4
$\begingroup$

If $\nu$ is a signed measure there exists unique positive measures $\nu^{+}$ and $\nu^{-}$ s.t. $\nu=\nu^{+}-\nu^{-}$ and $\nu^{+}\perp\nu^{-}$

Would appreciate if someone can guide me through the proof.

Proof: let $X=P\cup N$ be Hahn decomposition for $\nu$, define $\nu^{+}(E):=\nu(E\cap P)$ and $\nu^{-}(E):=-\nu(E\cap N)$ then both $\nu^{-}$ and $\nu^{+}$ are positive. Also $\nu=\nu^{+}-\nu^{-}$ and $\nu^{+}\perp\nu^{-}$

1) why is $\nu=\nu^{+}-\nu^{-}$ by the above definition we have $\nu(E)=\nu^{+}(E)-\nu^{-}(E)=\nu(E\cap P)+\nu(E\cap N)$ for some $E\in\mathcal{M}$ and $E\neq \emptyset$

2) Also why $\nu^{+}\perp\nu^{-}$? Is it because $P$ is null for $\nu^{-}$ and $N$ is null for $\nu^{+}$? and $P\cup N=X$,$P\cap N=\emptyset$

As for the uniqueness part. Let $\nu=\mu^{+}-\mu^{-}$ be another decomposition. Let $E,F$ be measurable s.t. $E\cup F=X$ and $F\cap E=\emptyset$ be another Hahn decompotion. Also $\mu^{+}(F)=\mu^{-}(E)=0$. Then $P\triangle E$ is $\nu$-null. Why is it $\nu$-null?

We need to show that $\nu^{-}=\mu^{-}$ so we pick $A\in\mathcal{M}$ then $\mu^{+}(A)=\mu^{+}(A\cap E)=\nu(A\cap E)=\nu(A\cap P)=\nu^{+}(A)$ How do we get the second equality? Does the third equality follows from the fact that $P\triangle E$ is $\nu$-null?

$\endgroup$
4
$\begingroup$

I think the idea of the proof is: Find a potential $\nu^+,\nu^-$ positive using Hahn dec; Via the construction it follows mutual singularity; (showing that is is unique is intuitively a consequence of $P\triangle E$) Get new pair of positive measures s.t. $\nu=\mu^+-\mu^-$, mutual singularity $\Rightarrow$ Hahn dec, but any difference in the two positive (negative) Hahn sets is a null set.

1) I think you have to do the steps the other way around using a Hahn dec. to obtain your two measures (one positive and one negative) which are a candidate to be proved to be the unique pair: $$ \nu(E)=\nu(E\cap (P\cup N))=\nu(E\cap P) +\nu(E\cap N) $$ 2) Yes, $\forall A\subset N$ $$ \nu^+(A)=\nu(A\cap P) = 0 $$ Similar steps for $\nu^-$.

3.?1) $P\triangle E$ is $\nu$-null is a result from Hahn Decomposition Thm when you have two Hahn dec for the same signed measure.

3.?2) $$\nu(A\cap E)=\nu^+(A\cap E)-\nu^-(A\cap E)=\nu^+(A\cap E)+\nu(A\cap E\cap F) $$ but $E\cap F=\emptyset$ because $E,F$ is a Hahn dec for $\nu$.

3.?3)I do not see it the way I prove it, $$ \nu(A\cap E\cap (P\cup N))= \nu(A\cap E\cap P)+\nu(A\cap E\cap N) $$ but the $E$ is positive and $N$ negative, then $$ \nu(A\cap E\cap N)\geq 0 \text{ because }A\cap E\cap N\subset E $$ $$ \nu(A\cap E\cap N)\leq 0 \text{ because }A\cap E\cap N\subset N $$ $$ \Rightarrow \nu(A\cap E\cap N)=0. $$ Similarly $$ \nu(A\cap P\cap (E\cup F))=\nu(A\cap P\cap E)+\nu(A\cap P\cap F)=\nu(A\cap P\cap E) $$ and so $$ \nu(A\cap P)=\nu(A\cap E) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.