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.
  • 1
    $\begingroup$ If I'm correct, $X$ is just a metrizable space, the choice of a distance playing no role. $\endgroup$
    – YCor
    Nov 6, 2022 at 12:34
  • 1
    $\begingroup$ @YCor Yeah I don't see we use any property of the metric here. $\endgroup$
    – Analyst
    Nov 6, 2022 at 12:35
  • 2
    $\begingroup$ Let $X=[0,1]$ and let $m$ be Lebesgue measure. Take $\mu = m+im$. $\endgroup$
    – Nik Weaver
    Nov 6, 2022 at 14:18
  • 1
    $\begingroup$ @NikWeaver Could you please have a check if my update is fine? $\endgroup$
    – Analyst
    Nov 6, 2022 at 21:46
  • 1
    $\begingroup$ Looks good. I guess the $\frac{1}{2}$ can be improved to $\frac{1}{\sqrt{2}}$. $\endgroup$
    – Nik Weaver
    Nov 6, 2022 at 23:17

1 Answer 1


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.


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