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We may restrict attention to an interval $I = (A,+\infty)$ on which $g(x) > 1$. Let $\varepsilon > 0$ be given.

The set $S$ of points $x$ in $I$ at which $g'(x) > (g(x))^{1 + \varepsilon}$ is open, hence is a disjoint union of countably many open intervals of the form $(a,b)$.

Given such an interval $(a,b)$, with $b$ finite, we have $g'(x) \geq 1$ on $[a,b]$. Thus the function $y = g(x)$ has a $C^1$ inverse on $[a,b]$. Moreover the inequality $dy/dx > y^{1 + \varepsilon}$ implies $dx/dy < y^{-1-\varepsilon}$. Integrating with respect to $y$, we find $b - a \leq \int_{g(a)}^{g(b)} y^{-1-\varepsilon}dy = \frac{1}{\varepsilon}[(g(a))^{-\varepsilon} - (g(b))^{-\varepsilon}]$.
(If $b = +\infty$, an obvious modification of this argument shows that $b - a$ is finite, a contradiction.)

Therefore, the length of the set $S$ is bounded above by $1/\varepsilon$ times the length of its image under the decreasing function $h(x) = (g(x))^{-\varepsilon}.$ But this image is contained in $(0,1)$.