Integral with respect to Dirac Measure/Proof 1

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: