Dirac measure

In mathematics, a Dirac measure is a measure &delta;x on a set X (with any &sigma;-algebra of subsets of X) that gives the singleton set {x} the measure 1, for a chosen element x &isin; X:
 * $$\delta_{x} \left( \{ x \} \right) = 1.$$

In general, the measure is defined by
 * $$\delta_{x} (A) = \begin{cases} 0, & x \not \in A; \\ 1, & x \in A. \end{cases}$$

for any measurable set A &sube; X.

The Dirac measure is a probability measure, and in terms of probability it represents the almost sure outcome x in the sample space X. We can also say that the measure is a single atom at x; however, treating the Dirac measure as an atomic measure is not correct when we consider the sequential definition of Dirac delta, as the limit of a delta sequence. The Dirac measures are the extreme points of the convex set of probability measures on X.

The name is a back-formation from the Dirac delta function, considered as a Schwartz distribution, for example on the real line; measures can be taken to be a special kind of distribution. The identity
 * $$\int_{X} f(y) \, \mathrm{d} \delta_{x} (y) = f(x),$$

which, in the form
 * $$\int_{X} f(y) \delta_{x} y \, \mathrm{d} (y) = f(x),$$

is often taken to be part of the definition of the "delta function", holds as a theorem of Lebesgue integration.

Properties of the Dirac measure
Let &delta;x denote the Dirac measure centred on some fixed point x in some measurable space (X, Σ).
 * &delta;x is probability measure, and hence a finite measure.

Suppose that (X, T) is a topological space and that &Sigma; is at least as fine as the Borel σ-algebra &sigma;(T) on X.
 * &delta;x is a strictly positive measure if and only if the topology T is such that x lies within every open set, e.g. in the case of the trivial topology {&empty;, X}.
 * Since &delta;x is probability measure, it is also a locally finite measure.
 * If X is a Hausdorff topological space with its Borel σ-algebra, then &delta;x satisfies the condition to be an inner regular measure, since singleton sets such as {x} are always compact. Hence, &delta;x is also a Radon measure.
 * Assuming that the topology T is fine enough that {x} is closed, which is the case in most applications, the support of &delta;x is {x}. (Otherwise, supp(&delta;x) is the closure of {x} in (X, T).) Furthermore, &delta;x is the only probability measure whose support is {x}.
 * If X is n-dimensional Euclidean space Rn with its usual σ-algebra and n-dimensional Lebesgue measure λn, &delta;x is a singular measure with respect to λn: simply decompose Rn as A = Rn \ {x} and B = {x} and observe that &delta;x(A) = λn(B) = 0.