If $g$ is integrable over $[a,b]$ then $f(x)=\int_{a}^{x}g(x)\,{\rm d}x$ is absolutely continuous on $[a,b]$

Question

From the definition this flows naturally up until near the end:

For all $\epsilon>0$, I am looking for a $\delta$ such that $\sum_{i=1}^{n}|x_{i+1}-x_i|<\delta \implies \sum_{i=1}^{n}|f(x_{i+1})-f(x_i)|<\epsilon$ for all subintervals $(x_{k+1},x_k)$ of $[a,b]$.

Now,

$$\sum_{i=1}^{n}|f(x_{i+1})-f(x_i)|=\sum_{i=1}^{n}\left|\int^{x_{i+1}}_{a}g(x)\,{\rm d}x-\int_{a}^{x_i}g(x)\,{\rm d}x\right|\\=\sum_{i=1}^{n}\left|\int^{x_{i+1}}_{x_i}g(x)\,{\rm d}x\right|\leq \sum_{i=1}^{n}\int^{x_{i+1}}_{x_i}|g(x)|\,{\rm d}x=\int^{b}_{a}|g(x)|\,{\rm d}x$$

I am not sure how to go on from here. How does this sequence of implications help me find a $\delta$? All help is appreciated.

Answer 1

$\sum_{i=1}^{n}|f(x_{i+1})-f(x_i)|=\sum_{i=1}^{n}|\int^{x_{i+1}}_{a}g-\int_{a}^{x_i}g|=\sum_{i=1}^{n}|\int^{x_{i+1}}_{x_i}g|\leq \sum_{i=1}^{n}\int^{x_{i+1}}_{x_i}|g|\leq ||g||_\infty\sum_{i=1}^{n}|x_{i+1}-x_i|<||g||_\infty\delta$

So take $\delta=\frac{\epsilon}{||g||_\infty}$

If $g$ is unbounded, notice that $\lim_{ M \to \infty} \int_{\{x | g(x) \ge M \}} |g| = 0$