Let \(s\) be a signed measure on a measurable space \(\Omega \), and let \(s = s^+ - s^-\) be its Jordan decomposition into a pair of (finite, mutually singular) positive measures \(s^+\) and \(s^-\). Then for every measurable set \(S \subseteq \Omega \),

where \(s^+(S)\) and \(s^-(S)\) denote the real-valued measures of \(S\) under the positive and negative parts.

Let \(\mu \) and \(\nu \) be probability measures on a measurable space \(\Omega \), and let \(\mu - \nu \) denote their signed measure difference. Then \((\mu - \nu )(\Omega ) = 0\).

Let \(\mu \) and \(\nu \) be probability measures on a measurable space \(\Omega \), and let \(\mu - \nu \) be their signed measure difference. Writing \((\mu -\nu )^{+}\) and \((\mu -\nu )^{-}\) for the positive and negative parts of the Jordan decomposition of \(\mu - \nu \), the two parts assign the same total real mass to the whole space:

Let \(s\) be a signed measure on a measurable space \(\Omega \), with Jordan decomposition \(s = s^+ - s^-\) into its positive and negative parts \(s^+\) and \(s^-\). Then the total mass of the total variation measure \(\left\lvert s\right\rvert \) of \(s\), taken as a real number, is the sum of the total masses of the two Jordan parts:

Let \(\mu \) and \(\nu \) be probability measures on a measurable space \(\Omega \), and let \(\mu -\nu \) denote their signed measure difference. Writing \(\left\lvert \mu -\nu \right\rvert \) for its total variation measure and \((\mu -\nu )^{+}\) for the positive part of its Jordan decomposition, one has

where both sides are evaluated as real numbers on the whole space \(\Omega \).

Let \(\mu \) and \(\nu \) be probability measures on a measurable space \(\Omega \), and let \(\mu - \nu \) denote their signed measure difference. Then for every measurable set \(S \subseteq \Omega \),

where \(\left\lvert \mu - \nu \right\rvert \) is the total variation measure of \(\mu - \nu \) and \(\left\lvert \mu - \nu \right\rvert (\Omega )\) is its total mass.

Let \(\mu \) and \(\nu \) be probability measures on a measurable space \(\Omega \), and let \(\mu - \nu \) denote their signed measure difference. Then for every measurable set \(S \subseteq \Omega \),

where \(\left\lvert \mu - \nu \right\rvert \) is the total variation measure of \(\mu - \nu \).

Let \(\mu \) and \(\nu \) be probability measures on a measurable space \(\Omega \), and let \(\mu - \nu \) denote their signed measure difference. Then there exists a measurable set \(S \subseteq \Omega \) at which

where \(\left\lvert \mu - \nu \right\rvert \) is the total variation measure of \(\mu - \nu \); that is, some event realizes exactly half the total variation of \(\mu - \nu \) over the whole space.

Let \(\mu \) and \(\nu \) be probability measures on a measurable space \(\Omega \), and let \(\mu - \nu \) denote their signed measure difference, with total variation measure \(\left\lvert \mu -\nu \right\rvert \). Then there exists a measurable set \(S \subseteq \Omega \) such that

Let \(\alpha \) be a finite type and let \(a : \alpha \to \mathbb {R}\) be a real-valued function. For every subset \(S \subseteq \alpha \),

Let \(a\colon \alpha \to \mathbb {R}\) be a real-valued function on a finite type \(\alpha \) whose values sum to zero, that is, \(\sum _{x\in \alpha } a(x) = 0\). Then the sum of the positive parts of \(a\) equals half the sum of its absolute values:

Let \(\alpha \) be a finite type and let \(a : \alpha \to \mathbb {R}\) be a real-valued function whose entries sum to zero, \(\sum _{x \in \alpha } a(x) = 0\). Then for every subset \(S \subseteq \alpha \),

Let \(\alpha \) be a finite type and let \(a : \alpha \to \mathbb {R}\) be a family of real numbers whose total sum vanishes, \(\sum _{x} a(x) = 0\). Then

Let \(\Omega \) be a finite discrete measurable space, and let \(\mu \) and \(\nu \) be probability measures on \(\Omega \). Then for every subset \(S \subseteq \Omega \),

where the sum ranges over all points of \(\Omega \) and each term contributes the singleton mass difference \(\mu (\{ \omega \} ) - \nu (\{ \omega \} )\) exactly when \(\omega \in S\) and \(0\) otherwise.

Let \(\Omega \) be a finite discrete measurable space, and let \(\mu \) and \(\nu \) be probability measures on \(\Omega \). Then the singleton mass differences sum to zero:

Let \(\mu \) be a probability measure on the two-point space \(\mathrm{Bool} = \{ \mathrm{true}, \mathrm{false}\} \). Then the mass at \(\mathrm{false}\) is one minus the mass at \(\mathrm{true}\):

Let \(a, b, c, d \in \mathbb {R}\) with \(b \ge 0\) and \(c \ge 0\). Then

Let \(\alpha \) and \(\beta \) be finite types, and let \(a, c : \alpha \to \mathbb {R}\) and \(b, d : \beta \to \mathbb {R}\) be real-valued weight functions. Suppose \(b\) and \(c\) are nonnegative and each sums to one, i.e. \(b(y) \ge 0\) for all \(y\), \(c(x) \ge 0\) for all \(x\), \(\sum _{y \in \beta } b(y) = 1\) and \(\sum _{x \in \alpha } c(x) = 1\). Then the \(\ell ^1\) gap between the two product weight functions is at most the sum of the marginal \(\ell ^1\) gaps:

Let \(\mu _1\) be a measure on a measurable space \(\alpha \) and \(\mu _2\) a \(\sigma \)-finite measure on a measurable space \(\beta \). Then for every \(x \in \alpha \) and \(y \in \beta \), the real-valued mass that the product measure \(\mu _1 \otimes \mu _2\) assigns to the singleton \(\{ (x,y)\} \) is the product of the singleton masses: