Riemann-Stieltjes Integral with Step Integrator

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Theorem

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

Let $f$ be a real function that is bounded on $\closedint a b$.

Let $\alpha$ be a real function on $\closedint a b$ such that:

$\forall x \in \hointr a c: \map \alpha x = \map \alpha a$
$\forall x \in \hointl c b: \map \alpha x = \map \alpha b$

Suppose that:

Either $f$ is left-continuous at $c$, or $\map \alpha c = \map \alpha a$
Either $f$ is right-continuous at $c$, or $\map \alpha c = \map \alpha b$


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

$\ds \int_a^b f \rd \alpha = \map f c \paren {\map \alpha b - \map \alpha a}$


Proof

Let $\epsilon > 0$ be arbitrary.

The construction of $P_\epsilon$ will depend on which of $f$ and $\alpha$ are continuous on which sides at $c$.


Case $f$ Left- and Right-Continuous

Suppose $f$ is both left- and right-continuous at $c$.

Define $\epsilon' := \dfrac \epsilon {\size {\map \alpha c - \map \alpha a} + \size {\map \alpha b - \map \alpha c} + 1} > 0$.

By Continuous at Point iff Left-Continuous and Right-Continuous:

$f$ is continuous at $c$

Then, by definition, there is some $\delta > 0$ such that:

$\forall x: \size {x - c} < \delta \implies \size {\map f x - \map f c} < \epsilon'$

By Existence of Subdivision with Small Norm, there exists some subdivision $P'_\epsilon$ of $\closedint a b$ such that:

$\norm P < \delta$

where $\norm P$ denotes the norm of $P$.

Define $P_\epsilon := P'_\epsilon \cup \set c$.


Let $P = \set {x_0, \dotsc, x_n}$ be finer than $P_\epsilon$.

Then, $c \in P$, so for some $r \in \set {1, \dotsc, n - 1}$:

$x_r = c$

By Norm of Refinement is no Greater than Norm of Subdivision:

$\norm P < \delta$

We have:

\(\ds \map S {P, f, \alpha}\) \(=\) \(\ds \sum_{k \mathop = 1}^n \map f {t_k} \paren {\map \alpha {x_k} - \map \alpha {x_{k - 1} } }\) Definition of Riemann-Stieltjes Sum
\(\ds \) \(=\) \(\ds \map f {t_r} \paren {\map \alpha {x_r} - \map \alpha {x_{r - 1} } } + \map f {t_{r + 1} } \paren {\map \alpha {t_{r + 1} } - \map \alpha {t_r} }\) For $k < r$ or $k > r + 1$, we have $\map \alpha {x_k} - \map \alpha {x_{k - 1} } = 0$
\(\ds \) \(=\) \(\ds \map f {t_r} \paren {\map \alpha c - \map \alpha a} + \map f {t_{r + 1} } \paren {\map \alpha b - \map \alpha c}\) by hypothesis on $\alpha$
\(\ds \leadsto \ \ \) \(\ds \size {\map S {P, f, \alpha} - \map f c \paren {\map \alpha b - \map \alpha a} }\) \(=\) \(\ds \size {\map f {t_r} \paren {\map \alpha c - \map \alpha a} + \map f {t_{r + 1} } \paren {\map \alpha b - \map \alpha c} - \map f c \paren {\map \alpha b - \map \alpha a} }\)
\(\ds \) \(=\) \(\ds \size {\paren {\map f {t_r} - \map f c} \paren {\map \alpha c - \map \alpha a} + \paren {\map f {t_{r + 1} } - \map f c} \paren {\map \alpha b - \map \alpha c} }\) $\map f c \map \alpha c - \map f c \map \alpha c = 0$
\(\ds \) \(\le\) \(\ds \size {\map f {t_r} - \map f c} \size {\map \alpha c - \map \alpha a} + \size {\map f {t_{r + 1} } - \map f c} \size {\map \alpha b - \map \alpha c}\) Triangle Inequality for Real Numbers and Absolute Value Function is Completely Multiplicative
\(\ds \) \(\le\) \(\ds \epsilon' \size {\map \alpha c - \map \alpha a} + \epsilon' \size {\map \alpha b - \map \alpha c}\) $c - t_r \le x_r - x_{r - 1} \le \norm P < \delta$, and likewise for $t_{r + 1}$
\(\ds \) \(=\) \(\ds \frac {\size {\map \alpha c - \map \alpha a} + \size {\map \alpha b - \map \alpha c} } {\size {\map \alpha c - \map \alpha a} + \size {\map \alpha b - \map \alpha c} + 1} \epsilon\) Definition of $\epsilon'$
\(\ds \) \(<\) \(\ds \epsilon\)

