Support (measure theory)

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In mathematics, the support of a measure $μ$ on a measurable topological space $(X,Borel(X))$ is a precise notion of where in the space $X$ the measure "lives". It is defined to be the largest (closed) subset of $X$ for which every open neighbourhood of every point of the set has positive measure.

1 Motivation
2 Definition
3 Properties
4 Examples
5 Signed and complex measures
6 Reference

[edit] Motivation

Recall that a (non-negative) measure $μ$ on a measurable space $(X, \mathcal{A})$ is really a function $\mu : \mathcal{A} \to [0, + \infty]$ . Therefore, in terms of the usual definition of support, the support of $μ$ is a subset of the sigma algebra $\mathcal{A}$ :

$\mathrm{supp} (\mu) := \overline{\{ A \in \mathcal{A} | \mu (A) > 0 \}}.$

However, this definition is somewhat unsatisfactory: we do not even have a topology on $\mathcal{A}$ ! What we really want to know is where in the space $X$ the measure $μ$ is non-zero. Consider two examples:

Lebesgue measure $λ$ on the real line $\mathbb{R}$ . It seems clear that $λ$ "lives on" the whole of the real line.
A Dirac measure $δ p$ at some point $p \in \mathbb{R}$ . Again, intuition suggests that the measure $δ p$ "lives at" the point $p$ , and nowhere else.

In light of these two examples, we can reject the following candidate definitions in favour of the one in the next section:

We could remove the points where $μ$ is zero, and take the support to be the remainder $X \setminus \{ x \in X | \mu ( \{ x \} ) = 0 \}$ . This might work for the Dirac measure $δ p$ , but it would definitely not work for $λ$ : since the Lebesgue measure of any point is zero, this definition would give $λ$ empty support.
By comparison with the notion of strict positivity of measures, we could take the support to be the set of all points with a neighbourhood of positive measure:

$\{ x \in X | \mbox{for some open } N_{x} \ni x, \mu(N_{x}) > 0 \}$

(or the closure of this). This is also too simplistic: by taking

N x = X

for all points $x \in X$ , this would make the support of every measure except the zero measure the whole of

X

However, the idea of "local strict positivity" is not too far from a workable definition:

[edit] Definition

Let $(X, \mathcal{T})$ be a topological space; let $Borel(X)$ denote the Borel σ-algebra on $X$ ,i.e. the smallest sigma algebra on $X$ that contains all open sets $U \in \mathcal{T}$ . Let $μ$ be a measure on $(X,Borel(X))$ . Then the support of $μ$ is defined to be the set of all points $x \in X$ for which every open neighbourhood of $x$ has positive measure:

$\mathrm{supp} (\mu) := \{ x \in X | \forall x \in N_{x} \in \mathcal{T}, \mu (N_{x}) > 0 \}.$

Some authors prefer to take the closure of the above set. However, this is not necessary: see "Properties" below. As such, an equivalent definition of the support is as the largest closed set $C \subseteq X$ (with respect to inclusion) such that

$U \in \mathcal{T} \mbox{ and } U \cap C \neq \emptyset \implies \mu (U \cap C) > 0,$

i.e. every open set that has non-trivial intersection with the support has positive measure.

[edit] Properties

A measure $μ$ on $X$ is strictly positive if and only if it has support $supp(μ) = X$ . If $μ$ is strictly positive and $x \in X$ is arbitrary, then any open neighbourhood of $x$ , since it is an open set, has positive measure; hence, $x \in \mathrm{supp} (\mu)$ , so $supp(μ) = X$ . Conversely, if $supp(μ) = X$ , then every open set is an open neighbourhhod of some point in its interior, which is also a point of the support, and so has positive measure; hence, $μ$ is strictly positive.
The support of a measure is closed in $X$ . Suppose that $x$ is a limit point of $supp(μ)$ , and let $N x$ be an open neighbourhood of $x$ . Since $x$ is a limit point of the support, there is some $y \in N_{x} \cap \mathrm{supp} (\mu)$ , $y \neq x$ . But $N x$ is also an open neighbourhood of $y$ , so $μ(N x) > 0$ , as required. Hence, $supp(μ)$ contains all its limit points, i.e. it is closed.
If $A$ is a measurable set outside the support, then $A$ has measure zero:

$A \subseteq X \setminus \mathrm{supp} (\mu) \implies \mu (A) = 0.$

The converse is not true in general: it fails if there exists $x \in \mathrm{supp} (\mu)$ such that $\mu \left( \{ x \} \right) = 0$ (e.g. Lebesgue measure).

One does not need to "integrate outside the support": for any measurable function $f : X \to \mathbb{R}$ or $\mathbb{C}$ ,

$\int_{X} f(x) \, \mathrm{d} \mu (x) = \int_{\mathrm{supp} (\mu)} f(x) \, \mathrm{d} \mu (x).$

[edit] Examples

[edit] Lebesgue measure

In the case of Lebesgue measure on the real line, consider an arbitrary point $x \in \mathbb{R}$ . Then any open neighbourhood $N x$ of $x$ must contain some open interval $(x - \varepsilon, x + \varepsilon)$ for some $\varepsilon > 0$ . This interval has Lebesgue measure $\varepsilon > 0$ , so $\mu (N_{x}) \geq \varepsilon > 0$ . Since $x \in \mathbb{R}$ was arbitrary, $\mathrm{supp} (\lambda) = \mathbb{R}$ .

[edit] Dirac measure

In the case of Dirac measure $δ p$ , let $x \in \mathbb{R}$ and consider two cases:

if $x = p$ , then every open neighbourhood $N x$ of $x$ contains $p$ , so $δ p (N x) = 1 > 0$ ;
on the other hand, if $x \neq p$ , then there exists a sufficiently small open ball $B$ around $x$ that does not contain $p$ , so $δ p (B) = 0$ .

We conclude that $supp(δ p)$ is the closure of the singleton set ${p}$ , which is ${p}$ itself.

In fact, a measure $μ$ on the real line is a Dirac measure $δ p$ for some point $p$ if and only if the support of $μ$ is the singleton set ${p}$ . Consequently, Dirac measure on the real line is the unique measure with zero variance [provided that the measure has variance at all].

[edit] A uniform distribution

Consider the measure $μ$ on the real line defined by

$\mu (A) := \lambda (A \cap (0, 1))$

i.e. a uniform measure on the open interval $(0,1)$ . A similar argument to the Dirac measure example shows that $supp(μ) = [0,1]$ . Note that the boundary points 0 and 1 lie in the support: any open set containing 0 (or 1) contains an open interval about 0 (or 1), which must intersect $(0,1)$ , and so must have positive $μ$ -measure.

[edit] Signed and complex measures

Suppose that $\mu : \mathcal{A} \to [- \infty, + \infty]$ is a signed measure. Use the Hahn decomposition theorem to write

μ = μ + - μ -,

where $\mu^{\pm}$ are both non-negative measures. Then the support of $μ$ is defined to be

$\mathrm{supp} (\mu) := \mathrm{supp} (\mu^{+}) \cup \mathrm{supp} (\mu^{-}).$

Similarly, if $\mu : \mathcal{A} \to \mathbb{C}$ is a complex measure, the support of $μ$ is defined to be the union of the supports of its real and imaginary parts.

[edit] Reference

Ambrosio, L., Gigli, N. & Savaré, G. (2005). Gradient Flows in Metric Spaces and in the Space of Probability Measures. ETH Zürich, Birkhäuser Verlag, Basel. ISBN 3-764-32428-7.

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Category: Measure theory