Prove that, $\int_a^b f(x)\ \text{d}\alpha=\sum_{n=1}^{\infty}\int_a^b f_n(x)\ \text{d}\alpha.$

Question

Let $ \{f_n \} $ be a sequence of Riemann-Stiletjes functions integrable with respect to a function $ \alpha $ in $ [a, b] $. Suppose that $$ f (x) = \sum_ {n = 1} ^ \infty f_n (x) $$ converges uniformly on $ [a, b] $.$$\int_a^b f(x)\ \text{d}\alpha=\sum_{n=1}^{\infty}\int_a^b f_n(x)\ \text{d}\alpha.$$

Try:

I really don't have much in this test, I just have an intuitive idea that I don't know how to conclude in discrete terms. First, note that, $$\int_a^b f(x)\ \text{d}\alpha=\int_a^b \sum_{n=1}^{\infty} f_n(x)\ \text{d}\alpha=\lim_{N\to \infty}\int_a^b\sum_{n=1}^{N}f_n(x)\ \text{d}\alpha.$$

Then, for all $\epsilon>0$, exists $n\in \Bbb N$ s.t, $$\left |f(x)-\sum_{n=1}^{N}f_n(x)\right |<\epsilon.$$ So, $$\left |\int_a^b f(x)\ \text{d}\alpha -\int_a^b \sum\limits_{n=1}^N f_n(x)\ \text{d}\alpha\right| \leq \int_a^b \left |f(x) - \sum\limits_{n=1}^N f_n(x)\right |\ \text{d}\alpha < \epsilon (\alpha(b)-\alpha(a))$$

I don't know how to finish the exercise. Any help me?

Answer 1

You showed that for all $\varepsilon > 0$ there exists $N_0 \in \mathbb N$ such that for all $N \ge N_0$ we have$$\left |\int_a^b f(x)\ \text{d}\alpha - \sum\limits_{n=1}^N\int_a^b f_n(x)\ \text{d}\alpha\right| = \left |\int_a^b f(x)\ \text{d}\alpha -\int_a^b \sum\limits_{n=1}^N f_n(x)\ \text{d}\alpha\right| < \varepsilon (\alpha(b)-\alpha(a)),$$ which is the definition of $$\int_a^b f(x)\ \text{d}\alpha=\sum_{n=1}^{\infty}\int_a^b f_n(x)\ \text{d}\alpha = \lim_{N \rightarrow \infty} \sum_{n=1}^{N}\int_a^b f_n(x)\ \text{d}\alpha.$$ There is nothing left to show. If you wanted to, you could replace $\varepsilon$ with $\frac{\varepsilon}{\alpha(b)-\alpha(a)}$ to have the first inequality be just $< \varepsilon$, but this typically isn't strictly necessary.

Answer 2

The proof isn’t really complete, you should prove that $f$ is integrable wrt to $\alpha$ on $[a,b]$

Here $I_u(f,\alpha)$ and $I_l(f,\alpha)$ are respectively the upper and the lower integral and let $s_n(x) := \sum_{i = 1}^nf_i(x)$

By the uniform convergence for a given $\epsilon > 0$ there exists $N$ such that for every $n > N$ it holds $$-\frac{\epsilon}{3(\alpha(b) - \alpha(a))} + s_n(x) < f(x) < \frac{\epsilon}{3(\alpha(b) - \alpha(a))} + s_n(x)$$ for every $x$, which gives $$I_l\Big(-\frac{\epsilon}{3(\alpha(b) - \alpha(a))} + s_n,\alpha\Big) < I_l(f,\alpha) < I_u(f,\alpha) < I_u\Big(\frac{\epsilon}{3(\alpha(b) - \alpha(a))} + s_n,\alpha\Big)$$ and by the integrability of $s_n$ $$\int_a^bs_nd\alpha - \frac{\epsilon}{3} < I_l(f,\alpha) < I_u(f,\alpha) < \int_a^bs_nd\alpha + \frac{\epsilon}{3}$$ therefore $$I_u(f,\alpha) - I_l(f,\alpha) < \frac{2}{3}\epsilon < \epsilon$$ by the arbitarity of $\epsilon$ we get $$I_u(f,\alpha) = I_l(f,\alpha)$$