Integral with respect to Dirac Measure/Proof 2

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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

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 Measure-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 Measure-Integrable Function
\(\ds \) \(=\) \(\ds 0 - \paren {-\map f x}\)
\(\ds \) \(=\) \(\ds \map f x\)

Hence the demand.

$\blacksquare$

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