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($\displaystyle E\bigcap F$)

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- Mar 11th 2014, 11:06 PMsawlehaMeasure 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 AMromsekRe: Measure theory: measure on a sigma algebra
$(E \cup F) = E \cup (F \cap \overline{E})$

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 PMsawlehaRe: Measure theory: measure on a sigma algebra
Thank you for showing each step. it really clarifies things.