# Thread: Measure theory: measure on a sigma algebra

1. ## 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( $E\bigcup F$)+u( $E\bigcap F$)

2. ## Re: Measure theory: measure on a sigma algebra

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

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

$u(E \cup F) + u(E \cap F)=u(E) + u(F)$

3. ## Re: Measure theory: measure on a sigma algebra

Thank you for showing each step. it really clarifies things.