I have been going through the proof that the total variation of a signed measure is a finite measure given in "Introduction to Measure Theory and Functional Analysis" by Piermarco Cannarsa and Teresa D'Aprile.

They begin by showing that the total variation is additive. I understand the $|\mu|(A\sqcup B)\leq|\mu|(A)+|\mu|(B)$ direction, however I don't understand one step in the opposite direction:

Let $L,M\in\mathbb{R}$ be real numbers such that $L<|\mu|(A)$ and $M<|\mu|(B)$, then there exist partitions $(A_n)$ of $A$ and $(B_n)$ of $B$ such that $$\sum_{n=1}^\infty |\mu(A_n)| \geq L\text{, }\sum_{n=1}^\infty |\mu(B_n)| \geq M.$$ Furthermore, $(A_n \sqcup B_n)$ (actually they write $(A_n)\cup (B_n)$ but this is the same thing right?) is a partition of $A\sqcup B$, so $$|\mu|(A\sqcup B)\geq \sum_{n=1}^\infty (|\mu(A_n)|+|\mu(B_n)|)\geq L+M. \phantom{asdasdasdsadasd}\text{(1)}$$

This is where my problem is, I don't see why the first inequality is true. Trying to work through it myself, I get $$|\mu|(A\sqcup B)\geq \sum_{n=1}^\infty |\mu(A_n \sqcup B_n)|=\sum_{n=1}^\infty|\mu(A_n)+\mu(B_n)|$$ and then I'm stuck.

I'd really appreciate it if anybody could help me out - I know its most likely a case of me just not spotting something obvious.

Thanks for reading.


I think I've sorted it - I was getting a bit hung up on using additivity of $\mu$ and the triangle inequality. From the line marked (1), if we write $$C_{ni}=\begin{cases}A_n&\text{ if }i=1 \\ B_n&\text{ if } i=2\end{cases}$$ then $$L+M\leq \sum_{n=1}^\infty \sum_{i=1}^2|\mu(C_{ni})|.$$ Write the double summation as $\sum_{t=1}^\infty|\mu(C_{n_t i_t})|$, note that $(C_{n_t i_t})_{t\in\mathbb{N}}$ is a partition of $A\sqcup B$. So $$L+M\leq\sum_{t=1}^\infty |\mu(C_{n_t i_t})|\leq|\mu|(A\sqcup B).$$

I think this is the correct reasoning, somebody please point it out if I've done something horribly wrong though!


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