Support (measure theory)

In mathematics, the support (sometimes topological support or spectrum) of a measure $\mu$ on a measurable topological space $(X, \operatorname(X))$ 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.

Motivation

A (non-negative) measure $\mu$ on a measurable space $(X, \Sigma)$ is really a function \mu : \Sigma \to , +\infty Therefore, in terms of the usual definition of support, the support of $\mu$ is a subset of the σ-algebra $\Sigma:$
$\operatorname (\mu) := \overline,$
where the overbar denotes set closure. However, this definition is somewhat unsatisfactory: we use the notion of closure, but we do not even have a topology on $\Sigma.$ What we really want to know is where in the space $X$ the measure $\mu$ is non-zero. Consider two examples:
# Lebesgue measure $\lambda$ on the real line $\Reals.$ It seems clear that $\lambda$ "lives on" the whole of the real line.
# A Dirac measure $\delta_p$ at some point $p \in \Reals.$ Again, intuition suggests that the measure $\delta_p$ "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 $\mu$ is zero, and take the support to be the remainder $X \setminus \.$ This might work for the Dirac measure $\delta_p,$ but it would definitely not work for $\lambda:$ since the Lebesgue measure of any singleton is zero, this definition would give $\lambda$ 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: $\$ (or the closure of this). It 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.

Definition

Let $(X, T)$ be a topological space; let $B(T)$ denote the Borel σ-algebra on $X,$ i.e. the smallest sigma algebra on $X$ that contains all open sets $U \in T.$ Let $\mu$ be a measure on $(X, B(T))$ Then the support (or spectrum) of $\mu$ is defined as the set of all points $x$ in $X$ for which every open neighbourhood $N_x$ of $x$ has positive measure:
$\operatorname (\mu) := \.$

Some authors prefer to take the closure of the above set. However, this is not necessary: see "Properties" below.

An equivalent definition of support is as the largest $C \in B(T)$ (with respect to inclusion) such that every open set which has non-empty intersection with $C$ has positive measure, i.e. the largest $C$ such that:
$(\forall U \in T)(U \cap C \neq \varnothing \implies \mu (U \cap C) > 0).$

Signed and complex measures

This definition can be extended to signed and complex measures.
Suppose that \mu : \Sigma \to \infty, +\infty/math> is a signed measure. Use the Hahn decomposition theorem to write
$\mu = \mu^+ - \mu^-,$
where $\mu^\pm$ are both non-negative measures. Then the support of $\mu$ is defined to be
$\operatorname (\mu) := \operatorname (\mu^+) \cup \operatorname (\mu^-).$

Similarly, if $\mu : \Sigma \to \Complex$ is a complex measure, the support of $\mu$ is defined to be the union of the supports of its real and imaginary parts.

Properties

$\operatorname (\mu_1 + \mu_2) = \operatorname (\mu_1) \cup \operatorname (\mu_2)$ holds.

A measure $\mu$ on $X$ is strictly positive if and only if it has support $\operatorname(\mu) = X.$ If $\mu$ 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, $x \in \operatorname(\mu),$ so $\operatorname(\mu) = X.$ Conversely, if $\operatorname(\mu) = X,$ then every non-empty open set (being an open neighbourhood of some point in its interior, which is also a point of the support) has positive measure; hence, $\mu$ is strictly positive.
The support of a measure is closed in $X,$as its complement is the union of the open sets of measure $0.$

In general the support of a nonzero measure may be empty: see the examples below. However, if $X$ is a Hausdorff topological space and $\mu$ is a Radon measure, a Borel set $A$ outside the support has measure zero:
$A \subseteq X \setminus \operatorname (\mu) \implies \mu (A) = 0.$
The converse is true if $A$ is open, but it is not true in general: it fails if there exists a point $x \in \operatorname(\mu)$ such that $\mu(\) = 0$ (e.g. Lebesgue measure). Thus, one does not need to "integrate outside the support": for any measurable function $f : X \to \Reals$ or $\Complex,$
$\int_X f(x) \, \mathrm \mu (x) = \int_ f(x) \, \mathrm \mu (x).$

The concept of ''support'' of a measure and that of spectrum of a self-adjoint linear operator on a Hilbert space are closely related. Indeed, if $\mu$ is a regular Borel measure on the line $\mathbb,$ then the multiplication operator $(Af)(x) = xf(x)$ is self-adjoint on its natural domain
$D(A) = \$
and its spectrum coincides with the essential range of the identity function $x \mapsto x,$ which is precisely the support of $\mu.$

Examples

Lebesgue measure

In the case of Lebesgue measure $\lambda$ on the real line $\Reals,$ consider an arbitrary point $x \in \Reals.$ Then any open neighbourhood $N_x$ of $x$ must contain some open interval $(x - \epsilon, x + \epsilon)$ for some $\epsilon > 0.$ This interval has Lebesgue measure $2 \epsilon > 0,$ so $\lambda(N_x) \geq 2 \epsilon > 0.$ Since $x \in \Reals$ was arbitrary, $\operatorname(\lambda) = \Reals.$

Dirac measure

In the case of Dirac measure $\delta_p,$ let $x \in \Reals$ and consider two cases:
# if $x = p,$ then every open neighbourhood $N_x$ of $x$ contains $p,$ so $\delta_p(N_x) = 1 > 0.$
# 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 $\delta_p(B) = 0.$
We conclude that $\operatorname(\delta_p)$ is the closure of the singleton set $\,$ which is $\$ itself.

In fact, a measure $\mu$ on the real line is a Dirac measure $\delta_p$ for some point $p$ if and only if the support of $\mu$ is the singleton set $\.$ Consequently, Dirac measure on the real line is the unique measure with zero variance (provided that the measure has variance at all).

A uniform distribution

Consider the measure $\mu$ on the real line $\Reals$ 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 $\operatorname(\mu) =$, 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 $\mu$-measure.

A nontrivial measure whose support is empty

The space of all countable ordinals with the topology generated by "open intervals" is a locally compact Hausdorff space. The measure ("Dieudonné measure") that assigns measure 1 to Borel sets containing an unbounded closed subset and assigns 0 to other Borel sets is a Borel probability measure whose support is empty. Measure theory, Example 7.1.3

A nontrivial measure whose support has measure zero

On a compact Hausdorff space the support of a non-zero measure is always non-empty, but may have measure $0.$ An example of this is given by adding the first uncountable ordinal $\Omega$ to the previous example: the support of the measure is the single point $\Omega,$ which has measure $0.$

References

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* (See chapter 2, section 2)
* (See chapter 3, section 2)

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