Support (measure theory)
# 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 σ-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:

* Lebesgue measure $\lambda$ on the real line $\mathbb {R} .$ It seems clear that $\lambda$ "lives on" the whole of the real line.
* 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:

* 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.
* 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 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:

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

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 $\mu :\Sigma \to \mathbb {C}$ 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 $f:X\to \mathbb {R}$ or $\mathbb {C} ,$

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

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 $\lambda$ on the real line $\mathbb {R} ,$ consider an arbitrary point $x\in \mathbb {R} .$ 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 $x\in \mathbb {R}$ was arbitrary, $\operatorname {supp} (\lambda )=\mathbb {R} .$

### Dirac measure

In the case of Dirac measure $\delta _{p},$ let $x\in \mathbb {R}$ 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 $\operatorname {supp} (\delta _{p})$ is the closure of the singleton set $\{p\},$ which is $\{p\}$ 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 $\{p\}.$ 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 $\mathbb {R}$ 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 $\operatorname {supp} (\mu )=[0,1].$ 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 $0.$

