Caratheodory's Theorem: Difference between revisions
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== Proof == | == Proof == | ||
First, observe that <math>\mathcal{M}</math> is closed under complements due to symmetry in the meaning of <math> \mu </math>-measurability. Now, we show if <math> A, B </math> then <math> A \cup B \in \mathcal{M} </math>. | First, observe that <math>\mathcal{M}</math> is closed under complements due to symmetry in the meaning of <math> \mu </math>-measurability. Now, we show if <math> A, B </math> then <math> A \cup B \in \mathcal{M} </math>. | ||
Suppoe <math>E \subseteq X </math>. Then | |||
<math> \mu^*(E) = \mu^*(E \cap A) = \mu^*(E \cap A^c) </math> | |||
<math> = \mu^*(E \cap A \cap B) + \mu^*(E \cap A \cap B^c) + \mu^*(E \cap A^c \cap B) + \mu^*(E \cap A^c \cap B^c) </math> | |||
and by subadditivity | |||
<math> \geq \mu^*(E \cap A^c \cap B^c) + \mu^*(E \cap (A \cup B)) </math> |
Revision as of 22:37, 16 December 2020
Statement
Consider an out measure on . Define
.
Then is a -algebra and is a measure on .
Proof
First, observe that is closed under complements due to symmetry in the meaning of -measurability. Now, we show if then .
Suppoe . Then
and by subadditivity