$\Box$


Case $f$ is Left-Continuous and $\map \alpha c = \map \alpha b$

Suppose $f$ is left-continuous at $c$, and $\map \alpha c = \map \alpha b$.

Define $\epsilon' := \dfrac \epsilon {\size {\map \alpha c - \map \alpha a} + 1} > 0$.

By definition of left-continuous, there is some $\delta > 0$ such that:

$\forall x: c - \delta < x \le c \implies \size {\map f x - \map f c} < \epsilon'$

By Existence of Subdivision with Small Norm, there exists some subdivision $P'_\epsilon$ of $\closedint a b$ such that:

$\norm P < \delta$

where $\norm P$ denotes the norm of $P$.

Define $P_\epsilon := P'_\epsilon \cup \set c$.


Let $P = \set {x_0, \dotsc, x_n}$ be finer than $P_\epsilon$.

Then, $c \in P$, so for some $r \in \set {1, \dotsc, n - 1}$:

$x_r = c$

By Norm of Refinement is no Greater than Norm of Subdivision:

$\norm P < \delta$

We have:

\(\ds \map S {P, f, \alpha}\) \(=\) \(\ds \sum_{k \mathop = 1}^n \map f {t_k} \paren {\map \alpha {x_k} - \map \alpha {x_{k - 1} } }\) Definition of Riemann-Stieltjes Sum
\(\ds \) \(=\) \(\ds \map f {t_r} \paren {\map \alpha {x_r} - \map \alpha {x_{r - 1} } }\) For $k < r$ or $k \ge r + 1$, we have $\map \alpha {x_k} - \map \alpha {x_{k - 1} } = 0$
\(\ds \) \(=\) \(\ds \map f {t_r} \paren {\map \alpha c - \map \alpha a}\) by hypothesis on $\alpha$
\(\ds \leadsto \ \ \) \(\ds \size {\map S {P, f, \alpha} - \map f c \paren {\map \alpha b - \map \alpha a} }\) \(=\) \(\ds \size {\map f {t_r} \paren {\map \alpha c - \map \alpha a} - \map f c \paren {\map \alpha b - \map \alpha a} }\)
\(\ds \) \(=\) \(\ds \size {\paren {\map f {t_r} - \map f c} \paren {\map \alpha c - \map \alpha a} }\) by hypothesis, $\map \alpha c = \map \alpha b$
\(\ds \) \(=\) \(\ds \size {\map f {t_r} - \map f c} \size {\map \alpha c - \map \alpha a}\) Absolute Value Function is Completely Multiplicative
\(\ds \) \(=\) \(\ds \size {\map f {t_r} - \map f {x_r} } \size {\map \alpha c - \map \alpha a}\) Definition of $r$
\(\ds \) \(=\) \(\ds \epsilon' \size {\map \alpha c - \map \alpha a}\) $x_r - t_r \le x_r - x_{r - 1} \le \norm P < \delta$
\(\ds \) \(=\) \(\ds \frac \epsilon {\size {\map \alpha c - \map \alpha a} + 1} \size {\map \alpha c - \map \alpha a}\) Definition of $\epsilon'$
\(\ds \) \(=\) \(\ds \frac {\size {\map \alpha c - \map \alpha a} } {\size {\map \alpha c - \map \alpha a} + 1} \epsilon\)
\(\ds \) \(<\) \(\ds \epsilon\)

$\Box$


Case $f$ is Right-Continuous and $\map \alpha c = \map \alpha a$

This case is symmetrical to the previous one.

Hence, the details are omitted whenever they are identical to those above.


Let $\epsilon' := \dfrac \epsilon {\size {\map \alpha b - \map \alpha c} + 1}$.

