1
$\begingroup$

Let $(X,F,\mu)$ be a finite measure space, i.e. $\mu(X)<\infty$. And let $x\in X$, and let $\delta_x$ be the Dirac measure with respect to $x$, i.e. $\delta_x(E)=1$ if $x\in E$ and $\delta_x(E)=0$ if $x\notin E$. Finally let $\nu$ be the signed measure $\mu-a\delta_x$, where $a=\mu(X)$. Find the Hahn decomposition of $\nu$.

I'm not really sure how to construct Hahn decompositions. I tried rewriting the proof of the Hahn-Jordan decomposition theorem for the case of this particular signed measure, but it didn't give me a concrete pair of sets.

$\endgroup$
1
$\begingroup$

Since the $\sigma$-algebra isn't specified, you cannot give an explicit choice for the Hahn-decomposition. (For example $F= \{X, \emptyset\}$ gives only a trivial decomposition. One other example, is $F=\{X, A,A^c,\emptyset\})$ with $X= [0,1]$, $A= [0,1/2]$ and $\mu = \delta_0$ and $x=1$. Then $X= A \cup A^c$ is the Hahn-decomposition.)

Assume that $\{x\} \in F$, then the decomposition can be determined. If $\mu(\{x\}) -a \ge 0$, then $\nu := \mu -a \delta_x$ is a non-negative measure and $\nu^{-} =0$ is trivial. (Thus the Hahn-decomposition is $X=P \cup N$ with $N = \emptyset$.) On the other hand, if $\mu(\{x\}) -a <0$. Then the Hahn-decompisition is $X =P \cup N$ with $N= \{x\}$ and $P = X \setminus \{x\}$.

$\endgroup$

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.