# Support (measure theory)

In mathematics, the support (sometimes topological support or spectrum) 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.

## 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 σ-algebra $\Sigma$ :

$\operatorname {supp} (\mu ):=\bigcap \{A={\overline {A}}\in \Sigma \mid \mu (A^{c})=0\},$

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 $\mathbb {R}$ . It seems clear that $\lambda$  "lives on" the whole of the real line.
2. A Dirac measure $\delta _{p}$  at some point $p\in \mathbb {R}$ . 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:
$\{x\in X\mid {\text{for some open }}N_{x}\ni x,\mu (N_{x})>0\}$
(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 (XT) 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 ∈ T. Let μ be a measure on (X, B(T)). Then the support (or spectrum) of μ is defined as the set of all points x in X for which every open neighbourhood Nx of x has positive measure:

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

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 ∈ 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:

$(\forall U\in T)(U\cap C\neq \varnothing \implies \mu (U\cap C)>0).$

## Properties

• $\operatorname {supp} (\mu _{1}+\mu _{2})=\operatorname {supp} (\mu _{1})\cup \operatorname {supp} (\mu _{2})$
• 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 ∈ X is arbitrary, then any open neighbourhood of x, since it is an open set, has positive measure; hence, x ∈ supp(μ), so supp(μ) = X. Conversely, if supp(μ) = 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, μ 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 topological Hausdorff space and μ is a Radon measure, a measurable 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 ∈ supp(μ) such that μ({x}) = 0 (e.g. Lebesgue measure).
Thus, one does not need to "integrate outside the support": for any measurable function f : X → R or C,
$\int _{X}f(x)\,\mathrm {d} \mu (x)=\int _{\operatorname {supp} (\mu )}f(x)\,\mathrm {d} \mu (x).$
• 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
$D(A)=\{f\in L^{2}(\mathbb {R} ,d\mu )\mid xf(x)\in L^{2}(\mathbb {R} ,d\mu )\}$
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 λ on the real line R, consider an arbitrary point x ∈ R. Then any open neighbourhood Nx of x must contain some open interval (x − εx + ε) for some ε > 0. This interval has Lebesgue measure 2ε > 0, so λ(Nx) ≥ 2ε > 0. Since x ∈ R was arbitrary, supp(λ) = R.

### Dirac measure

In the case of Dirac measure δp, let x ∈ R and consider two cases:

1. if x = p, then every open neighbourhood Nx of x contains p, so δp(Nx) = 1 > 0;
2. on the other hand, if x ≠ p, then there exists a sufficiently small open ball B around x that does not contain p, so δp(B) = 0.

We conclude that supp(δp) 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 δp 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].

### A uniform distribution

Consider the measure μ on the real line R 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.

### 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 Ω to the previous example: the support of the measure is the single point Ω, which has measure 0.

## Signed and complex measures

Suppose that μ : Σ → [−∞, +∞] is a signed measure. Use the Hahn decomposition theorem to write

$\mu =\mu ^{+}-\mu ^{-},$

where μ± are both non-negative measures. Then the support of μ is defined to be

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

Similarly, if μ : Σ → C is a complex measure, the support of μ is defined to be the union of the supports of its real and imaginary parts.