Integral with respect to Dirac Measure/Proof 1

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

Define the constant function $g : X \to \overline \R$ by:

$\map g {x'} = \map f x$

for each $x' \in X$.

From Constant Function is Measurable, we have:

$g$ is $\Sigma$-measurable.

From Measurable Functions Determine Measurable Sets:

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

Further:

$x \not \in \set {x' \in X : \map g {x'} \ne \map f {x'} }$

So from the definition of the Dirac measure, we have:

$\map {\delta_x} {\set {x' \in X : \map g {x'} \ne \map f {x'} } } = 0$

So:

$g = f$ $\delta_x$-almost everywhere.

From A.E. Equal Positive Measurable Functions have Equal Integrals: Corollary 1, we have:

$\ds \int g \rd \delta_x = \int f \rd \delta_x$

We finally have:

\(\ds \int g \rd \delta_x\)

\(=\)

\(\ds \map f x \int \chi_X \rd \delta_x\)

Integral of Integrable Function is Homogeneous

\(\ds \)

\(=\)

\(\ds \map f x \map {\delta_x} X\)

Integral of Characteristic Function: Corollary

\(\ds \)

\(=\)

\(\ds \map f x\)

$\blacksquare$