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 subscript) 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 equation 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 $m$