Suppose $\alpha$ increases monotonically on $[a,b]$, $g$ is continuous, and $g(x)=G'(x)$ for $a\le x \le b$. Prove that $$\int \limits_{a}^{b}\alpha(x)g(x)dx=G(b)\alpha(b)-G(a)\alpha(a)-\int \limits_{a}^{b}Gd\alpha.$$
I thought on above problem more than two days and I think that I coped. I would be really greatful to anyone who check my solution!
Proof: Let $\epsilon>0$ be given. Since $g\in C[a,b]$ then $g$ is uniformly continuous on $[a,b]$. Then $\exists \delta>0$ such that for any $x,y\in [a,b]$ with $|x-y|<\delta$ we have $|g(x)-g(y)|<\frac{\epsilon}{b-a}$. Let's $P_3$ be a partition of $[a,b]$ such that $\lVert P_3\rVert<\frac{\delta}{2}$. Since $\alpha g\in \mathscr{R}$ and $G\in \mathscr{R}(\alpha)$ on $[a,b]$ then exists partitions $P_1, P_2$ such that $$\int \limits_{a}^{b}\alpha(x)g(x)dx+\frac{\epsilon}{2}>U(P_1,\alpha g)\ge U(P^*,\alpha g), \int \limits_{a}^{b}G(x)d\alpha+\frac{\epsilon}{2}>U(P_2,G,\alpha)\ge U(P^*,G,\alpha)$$ where $P^*=P_1\cup P_2\cup P_3$ with $\lVert P^*\rVert<\frac{\delta}{2}$ and $P^*=\{x_0, x_1, \dots, x_n\}$.
Applying MVT to $G$ on every $[x_{i-1},x_i]$ we have: $G(x_i)-G(x_{i-1})=g(t_i)\Delta x_i$ for $t_i\in (x_{i-1},x_i)$ and $i=\overline{1,n}.$ Making some algebraic manipulations (I omit them) we get: $$\sum \limits_{i=1}^{n}\alpha(x_i)g(t_i)\Delta x_i=G(b)\alpha(b)-G(a)\alpha(a)-\sum \limits_{i=1}^{n}G(x_{i-1})\Delta\alpha_i \qquad (1)$$ It's easy to show that RHS of $(1)$ is $\ge G(b)\alpha(b)-G(a)\alpha(a)-U(P^*,G,\alpha)$. Let's take a look at new sum: $$\sum \limits_{i=1}^{n}\alpha(x_i)g(x_i)\Delta x_i \color{blue}{\le} U(P^*,\alpha g).$$ Furhermore: $$\left |\sum \limits_{i=1}^{n}\alpha(x_i)g(t_i)\Delta x_i-\sum \limits_{i=1}^{n}\alpha(x_i)g(x_i)\Delta x_i\right|\le \sum \limits_{i=1}^{n}|\alpha(x_i)||g(t_i)-g(x_i)|\Delta x_i \le$$$$\le\sum \limits_{i=1}^{n}\max\{|\alpha(a)|,|\alpha(b)|\}\frac{\epsilon}{b-a}\Delta x_i=\underbrace{\max\{|\alpha(a)|,|\alpha(b)|\}}_{L}\cdot \epsilon=\epsilon L$$ Substracting from both sides of $(1)$ new sum and using estimate that we got above we have: $$\epsilon L \ge \sum \limits_{i=1}^{n}\alpha(x_i)(g(t_i)-g(x_i))\Delta x_i\ge G(b)\alpha(b)-G(a)\alpha(a)-U(P^*,G,\alpha)-U(P^*,\alpha g) \ge$$$$\ge G(b)\alpha(b)-G(a)\alpha(a)-\int \limits_{a}^{b}\alpha(x)g(x)dx-\int \limits_{a}^{b}G(x)d\alpha-\epsilon.$$ Since $\epsilon>0$ was arbitrarily then: $$\int \limits_{a}^{b}\alpha(x)g(x)dx\ge G(b)\alpha(b)-G(a)\alpha(a)-\int \limits_{a}^{b}Gd\alpha.$$ Converse inequality can be obtained by considering little bit changed $(1)$ sum $$\sum \limits_{i=1}^{n}\alpha(x_{i-1})g(t_i)\Delta x_i=G(b)\alpha(b)-G(a)\alpha(a)-\sum \limits_{i=1}^{n}G(x_{i})\Delta\alpha_i \qquad (1)$$ and lower sums.
Thank you for attention! :)
