On the example of the The Hahn decomposition

Question

Suppose $d\nu = fdm$ is the measure for the following function $f$, $$ f(x)= \begin{cases} x& \text{if}~x\geq 1;\\ 0& \text{if}~-1<x\leq 1; \\ -x^2& \text{if}~x\leq -1;\\ \end{cases} $$ I am trying to construct an elaborative example to understand the Hahn decomposition and the Jordan decomposition of $\nu$ in the following manner:

The positive set $P=\{x\in\mathbb{R}; \nu(E)=\int_1^{+\infty}xdx\}=+\infty>0$

The negative set $N=\{x\in\mathbb{R}; \nu(E)=\int_{-\infty}^{-1}-x^2dx\}=-\infty<0$

and the null set $M=\{x\in\mathbb{R}; \nu(E)=\int_{-1}^{1}0dx\}=0$

The Jordan decomposition of $\nu$ is the set $\{-\nu,+\nu\}$

I thought The Hahn decomposition is the set $\{P,N\}$ since $P\cap N=\emptyset$. However, $\mathbb{R}\neq P\cup N$.

Please, help me rectify all the shortcomings in my attempt on the Hahn decomposition, Jordan decomposition and compute the total variation of $\nu$.

Answer 1

Recall that a positive subset of $(X,\nu)$ is a $\nu$-measurable set $P$ such that $\nu(E) \ge 0$ for any $\nu$-measurable $E\subseteq P$. Note that this includes the possibility that $\nu(E) = 0$. Thus every $\nu$-null set is positive. Similarly, every $\nu$-null set is negative.

A Hahn decomposition of $(X,\nu)$ consists of two sets $P$ and $N$ such that

1. $P \cup N = X$,
2. $P \cap N = \emptyset$,
3. $P$ is a positive set, and $N$ is a negative set.

The Hahn decomposition theorem further notes that such a decomposition is not unique, but that any two such decompositions will differ by a set of measure zero. In your example, absorb your set $M$ into one of the two sets (i.e. take your positive set to be $P \cup M$, or your negative set to be $N \cup M$).

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Next, recall the Jordan Decomposition Theorem, which states that if $\nu$ is a signed measure, then there are unique nonnegative measures $\nu^{+}$ and $\nu^{-}$ such that

1. $\nu = \nu^{+} - \nu^{-}$, and
2. $\nu^{+} \perp \nu^{-}$.

Given $(X,\nu)$ and a Hahn decomposition $(P,N)$ with respect to $\nu$, note that both $$ \nu^{+}(E) := \nu(E\cap P) \qquad\text{and}\qquad \nu^{-}(E) := -\nu(E\cap N) $$ are nonnegative measures. Further note that these satisfy the requirements of a Jordan decomposition (they are mutually singular, and $\nu = \nu^{+} - \nu^{-}$).

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Note that in all of the above, we have assumed that we already have our hands on a signed measure $\nu$. The Hahn and Jordan decompositions are defined in terms of the preexisting $\nu$. In your example, you have not stated what $\nu$ is supposed to be. From context, I assume you want to define $\nu$ by the integral, i.e. $$ \nu(E) = \int_{E} f \,\mathrm{d}m, $$ where $m$ is the usual Lebesgue measure. A possible Hahn decomposition of this measure is given by $(P=[0,\infty), N=(-\infty,0))$. Other decompositions are possible, but any two Hahn decomposition differ by at most a $\nu$-null set. Then we can define $$ \nu^{+}(E) := \nu(E\cap P) = \int_{E\cap P} x\,\mathrm{d}m = \int_{E\cap[0,\infty)} x\,\mathrm{d}m, $$ and $$ \nu^{-}(E) := -\nu(E\cap N) = \int_{E\cap N} x^2\,\mathrm{d}m = \int_{E\cap (-\infty,0)} x^2\,\mathrm{d}m. $$