According to wikipedia, there are two definitions of total varation for the case of real-valued signed measure $\mu$:
- the first is classical total variation definition, where the total variation $|\mu|$ is given by $|\mu|(A)=\sup_B\mu(B)-\inf_B\mu(B)$ for all $B\in\mathcal F$ and $B\subset A$. If we denote $\sup_B\mu(B)$ and $-\inf_B\mu(B)$ by $\mu^+$ and $\mu^-$ respectively, then we have $|\mu|=\mu^++\mu^-$;
- the second one is given by Rudin that $|\mu|(A)=\sup_\pi\sum_{B\in\pi}|\mu(B)|$, where the supremum is taken over all partitions $\pi$ of a measurable set $A$ into a countable number of disjoint measurable subsets.
Wiki concludes that this definition (by Rudin) coincides with the above definition $|\mu |=\mu ^{+}+\mu ^{-}$ for the case of real-valued signed measures, but no more details for why. I have no clue to prove this conclusion. Can you give me any hints?
Best Answer
In this answer, assume that $\mu^+$ and $\mu^-$ are both finite.
The first step is to show that for each $\pi$ you have $\displaystyle\sum_{B\in\pi}|\mu(B)|\leq\mu^++\mu^-$.
To do this, let $A_1$ be the union of the sets $C\in\pi$ such that $\mu(C)>0$, and $A_2$ be the union of the sets $E\in\pi$ such that $\mu(E)\leq0$.
Note that $\mu(A_1)\leq\mu^+$ and that $-\mu(A_2)\leq\mu^-$.
The second step is to show that for each $\epsilon>0$ you can find $\pi$ such that $\displaystyle\sum_{B\in\pi}|\mu(B)|\geq\mu^++\mu^--2\epsilon$.
To do this, find $G_1$ and $G_2$ such that $\mu(G_1)\geq\mu^+-\epsilon$ and that $-\mu(G_2)\geq\mu^--\epsilon$.
If $G_1$ and $G_2$ are already disjoint, then that can be your partition.
If $G_1$ and $G_2$ are not already disjoint, show that $|\mu(G_1\setminus G_2)|+|\mu(G_1\cap G_2)|+|\mu(G_2\setminus G_1)|\geq \mu(G_1)-\mu(G_2)$. I'll leave it as an exercise; comment if you want elaboration. Now, these three sets can also be your partition.