# Measure theory: measure on a sigma algebra

• Mar 11th 2014, 11:06 PM
sawleha
Measure theory: measure on a sigma algebra
Let u be a measure on a sigma algebra A of subsets of X. Prove that u(E)+u(F)=u($\displaystyle E\bigcup F$)+u($\displaystyle E\bigcap F$)
• Mar 12th 2014, 02:59 AM
romsek
Re: Measure theory: measure on a sigma algebra
Quote:

Originally Posted by sawleha
Let u be a measure on a sigma algebra A of subsets of X. Prove that u(E)+u(F)=u($\displaystyle E\bigcup F$)+u($\displaystyle E\bigcap F$)

$(E \cup F) = E \cup (F \cap \overline{E})$

These are disjoint so

$u(E \cup F) =u(E) + u(F \cap \overline{E})$

$u(E \cup F) + u(E \cap F) = u(E) + u(F \cap \overline{E}) + u(E \cap F)$

$F=(F \cap E) \cup (F \cap \overline{E})$

and both of these are disjoint so

$u(F) = u(F \cap \overline{E}) + u(E \cap F)$

so

$u(E \cup F) + u(E \cap F)=u(E) + u(F)$
• Mar 12th 2014, 10:56 PM
sawleha
Re: Measure theory: measure on a sigma algebra
Thank you for showing each step. it really clarifies things.