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