I have been learning about absolute continuity and I was able to prove that the product of two absolutely continuous functions is also absolutely continuous, but I am not sure about this one.
My question: if $g: [a,b] \rightarrow \mathbb{R}$ is absolutely continuous and we are further given that $g>0$ in the interval $[a,b]$ to prevent division by zero, then is $1/g$ also absolutely continuous?
by continuity $g(x) \geq c > 0$ for every $x \in [a,b]$. We know that $$\forall \epsilon > 0 \: \exists \delta(\epsilon)>0 \: \text{s.t.}\: \sum_i (b_i - a_i) < \delta \Rightarrow \sum_i |g(b_i) - g(a_i)| < \epsilon$$ where $\{(a_i,b_i)\}_{i \in \Bbb N}$ is a countable collection of disjoint open intervals in $[a,b]$ but now $$\sum_{i = 1}^n \Big|\frac{1}{g(b_i)} - \frac{1}{g(a_i)}\Big| = \sum_{i = 1}^n \Big|\frac{g(b_i) - g(a_i)}{g(a_i)g(b_i)}\Big| \leq \frac{1}{c^2}\sum_{i = 1}^n|g(b_i) - g(a_i)| \leq \frac{1}{c^2}\sum_{i=1}^n| g(b_i) - g(a_i)| < \frac{\epsilon}{c^2}$$ for every $n$