This is paraphrased from an excerpt on a proof from Stein and Shakarchi's $\textit{Real Analysis}$ on page 52,

Suppose we have $\bigcup_{k=1}^N E_k$, where each $E_k$ is measurable and of finite measure, and all the $E_k$'s are disjoint. Then they claim we can find a "refinement" of $\bigcup_{k=1}^N E_k$, consisting of sets $E_1^*$,$E_2^*\dots ,E_n^*$ such that:

1) $\bigcup_{k=1}^N E_k = \bigcup_{j=1}^n E_j^*$

2) The sets $E_j^*$ ($j = 1,\dots ,n$) are mutually disjoint, and

3) For each $k$, $E_k = \bigcup E_j^*$, where the union is taken over those $E_j^*$ that are contained in $E_k$

My thoughts: I begin with the construction of $E_j^*$'s by defining $E_1^* = E_1$, $E_2^*=E_2^* -(E_1\cup E_2)$, etc., where $E_j^*=E_j^* - (\bigcup_{n=1}^j E_n)$ But does this construction work? Is it even valid?

Much thanks, Dom


I think you might want a slight change. In your construction, you can't say $E^*_2 = E^*_2 - (E_1 \cup E_2)$, because you have $E_2^*$ on both sides.

But, you're close. Define

$E_k^* = E_k - [\cup_{j=1}^{k-1} E_j] = E_k \cap [ \cup_{j=1}^{k-1} E_j]^c$, where the $^c$ denotes taking the complement. Then, each $E_k^*$ is disjoint, and the union is the same. Basically, at each step, you're using $E_k$, but removing all pieces from the previous sets. This gives us disjointedness. The sets are measurable, because we are just using unions, complements, and intersections. The union of the $E_k$ and the $E_k^*$ is the same, as any point in an $E_k$ is in at least one of the $E_j^*$ sets.

  • $\begingroup$ Thanks! Fixing that one piece makes it all come together. $\endgroup$ – madisonfly Oct 24 '12 at 5:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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