I stumbled upon the following problem:
$\text{Consider } f,g:[a,b]\to \mathbb{R} , \ k>0, c\geq 1, \text{ with the following property:}$ $$|f(x)-f(y)|\leq k|g(x)-g(y)|^c, \ \forall x,y\in[a,b]$$ $\text{Prove that if } g \text{ is Riemann integrable, then } f\text{ is also Riemann integrable.}$
I think I am on the right path trying to prove this integrability using Lebesgue's properties for Riemann integrability ($f$ is integrable if it is bounded and continuous almost everywhere) in the following way:
$\text{If } f \text{ is defined on the closed interval }[a,b], \text{then it must return finite values for any } x \Rightarrow f \text{ is bounded between two values: } m \leq f(x)\leq M, \ \forall x \in [a,b], \ M,n\in\mathbb{R}$
What's left to prove is that $f$ is continuous almost everywhere. Here I tried:
$g \text{ is integrable} \Rightarrow g \text{ continuous almost everywhere} \\ \text{Let }x \in[a,b] \Rightarrow y = x+p, \text{ where } p \in[a-x, b-x] \Rightarrow |f(x)-f(x+p)|\leq k|g(x)-g(x+p)|^c \\[2ex] \text{I. Suppose } g \text{ is continuous in } x \Rightarrow \lim_{p\to 0}k|g(x)-g(x+p)|^c = 0 \geq \lim_{p\to 0}|f(x)-f(x+p)|\Rightarrow \\ \Rightarrow lim_{p\to 0}|f(x)-f(x+p)| = 0 \Rightarrow f(x) = \lim_{p\to 0} f(x+p) \Rightarrow f \text{ is continuous in } x \\[2ex] \text{II. Suppose }g \text{ isn't continuous in } x \\[1ex] \text{1. Suppose } g \text{ isn't continuous to the left in }x \Rightarrow \lim_{p\nearrow 0}k|g(x)-g(x+p)|^c=l \\ \text{Where } l\in\mathbb{R} \ \text{because } g \text{ returns only real values, so the difference right where the } \\ \text{discontinuity happens must be real too.} \\ \Rightarrow lim_{p\nearrow 0}|f(x)-f(x+p)|\leq l \Rightarrow f \text{ has a discontinuity to the left in } x. \\[1ex] \text{2. Suppose }g \text{ isn't continuous to the right in }x \\ \Rightarrow f \text{ has a discontunity to the right of }x \text{ (by the same reasoning)} \\[1ex] \text{3. Suppose g isn't continuous to the right nor to the left of }x \\ \Rightarrow \text{There exist } l_1, l_2\in\mathbb{R} \text{ for each discontinuity} \\ \Rightarrow f \text{ has a discontinuity to the left and right of } x \\ \text{ (} l_1 \text{ and } l_2 \text{ might be equal, but I don't think it matters)} \\[1ex] g\text{ is continuous almost everywhere and } f \text{ is not continuous for the same } x\text{'s} \\ \Rightarrow f \text{ is continuous almost everwhere} \\[2ex] \text{Thus, } f \text{ is Riemann integrable}.$
Is this correct? Am I missing any steps? Is there any better/smarter approach?
Your proof can be substantially shortened and made more rigorous in some intermediate steps.
Let $A$ denote the set of discontinuities of $f$, $B$ denote the set of discontinuities of $g$. It is then sufficient to show $A \subset B$. Having shown that, the integrability of $g$ implies $\lambda(B) = 0$, whence $\lambda(A) = 0$, i.e., $f$ is Riemann integrable.
Given $x_0 \in A$, then there exist $\varepsilon_0 > 0$ and a sequence $x_n \to x_0$ such that $|f(x_n) - f(x_0)| \geq \varepsilon_0$ for all $n$. By assumption, it then follows that \begin{align} |g(x_n) - g(x_0)| \geq k^{-1/c}|f(x_n) - f(x_0)|^{1/c} \geq (k^{-1}\varepsilon_0)^{1/c} > 0, \end{align} which shows that $x_0 \in B$. This completes the proof.