Support (measure theory)

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

Motivation
A (non-negative) measure $$\mu$$ on a measurable space $$(X, \Sigma)$$ is really a function $$\mu : \Sigma \to [0, +\infty].$$ Therefore, in terms of the usual definition of support, the support of $$\mu$$ is a subset of the &sigma;-algebra $$\Sigma:$$ $$\operatorname{supp} (\mu) := \overline{\{A \in \Sigma \,\vert\, \mu(A) \neq 0\}},$$ 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:
 * 1) Lebesgue measure $$\lambda$$ on the real line $$\Reals.$$ It seems clear that $$\lambda$$ "lives on" the whole of the real line.
 * 2) 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:
 * 1) We could remove the points where $$\mu$$ is zero, and take the support to be the remainder $$X \setminus \{x \in X \mid \mu(\{x\}) = 0\}.$$ 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.
 * 2) 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: $$\{x \in X \mid \exists N_x \text{ open} \text{ such that } (x \in N_x \text{ and } \mu(N_x) > 0)\}$$ (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 &sigma;-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: $$\operatorname{supp} (\mu) := \{x \in X \mid \forall N_x \in T \colon (x \in N_x \Rightarrow \mu (N_x) > 0)\}.$$

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 $$\mu : \Sigma \to [-\infty, +\infty]$$ 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 $$\operatorname{supp} (\mu) := \operatorname{supp} (\mu^+) \cup \operatorname{supp} (\mu^-).$$

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

A measure $$\mu$$ on $$X$$ is strictly positive if and only if it has support $$\operatorname{supp}(\mu) = X.$$ 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, $$x \in \operatorname{supp}(\mu),$$ so $$\operatorname{supp}(\mu) = X.$$ Conversely, if $$\operatorname{supp}(\mu) = X,$$ 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: $$A \subseteq X \setminus \operatorname{supp} (\mu) \implies \mu (A) = 0.$$ The converse is true if $$A$$ is open, but it is not true in general: it fails if there exists a point $$x \in \operatorname{supp}(\mu)$$ such that $$\mu(\{x\}) = 0$$ (e.g. Lebesgue measure). Thus, one does not need to "integrate outside the support": for any measurable function $$f : X \to \Reals$$ or $$\Complex,$$ $$\int_X f(x) \, \mathrm{d} \mu (x) = \int_{\operatorname{supp} (\mu)} f(x) \, \mathrm{d} \mu (x).$$

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 $$\mathbb{R},$$ then the multiplication operator $$(Af)(x) = xf(x)$$ is self-adjoint on its natural domain $$D(A) = \{f \in L^2(\Reals, d\mu) \mid xf(x) \in L^2(\Reals, d\mu)\}$$ and its spectrum coincides with the essential range of the identity function $$x \mapsto x,$$ which is precisely the support of $$\mu.$$

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, $$\operatorname{supp}(\lambda) = \Reals.$$

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

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 $$\{p\}.$$ 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 $$\operatorname{supp}(\mu) = [0, 1].$$ 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 $$\mu$$-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.

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 $$0.$$