Let $X$ be a metric space and $\mathcal B$ its Borel $\sigma$-algebra. For $B \in \mathcal B$ we denote by $\Pi(B)$ the collection of all finite measurable partitions of $B$, i.e., $$ \Pi(B)=\left\{\left(B_{1}, \ldots, B_{n}\right) \,\middle\vert\, n \in \mathbb{N^*}, B_{i} \in \mathcal B, B_{i} \cap B_{j}=\varnothing \text { for } 1 \leq i \neq j \leq n, \bigcup_{i=1}^{n} B_{i}=B\right\} . $$

Let $\mu$ be a complex Borel measure on $X$. The variation $|\mu|$ of $\mu$ is defined by $$ |\mu|(B) := \sup \left\{\sum_{i=1}^{n}\left|\mu\left(B_{i}\right)\right| \,\middle\vert\, \left(B_{1}, \ldots, B_{n}\right) \in \Pi(B)\right\} \quad \forall B \in \mathcal B. $$

Let $[\mu] :=|\mu|(X)$ be the total variation norm of $\mu$. Let $\mu_1, \mu_2$ be the real and imaginary parts of $\mu$ respectively, i.e., $\mu = \mu_1 + i\mu_2$. Then $\mu_1, \mu_2$ are finite signed Borel measures on $X$. So for each $B \in \mathcal B$ we have $$ \begin{align} |\mu|(B) &= \sup \left\{\sum_{i=1}^{n} \sqrt{|\mu_1(B_{i})|^2 + |\mu_2(B_{i})|^2} \,\middle\vert\, \left(B_{1}, \ldots, B_{n}\right) \in \Pi(B)\right\} \\ &\le \sup \left\{\sum_{i=1}^{n} (|\mu_1(B_{i})| + |\mu_2(B_{i})|) \,\middle\vert\, \left(B_{1}, \ldots, B_{n}\right) \in \Pi(B)\right\} \\ &\le \sup \left\{\sum_{i=1}^{n} |\mu_1(B_i)| \,\middle\vert\, \left(B_{1}, \ldots, B_{n}\right) \in \Pi(B)\right\} \\ &\quad + \sup \left\{\sum_{i=1}^{n} |\mu_2(B_i)| \,\middle\vert\, \left(B_{1}, \ldots, B_{n}\right) \in \Pi(B)\right\} \\ &= |\mu_1| (B) + \mu_2(B). \end{align} $$

As such, $|\mu| \le |\mu_1| + |\mu_2|$. In particular, $[\mu] \le [\mu_1] + [\mu_2]$.

I would like to ask if either $|\mu| \ge |\mu_1| + |\mu_2|$ or $[\mu] \ge [\mu_1] + [\mu_2]$ is true.

Thank you so much for your elaboration!

**Update:** Let's define a new variation on the space of complex Borel measures on $X$.

For a finite signed Borel measure $\mu$, its new variation is $|\mu|' := |\mu|$.

For a complex Borel measure $\mu = \mu_1 + i\mu_2$ with $\mu_1, \mu_2$ being its real and imaginary parts, its new variation is $|\mu|' := |\mu_1| + |\mu_2|$.

Then for any complex Borel measure $\mu$, we have $$ \frac{1}{2} |\mu|' \le |\mu| \le |\mu|'. $$

We define $[\mu]':= |\mu|' (X)$. Then $[\cdot]'$ is a norm on the space of complex Borel measures such that $$ \frac{1}{2} [\cdot]' \le [\cdot] \le [\cdot]'. $$

It follows that

- $|\mu|' = |\mu_1|' + |\mu_2|'$ and thus $[\mu]' = [\mu_1]' + [\mu_2]'$ for every complex Borel measure $\mu$ whose real and imaginary parts are $\mu_1$ and $\mu_2$ respectively.
- $[\cdot]$ and $[\cdot]'$ are equivalent norms on the space of complex Borel measures.
- $[\cdot]$ and $[\cdot]'$ coincide on the subspace of finite signed Borel measures.

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