A measure on a measurable space is really a function. Therefore, in terms of the usual definition of support, the support of is a subset of the σ-algebra : 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. What we really want to know is where in the space the measure is non-zero. Consider two examples:
A Dirac measure at some point. Again, intuition suggests that the measure "lives at" the point, 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 is zero, and take the support to be the remainder. This might work for the Dirac measure, but it would definitely not work for : since the Lebesgue measure of any singleton is zero, this definition would give 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:
However, the idea of "local strict positivity" is not too far from a workable definition:
Definition
Let be a topological space; let B 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. Then the support of μ is defined as the set of all points x in X for which every open neighbourhood Nx of x has positive measure: 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 such that every open set which has non-empty intersection with C has positive measure, i.e. the largest C such that:
Properties
A measure μ on X is strictly positive if and only if it has support supp = 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, so supp = X. Conversely, if supp = X, then every non-empty open set has positive measure; hence, μ is strictly positive.
The support of a measure is closed inX 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 setA outside the support has measure zero:
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 for some ε > 0. This interval has Lebesgue measure 2ε > 0, so λ ≥ 2ε > 0. Since x ∈ R was arbitrary, supp = R.
Dirac measure
In the case of Dirac measure δp, let x ∈ R and consider two cases:
if x = p, then every open neighbourhood Nx of x contains p, so δp = 1 > 0;
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 = 0.
We conclude that supp is the closure of the singleton set, which is 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. Consequently, Dirac measure on the real line is the unique measure with zero variance .
Consider the measure μ on the real line R defined by i.e. a uniform measure on the open interval. A similar argument to the Dirac measure example shows that supp = . Note that the boundary points 0 and 1 lie in the support: any open set containing 0 contains an open interval about 0, which must intersect, and so must have positive μ-measure.
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.
