Integration by Substitution/Riemann-Stieltjes Integral

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Theorem

Let $g$ be a real function that is continuous and strictly monotone on $\closedint a b$.

Let $\Bbb I = g \closedint a b$ be the image of $g$ under $\closedint a b$.

Let $f, \alpha$ be real functions that are bounded on $\Bbb I$.

Further suppose that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $\Bbb I$.


Let $h, \beta : \Bbb I \to \R$ be defined as:

$\map h x = \map f {\map g x}$
$\map \beta x = \map \alpha {\map g x}$

Then, $h$ is Riemann-Stieltjes integrable with respect to $\beta$ on $\Bbb I$ and:

$\ds \int_a^b h \rd \beta = \int_{\map g a}^{\map g b} f \rd \alpha$


Proof

We have the following special cases:

Increasing

Let $g$ be a real function that is continuous and strictly increasing on $\closedint a b$.

Let $f, \alpha$ be real functions that are bounded on $\closedint {\map g a} {\map g b}$.

Further suppose that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $\closedint {\map g a} {\map g b}$.


Let $h, \beta : \closedint a b \to \R$ be defined as:

$\map h x = \map f {\map g x}$
$\map \beta x = \map \alpha {\map g x}$

Then, $h$ is Riemann-Stieltjes integrable with respect to $\beta$ on $\closedint a b$ and:

$\ds \int_a^b h \rd \beta = \int_{\map g a}^{\map g b} f \rd \alpha$

$\Box$


Decreasing

Let $g$ be a real function that is continuous and strictly decreasing on $\closedint a b$.

Let $f, \alpha$ be real functions that are bounded on $\closedint {\map g b} {\map g a}$.

Further suppose that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $\closedint {\map g b} {\map g a}$.


Let $h, \beta : \closedint a b \to \R$ be defined as:

$\map h x = \map f {\map g x}$
$\map \beta x = \map \alpha {\map g x}$

Then, $h$ is Riemann-Stieltjes integrable with respect to $\beta$ on $\closedint a b$ and:

$\ds \int_a^b h \rd \beta = - \int_{\map g b}^{\map g a} f \rd \alpha$

$\Box$


By definition of strictly monotone, $g$ is either strictly increasing or strictly decreasing.

First, suppose $g$ is strictly increasing.

By the Intermediate Value Theorem, it is clear that:

$\Bbb I = \closedint {\map g a} {\map g b}$

Then, the result follows from the special case Increasing above.

$\Box$


Now, suppose $g$ is strictly decreasing.

Once again, by the Intermediate Value Theorem, it is clear that:

$\Bbb I = \closedint {\map g b} {\map g a}$

Then:

\(\ds \int_a^b h \rd \beta\) \(=\) \(\ds - \int_{\map g b}^{\map g a} f \rd \alpha\) Special case Decreasing above
\(\ds \) \(=\) \(\ds \int_{\map g a}^{\map g b} f \rd \alpha\) Convention for General Limits of Integration

$\Box$


In each case, the result holds, so the theorem follows from Proof by Cases.

$\blacksquare$


Sources

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