Sum of Riemann-Stieltjes Integrals on Adjacent Intervals

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Theorem

Let $a, b, c \in \R$.

Let $f, \alpha$ be real functions that are bounded on some closed interval containing $a, b, c$.

Suppose that two of the following three Riemann-Stieltjes integrals exist, with the limits of integration interpreted in the general sense:

$\ds \int_a^b f \rd \alpha$
$\ds \int_b^c f \rd \alpha$
$\ds \int_c^a f \rd \alpha$

Then, the third exists as well, and:

$\ds \int_a^b f \rd \alpha + \int_b^c f \rd \alpha + \int_c^a f \rd \alpha = 0$


Proof

We have the following special cases:

Part to Whole

Let $a < c < b$ be real numbers.

Let $f, \alpha$ be real functions that are bounded on $\closedint a b$

Suppose that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $\closedint a c$, and also on $\closedint c b$.


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

$\ds \int_a^b f \rd \alpha = \int_a^c f \rd \alpha + \int_c^b f \rd \alpha$

$\Box$


Whole to Part

Let $a < c < b$ be real numbers.

Let $f, \alpha$ be real functions that are bounded on $\closedint a b$

Suppose that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $\closedint a b$, and also on one of the two intervals $\closedint a c$ and $\closedint c b$.


Then, $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on the other interval as well, and:

$\ds \int_a^c f \rd \alpha + \int_c^b f \rd \alpha = \int_a^b f \rd \alpha$

$\Box$


To prove the full theorem, we will consider separate cases according to how many of $a, b, c$ are equal to one another.


Case: All Equal

First, suppose $a = b = c$.

Then, all three integrals are equal to $0$ by convention, and the result is trivial.

$\Box$


Case: Two Equal

Next, suppose two of $a, b, c$ are equal to each other.

Without loss of generality, let them be $a = c$.

Then:

$\ds \int_c^a f \rd \alpha = 0$

We are given the existence of two different integrals, so at least one of the following is among them:

$\ds \int_a^b f \rd \alpha$
$\ds \int_b^c f \rd \alpha = \int_b^a f \rd \alpha$

Therefore, the other is the negative of it by convention, and the result once again follows.

$\Box$


Case: All Distinct

Finally, suppose $a, b, c$ are all distinct.

Without loss of generality, suppose:

$\ds \int_a^b f \rd \alpha$
$\ds \int_b^c f \rd \alpha$

are the two integrals known to exist.

If necessary, we can reverse the limits of integration on all three integrals and swap the roles of $a$ and $c$ to ensure:

$a < c$

There are now three cases:

$\paren 1 \quad a < b < c$
$\paren 2 \quad b < a < c$
$\paren 3 \quad a < c < b$


Suppose $\paren 1$ holds.

Then, by Part to Whole above:

$\ds \int_a^c f \rd \alpha = \int_a^b f \rd \alpha + \int_b^c f \rd \alpha$

Therefore, by convention:

$\ds \int_c^a f \rd \alpha = - \int_a^b f \rd \alpha - \int_b^c f \rd \alpha$

from which the result follows.

$\Box$


Suppose $\paren 2$ holds.

Then, by convention:

$\ds \int_a^b f \rd \alpha = - \int_b^a f \rd \alpha$

By Whole to Part above, we know that $\ds \int_a^c f \rd \alpha$ exists and:

$\ds \int_b^a f \rd \alpha + \int_a^c f \rd \alpha = \int_b^c f \rd \alpha$

Again by convention:

$\ds \int_c^a f \rd \alpha = - \int_a^c f \rd \alpha$

Therefore:

$\ds - \int_a^b f \rd \alpha - \int_c^a f \rd \alpha = \int_b^c f \rd \alpha$

from which the result follows.

$\Box$


Finally, suppose that $\paren 3$ holds.

Then, by convention:

$\ds \int_b^c f \rd \alpha = - \int_c^b f \rd \alpha$

By lemma 2, we know that $\ds \int_a^c f \rd \alpha$ exists and:

$\ds \int_a^c f \rd \alpha + \int_c^b f \rd \alpha = \int_a^b f \rd \alpha$

Again by convention:

$\ds \int_c^a f \rd \alpha = - \int_a^c f \rd \alpha$

Therefore:

$\ds - \int_c^a f \rd \alpha + - \int_b^c f \rd \alpha = \int_a^b f \rd \alpha$

from which the result follows.

$\Box$


Since the result holds in every case, it follows from Proof by Cases.

$\blacksquare$


Sources

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