---
tag: E76H86NN6Y
kind: definition
name: Total variation of a measure
msc: [28A10, 28A33]
defines: EPBND51CR5
uses: [VG2P7R91Z7, NY6TYHRGMS, 0BTMQ04556]
sources:
  - url: https://en.wikipedia.org/wiki/Total_variation
---

## Definition

Let $(X, \Sigma)$ be a measurable space and let $\nu$ be a [[signed measure|VG2P7R91Z7]] or a [[complex measure|NY6TYHRGMS]] on it.
The *total variation* of $\nu$ is the map $\abs{\nu} \colon \Sigma \to [0, \infty]$ given by
$$\abs{\nu}(E) = \sup \set{ \sum_{i=1}^{n} \abs{\nu(E_i)} : n \ge 1, \ E_1, \ldots, E_n \in \Sigma \text{ pairwise disjoint}, \ E = E_1 \cup \cdots \cup E_n },$$
where the [[supremum|0BTMQ04556]] is over all finite partitions of $E$ into measurable sets.
The *total variation norm* of $\nu$ is $\norm{\nu} = \abs{\nu}(X)$.

## Remarks

The map $\abs{\nu}$ is a measure, and $\abs{\nu(E)} \le \abs{\nu}(E)$ for all $E \in \Sigma$.
For a complex measure, $\abs{\nu}$ is a finite measure.

Example: let $\nu = \delta_0 - \delta_1$ on the subsets of $\R$, where $\delta_a$ is the Dirac measure at $a$.
For $E = \set{0, 1}$, one has $\nu(E) = 0$, but the partition $\set{0}, \set{1}$ gives $\abs{1} + \abs{-1} = 2$.
No partition gives more, so $\abs{\nu}(E) = 2$ and $\norm{\nu} = 2$.

Non-example: the map $E \mapsto \abs{\nu(E)}$ is not the total variation in general.
For the same $\nu$ it gives $0$ on $\set{0, 1}$, and it is not even additive: $\abs{\nu(\set{0})} + \abs{\nu(\set{1})} = 2 \ne 0$.

For a signed measure with Jordan decomposition $\nu = \nu^+ - \nu^-$, one has $\abs{\nu} = \nu^+ + \nu^-$.
If $\nu(E) = \int_E f \, d\mu$ with $f \in L^1(\mu)$, then $\abs{\nu}(E) = \int_E \abs{f} \, d\mu$.