# question about total variation (measure theory)

suppose $v$ is a signed measure on $(X,M)$ and $E\in M$

how do i go about showing that $|v|(E)=sup\{\sum_1^n |v(E_j)|: E_j\cap E_i=0 \forall i\neq j, \cup_1^n E_j=E\}$

sorry it took me awhile to fix the latex

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Hint: Consider the Hahn Decomposition Theorem, which allows us to split $X$ into a positive and negative set for $\nu$. That is, we may write $X=P\cup N$, $P\cap N=\emptyset$ where $P$ is a positive set for $\nu$ and $N$ is a negative set for $\nu$.

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so $|v|(E)=v(E \cap P)-v(E \cap N)$, so we can let $E_1=E \cap P$ and $E_2=E \cap N$, clearly they are disjoint and union to $E$. Is this all I have to say? –  jack Mar 5 '11 at 15:41