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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.

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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 $\varepsilon' = \varepsilon +\sum_n|\varepsilon_n|$.

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