The Dirac Measure at x

# The Dirac Measure at x

Recall from the General Measurable Spaces and Measure Spaces page that if $X$ is a set and $\mathcal A$ is a $\sigma$-algebra on $X$ then the pair $(X, \mathcal A)$ is called a measurable space. We said that the sets $E \in \mathcal A$ are called measurable sets.

We said that a function $\mu : \mathcal A \to [0, \infty]$ is called a measure on $\mathcal A$ if $\mu (\emptyset) = 0$ and if for every countable collection of disjoint measurable sets $(E_n)$ we have that $\displaystyle{\mu \left ( \bigcup_{n=1}^{\infty} E_n \right ) = \sum_{n=1}^{\infty} \mu (E_n)}$, and we defined the triple $(X, \mathcal A, \mu)$ to be a measure space.

We now look a special measure.

 Definition: Let $X$ be any set, and let $\mathcal A$ be any $\sigma$-algebra on $X$. Let $x \in X$. Then the Dirac Measure at $x$ is the measure $\delta_x : \mathcal A \to \{ 0, 1 \}$ defined for all $E \in \mathcal A$ by $\left\{\begin{matrix} 1 & \mathrm{if} \: x \in E \\ 0 & \mathrm{if} \: x \not \in E \end{matrix}\right.$.

Let $x \in X$. Let's verify that $\delta_x$ is indeed a measure.

Clearly, since $x \not \in \emptyset$ we have that:

(1)
\begin{align} \quad \delta_x (\emptyset) = 0 \end{align}

Now let $(E_n)_{n=1}^{\infty}$ be a countable collection of mutually disjoint measurable sets. Then $\displaystyle{\bigcup_{n=1}^{\infty} E_n}$ contains $x$ if and only if there exists an $n \in \mathbb{N}$ such that $x \in E_n$. From this observation we immediately get that:

(2)
\begin{align} \quad \delta_x \left ( \bigcup_{n=1}^{\infty} E_n \right ) = \sum_{n=1}^{\infty} \delta_x (E_n) \end{align}

So indeed $\delta_x$ is a measure and $(X, \mathcal A, \delta_x)$ is a measure space.