Uniform Conv. of Sequences of Functions and R-S Integration Part 1

# Uniform Convergence of Sequences of Functions and Riemann-Stieltjes Integration Part 1

Recall from the Uniform Convergence of Sequences of Functions page that we say the sequence of functions $(f_n(x))_{n=1}^{\infty}$ converges uniformly to the limit function $f$ if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then for all $x \in [a, b]$ we have that:

(1)\begin{align} \quad \mid f_n(x) - f(x) \mid < \epsilon \end{align}

We will now look at a very important theorem connecting the concept of sequences of functions and Riemann-Stieltjes integrals. The proof is rather lengthy, so we prove the first part here and the second part on the Uniform Convergence of Sequences of Functions and Riemann-Stieltjes Integration Part 2 page.

Theorem 1: Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of real-valued functions on $[a, b]$ that uniformly converges to the limit function $f$ and let $\alpha$ be a function of bounded variation on $[a, b]$. Furthermore, for each $n \in \mathbb{N}$ let $f_n$ be Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. Then:a) $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$.b) $\displaystyle{\lim_{n \to \infty} \int_a^b f_n(x) \: d \alpha (x) = \int_a^b f(x) \: d \alpha (x)}$. |

**Proof of a)**It suffices to show that Theorem 1 holds for increasing functions since if so, then for any integrator $\alpha$ of bounded variation on $[a, b]$ we can express $\alpha$ as the difference of two increasing functions, $\alpha = \alpha_1 - \alpha_2$ and so if $f$ is Riemann-Stieltjes integrable with respect to $\alpha_1$ and $\alpha_2$ on $[a, b]$, we have that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$.

- So, assume that $\alpha$ is an increasing function on $[a, b]$. Further assume that $\alpha(a) < \alpha(b)$ since if $\alpha(a) = \alpha(b)$ then $\alpha$ is constant on $[a, b]$ and any function is Riemann-Stieltjes integrable with respect to a constant integrator as we saw on the Riemann Stieltjes-Integrals with Constant Integrators page. Let $\epsilon > 0$ be given. For any partition $P \in \mathscr{P}[a, b]$ consider the difference of the upper Riemann-Stieltjes sum and the lower Riemann-Stieltjes sum:

\begin{align} \quad U(P, f, \alpha) - L(P, f, \alpha) \end{align}

- Recall that for an increasing function $\alpha$ we have that for functions $f$ and $g$ defined on $[a, b]$ that:

\begin{align} \quad L(P, f, \alpha) + L(P, g, \alpha) \leq L(P, f + g, \alpha) \leq U(P, f + g, \alpha) \leq U(P, f, \alpha) + U(P, g, \alpha) \end{align}

- This implies that:

\begin{align} \quad U(P, f + g, \alpha) - L(P, f + g, \alpha) \leq [U(P, f, \alpha) - L(P, f, \alpha)] + [U(P, g, \alpha) - L(P, g, \alpha)] \end{align}

- Now, for any $n \in \mathbb{N}$ we can write $f = f_n + (f - f_n)$. Then, by using the inequality above we have that:

\begin{align} \quad U(P, f, \alpha) - L(P, f, \alpha) \leq [U(P, f_n, \alpha) - L(P, f_n, \alpha)] + [U(P, f - f_n, \alpha) - L(P, f - f_n, \alpha)] \end{align}

- Now, since the sequence of functions $(f_n(x))_{n=1}^{\infty}$ is uniformly continuous on $[a, b]$ we have that for $\epsilon_1 = \frac{\epsilon}{3[\alpha(b) - \alpha(a)]} > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then for all $x \in [a, b]$ we have that:

\begin{align} \quad \mid f_n(x) - f(x) \mid < \epsilon_1 = \frac{\epsilon}{3[\alpha(b) - \alpha(a)]} \quad (*) \end{align}

- Therefore we see that for any partition $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ and for all $n \geq N$ we have by $(*)$ that:

\begin{align} \quad \mid U(P, f - f_n, \alpha) \mid &= \sum_{k=1}^{n} \mid \sup \{ f_n(x) - f(x) : x \in [x_{k-1}, x_k] \} \mid \Delta \alpha_k \leq \sum_{k=1}^{n} \epsilon_1 \Delta \alpha_k =\epsilon_1 \sum_{k=1}^{n} \Delta \alpha_k = \frac{\epsilon}{3[\alpha(b) - \alpha(a)]} \cdot [\alpha(b) - \alpha(a)] = \frac{\epsilon}{3} \quad (**) \end{align}

- Similarly, for all $n \geq N$ and using $(*)$ that:

\begin{align} \quad \mid L(P, f - f_n, \alpha) \mid &= \sum_{k=1}^{n} \mid \inf \{ f_n(x) - f(x) : x \in [x_{k-1}, x_k] \} \mid \Delta \alpha_k \leq \sum_{k=1}^{n} \epsilon_1 \Delta \alpha_k = \epsilon_1 \sum_{k=1}^{n} \Delta \alpha_k = \frac{\epsilon}{3[\alpha(b) - \alpha(a)]} \cdot [\alpha(b) - \alpha(a)] = \frac{\epsilon}{3} \quad (***) \end{align}

- Using $(**)$ and $(***)$ together with the triangle inequality yields:

\begin{align} \quad \: U(P, f - f_n, \alpha) - L(P, f - f_n, \alpha) \leq \mid U(P, f - f_n, \alpha) \mid + \mid L(P, f - f_n, \alpha) \mid \leq \frac{\epsilon}{3} + \frac{\epsilon}{3} = \frac{2\epsilon}{3} \end{align}

- Now consider $f_N$. We are given that $f_N$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ so for $\epsilon_1 = \frac{\epsilon}{3} > 0$ there exists a partition $P_{\epsilon_1} \in \mathscr{P}[a, b]$ such that for all partitions $P$ finer than $P_{\epsilon_1}$ ($P_{\epsilon_1} \subseteq P$) we have by Riemann's condition that:

\begin{align} \quad U(P, f_N, \alpha) - L(P, f_N, \alpha) < \epsilon_1 = \frac{\epsilon}{3} \end{align}

- Let $P_{\epsilon} = P_{\epsilon_1}$. Then for all any partition $P$ finer than $P_{\epsilon}$ ($P_{\epsilon} \subseteq P$) we have that:

\begin{align} \quad U(P, f, \alpha) - L(P, f, \alpha) & \leq [U(P, f_N, \alpha) - L(P, f_N, \alpha)] + [U(P, f - f_N, \alpha) - L(P, f - f_N, \alpha)] \\ \quad & < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} \\ \quad & < \epsilon \end{align}

- Therefore Riemann's condition is satisfied and so $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. $\blacksquare$