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 ${\delta}_{x}:\mathcal{P}(X)\to \{0,1\}$ defined by
$${\delta}_{x}(E)=\{\begin{array}{cc}1\hfill & \text{if}x\in E\hfill \\ 0\hfill & \text{if}x\notin E.\hfill \end{array}$$ 
Note that the Dirac measure ${\delta}_{x}$ is indeed a measure^{}:

1.
Since $x\notin \mathrm{\varnothing}$, we have ${\delta}_{x}(\mathrm{\varnothing})=0$.

2.
If ${\{{A}_{n}\}}_{n\in \mathbb{N}}$ is a sequence of pairwise disjoint subsets of $X$, then one of the following must happen:

–
$x\notin {\displaystyle \bigcup _{n\in \mathbb{N}}}{A}_{n}$, in which case ${\delta}_{x}\left({\displaystyle \bigcup _{n\in \mathbb{N}}}{A}_{n}\right)=0$ and ${\delta}_{x}({A}_{n})=0$ for every $n\in \mathbb{N}$;

–
$x\in {\displaystyle \bigcup _{n\in \mathbb{N}}}{A}_{n}$, in which case $x\in {A}_{{n}_{0}}$ for exactly one ${n}_{0}\in \mathbb{N}$, causing ${\delta}_{x}\left({\displaystyle \bigcup _{n\in \mathbb{N}}}{A}_{n}\right)=1$, ${\delta}_{x}({A}_{{n}_{0}})=1$, and ${\delta}_{x}({A}_{n})=0$ for every $n\in \mathbb{N}$ with $n\ne {n}_{0}$.

–
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 $f:X\to \overline{\mathbb{R}}$, the integral of $f$ with respect to the Dirac measure ${\delta}_{x}$ is
$$\underset{X}{\int}f\mathit{d}{\delta}_{x}=f(x).$$ 
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^{} $f:\mathbb{R}\to \mathbb{R}$, we have
$$\underset{A}{\int}\delta (tx)f(t)\mathit{d}m(t)=\underset{A}{\int}f\mathit{d}{\delta}_{x}=f(x){\delta}_{x}(A).$$ 
Moreover, if $f$ is defined so that $f(t)=1$ for all $t\in A$, the above becomes
$$\underset{A}{\int}\delta (tx)\mathit{d}m(t)=\underset{A}{\int}\mathit{d}{\delta}_{x}={\delta}_{x}(A).$$ 
In other words, the function $\delta (tx)$ (with $x\in \mathbb{R}$ fixed and $t$ a real variable) behaves like a RadonNikodym 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 RadonNikodym 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  20130322 17:19:40 
Last modified on  20130322 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 