Measure Theory: Prove Set is Measurable Question

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  • #1

Homework Statement

The question is from Stein, "Analysis 2", Chapter 1, Problem 5:

Suppose E is measurable with m(E) < ∞, and E = E1 ∪ E2, E1 ∩ E2 = ∅.


a) If m(E) = m∗(E1) + m∗(E2), then E1 and E2 are measurable.

b) In particular, if E ⊂ Q, where Q is a finite cube, then E is measurable if and only if m(Q) = m∗(E) + m∗(Q − E).

Homework Equations

The definition of a 'measurable set' given in the book is that for any ε > 0 there exists an open set O with E ⊂ O and m∗(O − E) ≤ ε, so I'm looking for a set of implications that lead me back to this definition.

The Attempt at a Solution

The problem seems suspiciously similar to the definition of a measurable set as one that satisfies the 'caratheodory criterion'. My attempt at a solution has been to try to show that what we are given in the problem must imply that the caratheodory criterion holds and from there show that if the caratheodory criterion holds then the set is measurable in the above sense. I'm having trouble knowing where to start filling in the details.

I also wonder though, if there is a simpler and neater way to solve the problem?

Thanks in advance for any help you can give me - it's very much appreciated. This one is doing my head in!

Answers and Replies

  • #2
One way that I have looked to prove (a) is to use the lemma that:

If E ⊂ Rd, then m∗(E) = inf m∗(O), where the infimum is taken over all open sets O containing E.

To construct an open set O1 that encompasses E1 and such that m∗(E1) = inf m∗(O1). And likewise and open set that encompasses E2 such that m∗(E2) = inf m∗(O2).

Then if I can show that m*(O1-E1)=0 and m*(O2-E2)=0, then E1 and E2 differ from an open set by measure 0 and so by a theorem they are also measurable.

However, I'm unsure about how to do this and get to m*(O1-E1)=0 and m*(O2-E2)=0?
  • #3
...Anyone out there?
  • #4
Hey, still stuck on this one. Let me know if you need me to clarify anything.