If $G_1$ and $G_2$ are open subset of [a,b] then $|G_1|+|G_2|=|G_1 \cup G_2|+|G_1 \cap G_2|$ I have some problem to understand complete theorem. in this theorem first case is finitely many intervals reference book using $\chi$ function which have 4 cases.whenever $x \in [a,b]$ then they will directly say that $\chi$ function is Riemann integrable i don't know why?please help me.thanks in advance.
If $G_1$ and $G_2$ are open subset of [a,b] then $|G_1|+|G_2|=|G_1 \cup G_2|+|G_1 \cap G_2|$
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real-analysis
measure-theory
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0It has to be o$n$ly a measure in a more general way. – 2012-08-18
2 Answers
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First note that $G_1\cup G_2 = G_1\cup(G_2 - (G_1\cap G_2).$ This union is disjoint so
$|G_1\cup G_2| = |G_1| + |G_2 - (G_1\cap G_2)|.$
Now suppose we have $B$ and $C$ with $C\subseteq B$. Then we have the disjoint union $B = C\cup (B-C). $ Using finite additivity, $|B| = |C| - |B - C|.$ so $|B - C| = |B| -|C|.$
Apply this principle to see that
$|G_1 - (G_1\cap G_2)| = |G_2| - |G_1\cap G_2|. $ Your result follows immediately.
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Here are two hints:
- For every sets $A$ and $B$, one has $\mathbf 1_{A\cup B}+\mathbf 1_{A\cap B}=\mathbf 1_{A}+\mathbf 1_{B}$.
- For every subset $A$ of $[a,b]$, one has $|A|=\int\limits_a^b\mathbf 1_{A}$.
The result follows. Additionally, this method shows that $G_1$ and $G_2$ can be any measurable subsets. (If the question is to know why every open set is measurable, recall that, by definition, the Borel sigma-algebra is generated by the collection of open sets.)