Caratheodory's Theorem: Difference between revisions
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<math> \mu^*(E) = \mu^*(E \cap A) = \mu^*(E \cap A^c) </math> | <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> | <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 | and by subadditivity | ||
<math> \geq \mu^*(E \cap A^c \cap B^c) + \mu^*(E \cap (A \cup B)) </math> | <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