What Is the Support of a Regular Borel Measure?

The support of a measure describes where the measure is present in a topological sense. It does not simply mean the set of points where a particular density happens to be nonzero. The key test is whether every open neighborhood of a point has positive measure.

For a regular Borel measure that is inner regular on open sets, the support is closed and its complement has measure zero. Consequently, it is the smallest closed set that carries the whole measure.

Definition of support

Let $X$ be a topological space and let $\mu$ be a Borel measure on it. Define

$$\operatorname{supp}(\mu) = \{x \in X : \mu(U) > 0 \text{ for every open neighborhood } U \text{ of } x\}.$$

Equivalently, a point is outside the support if it has at least one open neighborhood of measure zero. Thus

$$X \setminus \operatorname{supp}(\mu) = \bigcup\{U \subseteq X : U \text{ is open and } \mu(U)=0\}.$$

This definition makes the support a topological object: it depends on the measure and the open sets of $X$.

Why the support is closed and has full measure

The complement of the support is a union of open sets, so it is open. Therefore $\operatorname{supp}(\mu)$ is closed.

To see why the complement has measure zero, set $V=X\setminus\operatorname{supp}(\mu)$. Each point of $V$ has an open neighborhood of measure zero. Let $K$ be any compact subset of $V$. Those zero-measure neighborhoods cover $K$, and compactness gives a finite subcover, say $U_1,\ldots,U_n$. Hence

$$\mu(K) \leq \sum_{i=1}^{n}\mu(U_i)=0.$$

So every compact subset of $V$ has measure zero. If $\mu$ is inner regular on open sets, then

$$\mu(V)=\sup\{\mu(K):K\subseteq V,\ K\text{ compact}\}=0.$$

This inner-regularity property holds for Radon measures, under the usual convention that a Radon measure is locally finite and regular on a suitable locally compact Hausdorff space. Terminology varies: when using the phrase “regular Borel measure,” check that inner regularity on open sets is part of the convention.

The support is also the smallest closed set of full measure. Indeed, if $F$ is closed and $\mu(X\setminus F)=0$, then every point outside $F$ has the open, measure-zero neighborhood $X\setminus F$. No such point can belong to the support, so $\operatorname{supp}(\mu)\subseteq F$.

Worked example: Lebesgue measure on an interval

Define a measure on $\mathbb{R}$ by

$$\mu(A)=\int_A \mathbf{1}_{[0,1]}(x)\,dx,$$

where $\mathbf{1}_{[0,1]}$ is 1 on $[0,1]$ and 0 elsewhere. Then

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

If $x\in(0,1)$, every open interval around $x$ contains a subinterval of positive length inside $[0,1]$, so it has positive measure. The same holds at the endpoints: every neighborhood of 0 or 1 contains an interval of positive length inside $[0,1]$. If $x<0$ or $x>1$, a sufficiently small open interval around $x$ misses $[0,1]$ and has measure zero. Thus exactly the points of $[0,1]$ meet the definition.

The support need not have positive measure with respect to a separate background measure: for example, a Dirac measure at a point has that singleton as its support, even though the singleton has Lebesgue measure zero. The relevant measure in the definition is the measure being supported, not some other background measure.

Common mistakes

• Using the pointwise values of a density. For a measure with density $f$ relative to Lebesgue measure, changes to $f$ on a Lebesgue-null set do not change the measure or its support. The support is determined by whether neighborhoods have positive integral, not whether $f(x)>0$ at one particular point.
• Assuming the support is open. It is closed by definition; it need not be open or compact. For example, the measure on $\mathbb{R}$ with density $\mathbf{1}_{[0,\infty)}$ has support $[0,\infty)$, which is neither open nor compact. Lebesgue measure on all of $\mathbb{R}$ has support $\mathbb{R}$, which is not compact.
• Skipping the regularity condition. The support is closed from its definition, but the argument that its complement has measure zero uses inner regularity on open sets. Do not infer that full-measure conclusion from the word “Borel” alone.
• Confusing full measure with containing every point where mass can be assigned. A closed full-measure set is not necessarily the support. The support is the smallest such closed set, and each of its points must have positive measure in every neighborhood.