Uniform Conv. of Sequences of Functions and R-S Integration Part 2

# Uniform Convergence of Sequences of Functions and Riemann-Stieltjes Integration Part 2

On the Uniform Convergence of Sequences of Functions and Riemann-Stieltjes Integration Part 2 page we partially proved at a very important theorem regarding a sequence $(f_n(x))_{n=1}^{\infty}$ of functions and Riemann-Stieltjes integration. We now complete the second part of the proof.

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) $\displaystyle{\lim_{n \to \infty} \int_a^b f_n(x) \: d \alpha (x) = \int_a^b f(x) \: d \alpha (x)}$. |

*Part (b) shows that under the conditions in Theorem 1 we have that $\displaystyle{\lim_{n \to \infty} \int_a^b f_n(x) \: d \alpha (x) = \int_a^b \lim_{n \to \infty} f_n(x) \: d \alpha (x)}$.*

**Proof of b)**Once again, we assume that $\alpha$ is an increasing function since if Theorem 1 holds for increasing functions it also holds for functions of bounded variation. We further assume that $\alpha (a) < \alpha(b)$ since if $\alpha(a) = \alpha(b)$ this implies that $\alpha$ is constant on $[a, b]$ and both integrals will be equal to $0$.

- To show that $\lim_{n \to \infty} \int_a^b f_n(x) \: d \alpha (x) = \int_a^b f(x) \: d \alpha(x)$ we will show that 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 $\biggr \lvert \int_a^b f_n(x) \: d \alpha(x) - \int_a^b f(x) \: d \alpha (x) \biggr \rvert < \epsilon$.

- Let $\epsilon > 0$ be given. Since $(f_n(x))_{n=1}^{\infty}$ is a uniformly convergent sequence of functions, for $\epsilon_1 = \frac{\epsilon}{2[\alpha(b) - \alpha(a)]}$ there exists an $N_1 \in \mathbb{N}$ such that if $n \geq N_1$ then for all $x \in [a, b]$ we have that:

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

- Let $N = N_1$ so that if $n \geq N$ then $(****)$ is satisfied. Now since $\alpha$ is an increasing function we have by the theorem on The Absolute Value of Riemann-Stieltjes Integrals with Increasing Integrators page that $\biggr \lvert \int_a^b g \: d \alpha \biggr \rvert \leq \int_a^b \mid g \mid \: d \alpha$ for any R-S integrable function $g$ with respect to $\alpha$ on $[a, b]$, and so, for $n \geq N$ we have that for all $x \in [a, b]$:

\begin{align} \quad \biggr \lvert \int_a^b f_n(x) \: d \alpha (x) - \int_a^b f(x) \: d \alpha(x) \biggr \rvert &= \biggr \lvert \int_a^b [f_n(x) - f(x)] \: d \alpha (x) \\ \quad & \leq \int_a^b \mid f_n(x) - f(x) \mid \: d \alpha (x) \\ \quad & < \int_a^b \epsilon_1 \: d \alpha (x) \\ \quad & < \epsilon_1 \int 1 \: d \alpha (x) \\ \quad & < \frac{\epsilon}{[\alpha(b) - \alpha(a)]}{ [\alpha(b) - \alpha(a)]} \\ \quad & < \epsilon \end{align}

- So for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $\biggr \lvert \int_a^b f_n(x) \: d \alpha(x) - \int_a^b f(x) \: d \alpha (x) \biggr \rvert < \epsilon$ and so for all $x \in [a, b]$:

\begin{align} \quad \lim_{n \to \infty} \int_a^b f_n(x) \: d \alpha (x) = \int_a^b f(x) \: d \alpha(x) \quad \blacksquare \end{align}