# Complex Borel measures: relation between the total variation norm of a measure and those of its real and imaginary parts

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

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.