Integral with respect to Dirac 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 $x$.


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


Then:

$\displaystyle \int f \, \mathrm d \delta_x = f \left({x}\right)$

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


Proof

Notice that the constant function $g$ defined as

$X \ni x' \mapsto g \left({x'}\right) := f \left({x}\right) \in \overline{\R}$

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:

$\left\{ {x' \in X : f \left({x}\right) = g \left({x'}\right) \ne f \left({x'}\right)}\right\}$

does not contain $x$.


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

Therefore:

\(\displaystyle \int f \, \mathrm d \delta_x\) \(=\) \(\displaystyle \int g \, \mathrm d \delta_x\)
\(\displaystyle \) \(=\) \(\displaystyle \delta_x \left({X}\right) f \left({x}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle f \left({x}\right)\)

$\blacksquare$


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