Dirac measure

Let $X$ be a nonempty set. Let $\mathcal{P}(X)$ denote the power set of $X$ . Then $(X,\mathcal{P}(X))$ is a measurable space.

Let $x\in X$ . The Dirac measure concentrated at $x$ is $\delta_{x}\colon\mathcal{P}(X)\to\{0,1\}$ defined by

\delta_{x}(E)=\begin{cases}1&\text{if }x\in E\\ 0&\text{if }x\notin E.\end{cases}

Note that the Dirac measure $\delta_{x}$ is indeed a measure:

1.

Since $x\notin\emptyset$ , we have $\delta_{x}(\emptyset)=0$ .
2.
If $\{A_{n}\}_{n\in\mathbb{N}}$ is a sequence of pairwise disjoint subsets of $X$ , then one of the following must happen:
- –
  
  $\displaystyle x\notin\bigcup_{n\in\mathbb{N}}A_{n}$ , in which case $\displaystyle\delta_{x}\left(\bigcup_{n\in\mathbb{N}}A_{n}\right)=0$ and $\delta_{x}(A_{n})=0$ for every $n\in\mathbb{N}$ ;
- –
  
  $\displaystyle x\in\bigcup_{n\in\mathbb{N}}A_{n}$ , in which case $x\in A_{n_{0}}$ for exactly one $n_{0}\in\mathbb{N}$ , causing $\displaystyle\delta_{x}\left(\bigcup_{n\in\mathbb{N}}A_{n}\right)=1$ , $\delta_{x}(A_{n_{0}})=1$ , and $\delta_{x}(A_{n})=0$ for every $n\in\mathbb{N}$ with $n\neq n_{0}$ .

Also note that $(X,\mathcal{P}(X),\delta_{x})$ is a probability space.

Let $\overline{\mathbb{R}}$ denote the extended real numbers. Then for any function $f\colon X\to\overline{\mathbb{R}}$ , the integral of $f$ with respect to the Dirac measure $\delta_{x}$ is

\int\limits_{X}f\,d\delta_{x}=f(x).

In other words, integration with respect to the Dirac measure $\delta_{x}$ amounts to evaluating the function at $x$ .

If $X=\mathbb{R}$ , $m$ denotes Lebesgue measure, $A$ is a Lebesgue measurable subset of $\mathbb{R}$ , and $\delta$ (no ) denotes the Dirac delta function, then for any measurable function $f\colon\mathbb{R}\to\mathbb{R}$ , we have

\int\limits_{A}\delta(t-x)f(t)\,dm(t)=\int\limits_{A}f\,d\delta_{x}=f(x)\delta% _{x}(A).

Moreover, if $f$ is defined so that $f(t)=1$ for all $t\in A$ , the above becomes

\int\limits_{A}\delta(t-x)\,dm(t)=\int\limits_{A}d\delta_{x}=\delta_{x}(A).

In other words, the function $\delta(t-x)$ (with $x\in\mathbb{R}$ fixed and $t$ a real variable) behaves like a Radon-Nikodym derivative of $\delta_{x}$ with respect to $m$ .

Note that, just as the Dirac delta function is a misnomer (it is not really a function), there is not really a Radon-Nikodym derivative of $\delta_{x}$ with respect to $m$ , since $\delta_{x}$ is not absolutely continuous with respect to $m$ .

Title	Dirac measure
Canonical name	DiracMeasure
Date of creation	2013-03-22 17:19:40
Last modified on	2013-03-22 17:19:40
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	18
Author	Wkbj79 (1863)
Entry type	Definition
Classification	msc 60A10
Classification	msc 26A42
Classification	msc 28A25
Classification	msc 28A12
Classification	msc 28A10
Related topic	Measure
Related topic	Integral2
Related topic	DiracDeltaFunction