User:Caliburn/s/mt/Darboux and Lebesgue Integral Coincide

Theorem

Let $\map \LL {\closedint a b}$ be the set of Lebesgue measurable subsets of $\closedint a b$.

Let $\lambda$ be the Lebesgue measure on $\struct {\closedint a b, \map \LL {\closedint a b} }$.

Let $f : \closedint a b \to \R$ be a Darboux integrable bounded function.

Then $f$ is Lebesgue integrable and:

$\ds \int_{\closedint a b} \map f x \map {\rd \lambda} x = \int_a^b \map f x \rd x$

Proof

From Riemann-Lebesgue Theorem: Proof 2: Lemma:

there exists Lebesgue integrable bounded functions $g, h : \closedint a b \to \R$ with $g = h$ $\lambda$-almost everywhere such that:

$\ds \int_{\closedint a b} \map g x \map {\rd \lambda} x = \int_{\closedint a b} \map h x \map {\rd \lambda} x = \int_a^b \map f x \rd x$

and:

$g \le f \le h$

Then there exists a $\lambda$-null set $N \subseteq \closedint a b$ such that:

if $x \in \closedint a b$ such that $\map g x \ne \map h x$ then $x \in N$.

Note that if $x \in \closedint a b$:

$\map g x = \map h x$

then:

$\map f x = \map g x = \map h x$

So, if:

$\map f x \ne \map g x$

then:

$\map g x \ne \map h x$

so $x \in N$.

So, we have:

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

From Lebesgue Measure is Complete, we have:

$\struct {\closedint a b, \map \LL {\closedint a b}, \lambda}$ is complete.

So by Function A.E. Equal to Measurable Function in Complete Measure Space is Measurable, we have:

$f$ is Lebesgue measurable.

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

$f$ is Lebesgue integrable

with:

$\ds \int_{\closedint a b} \map f x \map {\rd \lambda} x = \int_{\closedint a b} \map g x \map {\rd \lambda} x$

That is:

$\ds \int_{\closedint a b} \map f x \map {\rd \lambda} x = \int_a^b \map f x \rd x$