By definition of right-continuous, choose $\delta > 0$ such that:

$\forall x: c \le x < c + \delta \implies \size {\map f x - \map f c} < \epsilon'$

Select $P'_\epsilon$ as above, such that:

$\norm {P'_\epsilon} < \delta$

Define $P_\epsilon = P'_\epsilon \cup \set c$ as above.


Let $P$ be finer than $P_\epsilon$.

Then, $\norm P < \delta$.

Let $x_r = c$ as above.

Thus:

\(\ds \map S {P, f, \alpha}\) \(=\) \(\ds \sum_{k \mathop = 1}^n \map f {t_k} \paren {\map \alpha {x_k} - \map \alpha {x_{k - 1} } }\) Definition of Riemann-Stieltjes Sum
\(\ds \) \(=\) \(\ds \map f {t_{r + 1} } \paren {\map \alpha {x_{r + 1} } - \map \alpha {x_r} }\) For $k \le r$ or $k > r + 1$, we have $\map \alpha {x_k} - \map \alpha {x_{k - 1} } = 0$
\(\ds \) \(=\) \(\ds \map f {t_{r + 1} } \paren {\map \alpha b - \map \alpha c}\) by hypothesis on $\alpha$
\(\ds \leadsto \ \ \) \(\ds \size {\map S {P, f, \alpha} - \map f c \paren {\map \alpha b - \map \alpha a} }\) \(=\) \(\ds \size {\paren {\map f {t_{r + 1} } - \map f c} \paren {\map \alpha b - \map \alpha c} }\) by hypothesis, $\map \alpha c = \map \alpha a$
\(\ds \) \(=\) \(\ds \size {\map f {t_{r + 1} } - \map f {x_r} } \size {\map \alpha b - \map \alpha c}\) Absolute Value Function is Completely Multiplicative and $x_r = c$
\(\ds \) \(\le\) \(\ds \epsilon' \size {\map \alpha b - \map \alpha c}\) $t_{r + 1} - x_r \le x_{r + 1} - x_r \le \norm P < \delta$
\(\ds \) \(<\) \(\ds \epsilon\) Definition of $\epsilon'$

$\Box$


Case $\map \alpha a = \map \alpha c = \map \alpha b$

Suppose $\map \alpha c = \map \alpha a$ and $\map \alpha c = \map \alpha b$.

It follows immediately that:

$\forall x \in \closedint a b: \map \alpha x = \map \alpha c$

Therefore:

$\forall x, y \in \closedint a b: \map \alpha x - \map \alpha y = \map \alpha c - \map \alpha c = 0$

Define $P_\epsilon = \set {a, b}$.

Let $P = \set {x_0, \dotsc, x_n}$ be a subdivision of $\closedint a b$ that is finer than $P_\epsilon$.

Then:

\(\ds \map S {P, f, \alpha}\) \(=\) \(\ds \sum_{k \mathop = 1}^n \map f {t_k} \paren {\map \alpha {x_k} - \map \alpha {x_{k - 1} } }\) Definition of Riemann-Stieltjes Sum
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n \map f {t_k} \cdot 0\) $\map \alpha {x_{k - 1} } = \map \alpha {x_k}$
\(\ds \) \(=\) \(\ds 0\) Indexed Summation of Zero
\(\ds \leadsto \ \ \) \(\ds \size {\map S {P, f, \alpha} - \map f c \paren {\map \alpha b - \map \alpha a} }\) \(=\) \(\ds \size {- \map f c \paren {\map \alpha b - \map \alpha a} }\)
\(\ds \) \(=\) \(\ds \size {- \map f c \cdot 0}\) $\map \alpha a = \map \alpha b$
\(\ds \) \(=\) \(\ds 0\)
\(\ds \) \(<\) \(\ds \epsilon\)

$\Box$


In every case, we have constructed a subdivision $P_\epsilon$ of $\closedint a b$ such that, for every $P$ finer than $P_\epsilon$:

$\size {\map S {P, f, \alpha} - \map f c \paren {\map \alpha b - \map \alpha a}} < \epsilon$

Therefore, by definition of the Riemann-Stieltjes integral:

$\ds \int_a^b f \rd \alpha = \map f c \paren {\map \alpha b - \map \alpha a}$

$\blacksquare$


Sources

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