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I am reading a proof about the statement that

Let $A$ be any set and $(E_k)_{k \leq n}$ a finite disjoint collection of measurable sets, then $$m^*\left( A \cap (\cup_{k\leq n} E_k\right) = \sum m^*(A \cap E_k).$$

The proof starts with induction and claims that $$A \cap (\cup_{k=1}^{n} E_k) \cap E_n = A \cap E_n.$$

So how can $A \cap (E_1 \cup E_2 \cup E_3) \cap E_3 = A\cap E_3$? Unless each $E_k$ is pairwise disjoint?

Added: \begin{align} A \cap (E_1 \cup E_2 \cup E_3) \cap E_3 &= A \cap E_3 \cap(E_1 \cup E_2 \cup E_3) \\ &= (A \cap E_3 \cap E_1) \cup (A \cap E_3 \cap E_2) \cup (A \cap E_3 \cap E_3) \\ &= (A \cap E_3 \cap E_1) \cup (A \cap E_3 \cap E_2) \cup (A \cap E_3) \end{align}

So the RHS can't be $(A \cap E_3) $ unless $(A \cap E_3 \cap E_1) = (A \cap E_3 \cap E_2) = \emptyset$

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  • $\begingroup$ You seem to be confused about the definition of intersection. Sets being pairwise disjoint or not has nothing to do with the equality you are asking about. $\endgroup$ – Andrés E. Caicedo Sep 11 '14 at 5:39
  • $\begingroup$ Don't think about it abstractly yet. First try $\{1,2,3\}\cap(\{1,2,3,4,5\}\cup\{1,2,3,7,8,9\}\cup\{3\})\cap\{3\}$. $\endgroup$ – Andrés E. Caicedo Sep 11 '14 at 5:44
  • $\begingroup$ Your sets are not disjoint, but i'll try for $E_k = \{ 10 \}$ $\endgroup$ – Hawk Sep 11 '14 at 5:47
  • $\begingroup$ No, try the example I suggested first, you are confused enough already. For instance, saying that each $E_k$ is pairwise disjoint is meaningless. Try the example, and contrast it with what you are saying must happen. Once you see what is really happening, then (but only then) go back to thinking about the setting at hand where some sets are disjoint, etc. $\endgroup$ – Andrés E. Caicedo Sep 11 '14 at 5:50
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In $\cup E_k$ the book is including the set $E_n$, so $(\cup E_k)\cap E_n=E_n$

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  • $\begingroup$ But the $E_k$s aren't pairwise disjoint, the whole collection is disjoint. $\endgroup$ – Hawk Sep 11 '14 at 5:21
  • $\begingroup$ Maybe not, but $E_n\subset \cup E_k$ and $ B\cap E_n\subset E_n$ for any $B$. $\endgroup$ – almagest Sep 11 '14 at 5:43

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