On the example of the The Hahn decomposition 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 $\nu$.
 A: 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


*

*$P \cup N = X$,

*$P \cap N = \emptyset$,

*$P$ 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 $N \cup M$).

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


*

*$\nu = \nu^{+} - \nu^{-}$, and

*$\nu^{+} \perp \nu^{-}$.


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 $\nu = \nu^{+} - \nu^{-}$).

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
$$ \nu^{-}(E) := -\nu(E\cap N) = \int_{E\cap N} x^2\,\mathrm{d}m = \int_{E\cap (-\infty,0)} x^2\,\mathrm{d}m. $$
