Dirac Measure is Measure

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Theorem

Let $\left({X, \Sigma}\right)$ be a measurable space.

Let $x \in X$, and let $\delta_x$ be the Dirac measure at $x$.


Then $\delta_x$ is a measure.


Proof

Let us verify in turn that $\delta_x$ satisfies the axioms for a measure.


Axiom $(1)$

By definition of the Dirac measure, $\delta_x \left({E}\right) \ge 0$ for all $E \in \Sigma$.

$\Box$


Axiom $(2)$

Let $\left({E_n}\right)_{n \in \N}$ be a sequence of pairwise disjoint sets.

It follows that if for some $m \in \N$, $x \in E_m$, it must be that $n \ne m$ implies $x \notin E_n$.


Now suppose $x \in E_m$ for some $m \in \N$.

Then by definition of set union, $x \in \displaystyle \bigcup_{n \mathop \in \N} E_n$.

Thus:

\(\displaystyle \delta_x \left({\bigcup_{n \mathop \in \N} E_n}\right)\) \(=\) \(\displaystyle 1\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop \in \N} \delta_x \left({E_n}\right)\)

because $\delta_x \left({E_n}\right) = 0$ iff $n \ne m$, and $1$ otherwise.


Finally, if $x \notin E_n$ for all $n \in \N$, then by definition of set union:

$x \notin \displaystyle \bigcup_{n \mathop \in \N} E_n$

so that:

\(\displaystyle \delta_x \left({\bigcup_{n \mathop \in \N} E_n}\right)\) \(=\) \(\displaystyle 0\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop \in \N} 0\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop \in \N} \delta_x \left({E_n}\right)\)


Hence, from Proof by Cases:

$\displaystyle \sum_{n \mathop \in \N} \delta_x \left({E_n}\right) = \delta_x \left({\bigcup_{n \mathop \in \N} E_n}\right)$

$\Box$


Axiom $(3)$

By definition of the Dirac measure, $\delta_x \left({X}\right) = 1$.

Hence there is an $E \in \Sigma$ such that $\delta_x \left({E}\right)$ is finite.

$\Box$


Thus, $\delta_x$, satisfying all the axioms, is a measure.

$\blacksquare$