '17 at 15:22
do you mean $E =E $? – Idonknow Mar 01 '17 at 15:44

set which has outer measure as infinity can be expressed as disjoint union of countable measurable sets

1 Answers1