1. **Problem statement:** (a) Given functions $f, g \in R(\alpha)$ on $[a,b]$, show that $f - g \in R(\alpha)$ on $[a,b]$. (b) Given $0 < M < f$ and $f \in R(\alpha)$, prove that $\frac{1}{f} \in R(\alpha)$. 2. **Recall the definition:** A function $h$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$ (denoted $h \in R(\alpha)$) if the Riemann-Stieltjes integral $\int_a^b h(x) \, d\alpha(x)$ exists. 3. **Part (a) proof:** - Since $f, g \in R(\alpha)$, both $\int_a^b f(x) \, d\alpha(x)$ and $\int_a^b g(x) \, d\alpha(x)$ exist. - The Riemann-Stieltjes integral is linear, so: $$\int_a^b (f(x) - g(x)) \, d\alpha(x) = \int_a^b f(x) \, d\alpha(x) - \int_a^b g(x) \, d\alpha(x)$$ - Both integrals on the right exist, so their difference exists. - Therefore, $f - g \in R(\alpha)$ on $[a,b]$. 4. **Part (b) proof:** - Given $f \in R(\alpha)$ and $f(x) > M > 0$ for all $x \in [a,b]$. - Since $f$ is bounded away from zero, $\frac{1}{f}$ is bounded and continuous on $[a,b]$. - The composition of a Riemann-Stieltjes integrable function with a continuous function is Riemann-Stieltjes integrable. - Hence, $\frac{1}{f} \in R(\alpha)$ on $[a,b]$. **Final answers:** (a) $f - g \in R(\alpha)$ on $[a,b]$. (b) $\frac{1}{f} \in R(\alpha)$ on $[a,b]$. 5. **Additional note:** The arithmetic $6 + 4 = 10$ is correct but unrelated to the integrability proofs.