# Dirac measure

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

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

Since $x\notin\emptyset$, we have $\delta_{x}(\emptyset)=0$.

2. 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