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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

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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.

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

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