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](https://en.wikipedia.org/wiki/Norm_(mathematics)#Equivalent_norms) on the space of complex Borel measures.
* $[\cdot]$ and $[\cdot]'$ coincide on the subspace of finite signed Borel measures.



Complex Borel measures: relation between the total variation norm of a measure and those of its real and imaginary parts
I represent below @NikWeaver's idea of improving $\frac{1}{2} [\cdot]' \le [\cdot]$ to get $\frac{1}{\sqrt 2} [\cdot]' \le [\cdot]$.




---


For complex number $z = x + iy$ with $x,y \in \mathbb R$, we have
$$
|z| \ge \frac{|x| +|y|}{\sqrt{2}} .
$$


Fix a Borel subset $B$ of $X$. Then
$$
\begin{align}
|\mu|(B) &= \sup \left\{\sum\_{i=1}^{n} \sqrt{|\mu\_1(B\_{i})|^2 + |\mu\_2(B\_{i})|^2} \,\middle\vert\, \left(B\_{1}, \ldots, B\_{n}\right) \in \Pi(B)\right\} \\
&\ge \frac{1}{\sqrt 2} \sup \left\{\sum\_{i=1}^{n} |\mu\_1(B\_{i})| + |\mu\_2(B\_{i})| \,\middle\vert\, \left(B\_{1}, \ldots, B\_{n}\right) \in \Pi(B)\right\}.
\end{align}
$$


It remains to prove that
$$
\begin{aligned}
&\sup \left\{ \sum\_{i=1}^{n} |\mu\_1(B\_{i})| + |\mu\_2(B\_{i})| \,\middle\vert\, (B\_{1}, \ldots, B\_{n}) \in \Pi(B)\right\} \\
= & \sup \left\{ \sum\_{i=1}^{n} |\mu\_1(B\_{i})| \,\middle\vert\, (B\_{1}, \ldots, B\_{n}) \in \Pi(B) \right\} + \sup \left\{\sum\_{i=1}^{n} |\mu\_2(B\_{i})| \,\middle\vert\, (B\_{1}, \ldots, B\_{n}) \in \Pi(B) \right\}.
\end{aligned}
$$


The direction $\le$ is obvious. Let's prove the reverse $\ge$. For each $n$, let


* $(B\_{1,1}, \ldots, B\_{1, \varphi\_n}) \in \Pi(B)$ such that $\sum\_{i=1}^{\varphi\_n} |\mu\_1(B\_{1, i})| \nearrow |\mu\_1| (B)$.
* $(B\_{2,1}, \ldots, B\_{2, \psi\_n}) \in \Pi(B)$ such that $\sum\_{i=1}^{\psi\_n} |\mu\_2(B\_{2, i})| \nearrow |\mu\_2| (B)$.
* $(B\_{3, 1}, \ldots, B\_{3, \lambda\_n}) :=\{B\_{1, i} \cap B\_{2, j} \mid i = 1, \ldots, \varphi\_n \text{ and } j = 1, \ldots, \psi\_n\} \in \Pi(B)$.


Then $(B\_{3, 1}, \ldots, B\_{3, \lambda\_n}) \in B$ is finer than both $(B\_{1,1}, \ldots, B\_{1, \varphi\_n})$ and $(B\_{2,1}, \ldots, B\_{2, \psi\_n})$. By triangle inequality, we have
$$
\sum\_{i=1}^{\varphi\_n} |\mu\_1(B\_{1, i})| \le \sum\_{i=1}^{\lambda\_n} |\mu\_1(B\_{3, i})|
\quad \text{and} \quad 
\sum\_{i=1}^{\psi\_n} |\mu\_2(B\_{2, i})| \le \sum\_{i=1}^{\lambda\_n} |\mu\_2(B\_{3, i})|.
$$


It follows that
$$
 \sum\_{i=1}^{\lambda\_n} |\mu\_1(B\_{3,i})| + |\mu\_2(B\_{3,i})| \ge \sum\_{i=1}^{\varphi\_n} |\mu\_1(B\_{1, i})| + \sum\_{i=1}^{\psi\_n} |\mu\_2(B\_{2, i})| .
$$


The claim then follows.


