Algebraic property of Riemann-Stieltjes integral

Question

I have to prove the following property:

If $f \in R(\alpha)$ in $[a,b]$ and $f \in R(\beta)$ in $[a,b]$, then $f \in R(\alpha +c \beta)$ in $[a,b]$ and $$ \int_a^b f \, d(\alpha + c\beta) = \int_a^b f \, d\alpha + c\int_a^b f \, d\beta $$

And what I tried to do is:

First let $$S(P,f,g)=\left\vert\sum_{k=1}^n f(t_k) (g(x_k)-g(x_{k-1}))-A \right\vert< \epsilon$$

By hypothesis, as $f \in R(\alpha)$ in $[a,b]$, then exist $A_1 \in \mathbb R$ such that for every $\epsilon>0$, exist $P_1 \in P([a,b])$ and for all $t_k \in[x_{k-1},x_k]$, $\vert S(f,P_1,\alpha)-A_1 \vert< \frac{\epsilon}{2}$. (The definition of Riemann-stieltjes)

Also we have for $f \in R(\beta)$ in $[a,b]$, then exist $A_2 \in \mathbb R$ such that for every $\epsilon>0$, exist $P_2 \in P([a,b])$ and for all $t_k' \in [x_{k-1},x_k]$, $\vert S(f,P_2,\beta)-A_2 \vert< \frac{\epsilon}{2}$.

Then I tried to analyze $\left\vert S(P,f,\alpha - c\beta)-\int_a^b f \, d\alpha + c\int_a^b f \, d\beta \right\vert$. (1)

The problem is that I don't know how to take the new partition (to then conclude that (1) is less than epsilon), maybe the intersection of $P_1$ and $P_2$? idk .

Answer 1

I am also doing analysis this semester. I think the proof is similar to the case where $f = f_1 + f_2$. Also, I think we need the condition $c > 0$ since we demand $a+c\beta$ to be increasing.

(1) $f \in \mathscr{R}(\alpha+c\beta)$ on $[a,b]$.

Pick any $\epsilon >0$. We construct a partition $P$ such that $U(P,f,\alpha+c\beta)- L(P,f,\alpha+c\beta) < \epsilon$.

There is a partition $P_1$ such that $U(P_1,f,\alpha)- L(P_1,f,\alpha) < \epsilon/2$.

There is a partition $P_2$ such that $U(P_2,f,\beta)- L(P_2,f,\beta) < \epsilon/(2c)$.

Let $P=P_1\cup P_2$. Then $$L(P_1,f,\alpha) \leq L(P,f,\alpha) \leq U(P,f,\alpha) \leq U(P_1,f,\alpha)$$ $$L(P_2,f,\beta) \leq L(P,f,\beta) \leq U(P,f,\beta) \leq U(P_2,f,\beta)$$ $$L(P,f,\alpha)+cL(P,f,\beta) =\sum m_i[\alpha(x_i)-\alpha(x_{i-1})]+c\sum m_i[\beta(x_i)-\beta(x_{i-1})]=\sum m_i[(\alpha+c\beta)(x_i)-(\alpha+c\beta)(x_{i-1})]=L(P,f,\alpha+c\beta)$$ $$\implies L(P_1,f,\alpha)+cL(P_2,f,\beta)\leq L(P,f,\alpha+c\beta)\leq U(P,f,\alpha+c\beta) \leq U(P_1,f,\alpha)+ cU(P_2,f,\beta)$$ $$\implies U(P,f,\alpha+c\beta)- L(P,f,\alpha+c\beta)\leq [U(P_1,f,\alpha)+ cU(P_2,f,\beta)]-[L(P_1,f,\alpha)+cL(P_2,f,\beta)]<\epsilon/2 + c(\epsilon/(2c))=\epsilon $$ Thus (1) is done.

(2) $$\int_a^b f d(\alpha+c\beta) = \int_a^b f d\alpha+ c \int_a^b f d\beta $$

Suppose $$\int_a^b f d(\alpha+c\beta) > \int_a^b f d\alpha+ c \int_a^b f d\beta$$ Then let $\epsilon =\int_a^b f d(\alpha+c\beta) - \int_a^b f d\alpha+ c \int_a^b f d\beta$.

For $f \in \mathscr{R}(\alpha)$, there is a partition $P_1$ such that $U(P_1,f,\alpha) - L(P_1,f,\alpha) < \epsilon/2$

For $f \in \mathscr{R}(\beta)$, there is a partition $P_2$ such that $ U(P_2,f,\beta) - L(P_2,f,\beta) < \epsilon/(2c)$

Again, take $P = P_1\cup P_2$.

Since we have $$\int_a^b f d(\alpha+c\beta) \leq U(P,f,\alpha+c\beta),\ L(P_1,f,\alpha) \leq \int_a^b f d\alpha, L(P_2,f,\beta)\leq\int_a^b f d\beta ,$$ $$U(P_1,f,\alpha) -\int_a^b f d\alpha \leq U(P_1,f,\alpha) - L(P_1,f,\alpha) < \epsilon/2 $$ $$ U(P_2,f,\beta) -\int_a^b f d\beta \leq U(P_2,f,\beta) - L(P_2,f,\beta) < \epsilon/(2c) $$ $$\implies \int_a^b f d(\alpha+c\beta) -\int_a^b f d\alpha-c \int_a^b f d\beta \leq U(P,f,\alpha+c\beta)-\int_a^b f d\alpha-c \int_a^b f d\beta < \epsilon $$ $$\implies \int_a^b f d(\alpha+c\beta) < \int_a^b f d(\alpha+c\beta) $$ Contradiction arises. Similar for the other side...