Function $f$ is continuous and strictly increasing on $[a,b]$ and $g$ is Riemann integrable such that $g \circ f$ is defined. Is $g \circ f$ Riemann integrable?
I was able to show that if $f$ additionally satisfies:
$x_{1}<x_{2} \in [a,b] \Rightarrow f(x_2)-f(x_1) \geq x_{2}-x_{1}$, then $g \circ f$ is Riemann integrable.
Use the following fact: a bounded function on a closed interval is Riemann integrable if and only if its set of discontinuities has Lebesgue measure zero.
Now, there is a set $N$ of Lebesgue measure zero such that $g_{|_{[f(a),f(b)]\setminus N}}$ is continuous. Then, $g\circ f$ is continuous on $[a,b]\setminus N_1$ where $N_1=\{x\in [a,b]:f(x)\in N\}$. Since $f$ is strictly increasing, $N_1$ also has Lebesgue measure zero. Thus, $g\circ f_{|_{[a,b]\setminus N_1}}$ is continuous and therefore Riemann integrable.
edit: my proof is wrong. The problem is that "Since $f$ is strictly increasing, $N_1$ also has Lebesgue measure zero" is not true. To find a couterexample, we need a strictly increasing function that maps a set of positive measure to a set of measure zero:
Let $C$ be the Cantor set, $f:[0,1]\to [0,1]$ the Cantor-Lebesgue function, and define $g:[0,1]\to [0,2]$ by $g(x)=f(x)+x$. Then, $g$ is strictly increasing, continuous, and maps $C$ to a set of positive measure. Then, $g^{-1}$ is strictly increasing, continuous and maps $g(C)$ to $C$, a set of measure zero.
To finish, take any function $f$ whose set of discontinuities occur at precisely the points of $C$, for example $\chi_C.$ Then, $f$ is Riemann integrable but $f\circ g^{-1}$ is discontinuous at the points of $g(C)$, and so is not Riemann integrable.