# Relation between the distance between two complex measures and their total variations.

Consider measure space $$(X,\mathcal{A})$$ and space of complex measures on $$X$$, denoted by $$\mathcal{M}_{C}(X)$$. For $$\mu \in \mathcal{M}_C(X)$$, we define total variation of $$\mu$$, as $$\vert \mu \vert(A) = \sup\big\{\sum_{n=1}^\infty \vert \mu(A_n) \vert: \{A_n\}_{n \in \mathbb{N}} \subset \mathcal{A}, \bigcup_{n=1}^\infty A_n = A \big\},$$ for $$A \in \mathcal{A}.$$ Total variation $$\vert\mu \vert$$ is a finite nonnegatve measure on $$X$$, moreover it defines norm on $$\mathcal{M}_{C}(X),$$ by putting $$\Vert \mu \Vert_{\mathcal{M}_C} = \vert \mu \vert(X),$$ and turns $$\mathcal{M}_{C}(X)$$ into a Banach space.

Clearly, for $$\mu \in \mathcal{M}_C(X)$$, also $$\vert \mu \vert \in \mathcal{M}_C(X)$$ and $$\Vert \mu \Vert_{\mathcal{M}_C} = \Vert \vert \mu \vert \Vert_{\mathcal{M}_C} = \vert \mu \vert(X).$$ Now let us take $$\nu \in \mathcal{M}_C(X)$$. I am trying to find relation between $$\Vert \mu - \nu \Vert_{\mathcal{M}_C}$$ and $$\Vert \vert \mu \vert - \vert\nu \vert \Vert_{\mathcal{M}_C}$$ e. g. $$\Vert \mu - \nu \Vert_{\mathcal{M}_C} \le \Vert \vert \mu \vert - \vert\nu \vert \Vert_{\mathcal{M}_C}.$$ Is there any relation of that kind? If so I would be grateful for some proof hints.

Relation opposite to proposed holds: $$\| |\mu| - |\nu| \| \leq \| \mu -\nu\|.$$ This could be thought of as something similar to follow up of triangle inequality $$||a|-|b||\leq |a-b|$$. Below I give sketch of a proof.
Let $$\{A_n\}$$ be disjoint measurable sets summing up to $$X$$, and for each $$n$$ let $$\{B_n^k\}$$ be measurable disjoint sets summing up to $$A_n$$. In following calculations $$\varepsilon, |\varepsilon_n|$$ can be made arbitrarily small by taking finer divisions $$A$$ and then finer $$B$$. \begin{align} \| |\mu| - |\nu| \| &= ||\mu| -|\nu||(X)\\ &=\sum_n ||\mu|(A_n) -|\nu|(A_n)|+\varepsilon\\ &=\sum_n\bigg| \sum_k |\mu(B_n^k)| -\sum_k |\nu(B_n^k)| +\varepsilon_n\bigg|+\varepsilon\\ &\leq\sum_n \bigg(\bigg| \sum_k|\mu(B_n^k)| - \sum_k|\nu(B_n^k)|\bigg|+|\varepsilon_n|\bigg)+\varepsilon\\ &\leq\sum_n \bigg(\sum_k\bigg| |\mu(B_n^k)| - |\nu(B_n^k)|\bigg|+|\varepsilon_n|\bigg)+\varepsilon\\ &=\sum_{n,k}\bigg| |\mu(B_n^k)| - |\nu(B_n^k)|\bigg|+\varepsilon'\\ &\leq\sum_{n,k}\bigg| \mu(B_n^k) - \nu(B_n^k)\bigg|+\varepsilon'\\ &\leq\| \mu -\nu\| +\varepsilon'\\ \end{align} Where $$\varepsilon' = \varepsilon +\sum_n|\varepsilon_n|$$.