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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, $gcirc f$ is continuous on $[a,b]setminus N_1$ where $N_1=xin [a,b]:f(x)in N$. Since $f$ is strictly increasing, $N_1$ also has Lebesgue measure zero. Thus, $gcirc f__[a,b]setminus N_1$ 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 $fcirc g^-1$ is discontinuous at the points of $g(C)$, and so is not Riemann integrable.