Total Variation Measure: Definition? The total variation according to Rudin is defined as:
$$|\mu|(E):=\sup_{\bigcup_{k\in\mathbb{N}}E_k=E}\sum_k|\mu(E_k)|$$
where the supremum is taken over all countable partitions.
Now I'm reading in my notes the slightly different definition:
$$|\mu|(E):=\sup_{\bigcup_{k=1}^N E_k=E}\sum_k|\mu(E_k)|$$
here instead explicitely only considering finite partitions.
I distrust this definition also since it refers for proofs to Rudin but I'm stuck finding a counterexample giving merely a finitely additive total variation. Is there some example?
 A: The definitions are equivalent. Indeed, take a  countable partition $E=\bigcup E_k$. By unconditional convergence both the series and its absolute converge. Thus for every $\epsilon>0$ there is $N$ such that:
$$\|\sum_k|\mu(E_k)|-\sum_{k\leq N}|\mu(E_k)|||<\frac{\epsilon}{2}$$
whereas since the measure is countably additive there is $N'$ such that:
$$\|\mu(\bigcup_{k>N}E_k)\|=\|\mu(\bigcup_k E_k)-\mu(\bigcup_{k\leq N}E_k)\|=\|\sum_k\mu(E_k)-\sum_{k\leq N}\mu(E_k)\|<\frac{\epsilon}{2}$$
Taking both together this implies that the finite partition $E_1,\ldots,E_{N_0},\bigcup_{k>N_0}E_k$ with $N_0:=\max\{N,N'\}$ gets arbitrarily close:
$$\|\sum_k|\mu(E_k)|-\sum_{k\leq N}|\mu(E_k)|-|\mu(\bigcup_{k>N}E_k)|\|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$
This shows the desired inequality: $$\sup_{\bigcup_{k\in\mathbb{N}} E_k=E}\sum_k |\mu(E_k)|\leq\sup_{\bigcup_{k=1}^N E_k=E}\sum_k |\mu(E_k)|$$
The reverse inequality is either vacuous if "countable" includes "finite", or easy since every finite partition can be made countably infinite by letting $E_k=\varnothing$ for all large $k$.
