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 1
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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \map f x\) |
$\blacksquare$
Proof 2
We first prove the result for positive simple functions.
Let $g : X \to \R$ be a positive simple function.
From Simple Function has Standard Representation, there exists:
- a finite sequence $a_1, \ldots, a_n$ of real numbers
- a partition $E_0, E_1, \ldots, E_n$ of $X$ into $\Sigma$-measurable sets
such that:
- $\ds g = \sum_{i \mathop = 0}^n a_i \chi_{E_i}$
Then, from the definition of the $\delta_x$-integral of a positive simple function:
- $\ds \int g \rd \delta_x = \sum_{i \mathop = 1}^n a_i \map {\delta_x} {E_i}$
Since $E_0, E_1, \ldots, E_n$ is a partition of $X$, precisely one contains $x$.
That is, $x \in E_i$ for precisely one $i$.
Then:
- $\map {\delta_x} {E_j} = 0$ for $i \ne j$
leaving:
- $\ds \int g \rd \delta_x = a_i$
Note that:
- $\ds \map g x = \sum_{j \mathop = 0}^n a_j \map {\chi_{E_j} } x$
and $\map {\chi_{E_j} } x = 1$ if and only if $j = i$, so we have:
- $\map g x = a_i$
So we have:
- $\ds \int g \rd \delta_x = \map g x$
for positive simple functions $g$.
Now consider a positive $\Sigma$-measurable function $f : X \to \overline \R_{\ge 0}$.
From Measurable Function is Pointwise Limit of Simple Functions, there exists an increasing sequence $\sequence {f_n}_{n \mathop \in \N}$ of positive simple functions such that:
- $\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$
for each $x \in X$.
For each $n \in \N$, we have:
- $\ds \int f_n \rd \delta_x = \map {f_n} x$
from our previous work.
From the Integral of Positive Measurable Function as Limit of Integrals of Positive Simple Functions, we have:
- $\ds \int f \rd \delta_x = \lim_{n \mathop \to \infty} \int f_n \rd \delta_x$
That is:
- $\ds \int f \rd \delta_x = \lim_{n \mathop \to \infty} \map {f_n} x = \map f x$
Now consider an arbitrary $\Sigma$-measurable function $f : X \to \overline \R$.
From Function Measurable iff Positive and Negative Parts Measurable, we have:
- $f^+$ and $f^-$ are $\Sigma$-measurable.
Now, if $\map f x \ge 0$, we have:
- $\map {f^+} x = \map f x$
and:
- $\map {f^-} x = 0$
So:
- $\ds \int f^+ \rd \delta_x = \map f x$
and:
- $\ds \int f^- \rd \delta_x = 0$
giving:
\(\ds \int f \rd \delta_x\) | \(=\) | \(\ds \int f^+ \rd \delta_x - \int f^- \rd \delta_x\) | Definition of Integral of Integrable Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \map f x - 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map f x\) |
If $\map f x < 0$, then we have:
- $\map {f^+} x = 0$
and:
- $\map {f^-} x = -\map f x$
so that:
- $\ds \int f^+ \rd \delta_x = 0$
and:
- $\ds \int f^- \rd \delta_x = -\map f x$
Then we have:
\(\ds \int f \rd \delta_x\) | \(=\) | \(\ds \int f^+ \rd \delta_x - \int f^- \rd \delta_x\) | Definition of Integral of Integrable Function | |||||||||||
\(\ds \) | \(=\) | \(\ds 0 - \paren {-\map f x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map f x\) |
Hence the demand.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $9.10 \ \text{(i)}$