Integral with respect to Dirac Measure

Theorem
Let $\struct {X, \Sigma}$ be a measurable space.

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

Let $f \in \MM _{\overline \R}, f: X \to \overline \R$ be a measurable function.

Then:


 * $\ds \int f \rd \delta_x = \map f x$

where the integral sign denotes the $\delta_x$-integral.

Proof
Notice that the constant function $g$ defined as


 * $X \ni x' \mapsto \map g {x'} := \map f x \in \overline \R$

is $\delta_x$-almost everywhere equal to $f$.

This follows from the fact that the set of elements of $X$ where $f$ and $g$ take different values, namely:


 * $\set {x' \in X : \map f x = \map g {x'} \ne \map f {x'} }$

does not contain $x$.

So, by the very definition of $\delta_x$, its $\delta_x$-measure is $0$.

Therefore: