Is the positive or negative variation of a signed measure finite? I'm studying measure theory and read about signed measure. A signed measure is a function $\nu:\mathcal{A}\to \mathbb{R}\cup\{\pm\infty\}$, where $\mathcal A$ is a certain $\sigma-$algebra, such that 


*

*$\nu(\varnothing)=0$

*$\nu$ is $\sigma-$aditive

*$\nu$ can take the $\infty$ value or the $-\infty$ value, but not both.


I manage the next definitions. The positive variation of $\nu$ is defined by $\nu^+(A):=\sup\{\nu(B): B\subseteq A,B\in\mathcal{A}\},\quad\forall A\in\mathcal A$, and the negative variation of $\nu$ is defined by $\nu^-:=(-\nu)^+$. I prove that both are positive measures, but I don't know if anyone of them is finite. I read on wikipedia (https://en.wikipedia.org/wiki/Signed_measure, Properties section) that one of them is finite, but it uses Hahn decomposition theorem to define the positive and negative variations and I am not acquainted with that theorem.    
 A: Suppose $\nu$ does not take the $-\infty$ value. Let us prove that $\nu^-$ is finite. 
First, note that, if for each $n\in \mathbb{N}$, there is $B_n\in \mathcal A$ such that $\nu(B_n)<-n$, then, for all $n\in \mathbb{N}$, $\nu^-(B_n)>n$ and so, for all $n\in \mathbb{N}$,
$$\nu^-\left (\bigcup_{n\in \mathbb{N}}B_n \right)\geq\nu^-(B_n)>n$$
So
$$\nu^-\left (\bigcup_{n\in \mathbb{N}}B_n \right)= +\infty$$
Now, let us define, for all $n\in \mathbb{N}$, $D_0=B_0$ and, if $n\geq 1$, 
$$ D_n=B_n -\bigcup_{k=0}^{n-1}B_k$$
Then, it is imediate that, for all $n,m\in \mathbb{N}$, if $n\neq m$ then $D_n \cap D_m =\emptyset$ and 
$$ \sum_{n\in \mathbb{N}} D_n = \bigcup_{n\in \mathbb{N}}B_n$$
So we have 
$$ \sum_{n\in \mathbb{N}} \nu^-(D_n) = \nu^-\left (\sum_{n\in \mathbb{N}} D_n \right)=\nu^-\left (\bigcup_{n\in \mathbb{N}}B_n\right)= +\infty$$
Now, from the definition of $\nu^-$ we have that for each $n\in \mathbb{N}$, there is $E_n\subseteq D_n$ such that 
$$\nu^-(D_n)-\frac{1}{2^{n+1}}\leq -\nu(E_n)\leq \nu^-(D_n) $$
So we have 
$$\sum_{n\in \mathbb{N}} -\nu(E_n) = +\infty$$
Since, for all $n,m\in \mathbb{N}$, if $n\neq m$ then $E_n \cap E_m =\emptyset$, we have 
$$\nu \left (\sum_{n\in \mathbb{N}} E_n \right) =\sum_{n\in \mathbb{N}} \nu(E_n) = -\infty$$
But this is a  contradiction, because $\nu$ does not take the $-\infty$ value. 
So, there is $n\in \mathbb{N}$ such that, for all $B\in\mathcal A$, 
$\nu(B)\geq -n$. It means that, for all $B\in\mathcal A$,  $-\nu(B)\leq n$ and then we have, for all $A\in\mathcal A$, $$\nu^-(A)=\sup\{-\nu(B): B\subseteq A,B\in\mathcal{A}\}\leq n$$ 
So $\nu^-$ is finite. 
In a similar way we can prove that if $\nu$ does not take the $\infty$ value, then $\nu^+$ is finite. 
A: It is not finite. $(0,\infty)$ with the standard measure has infinite variation.
