Why are the functions inside the integral integrable?

Question

Here is the question :

Let $f$ and $g$ be absolutely continuous functions on $[a,b]$. Show that $$\int_{a}^{b} fg^{'} = f(b)g(b) - f(a)g(a) - \int_{a}^{b} f^{'}g.$$

My trial:

Let $f$ and $g$ be absolutely continuous functions on $[a,b].$ since the product of absolutely continuous functions is an absolutely continuous function(I know how to prove this ). Therefore, by theorem 10 on pg.124 in Royden and Fitzpatrick (4th edition)$$\int_{a}^{b} (fg)^{'} = f(b)g(b) - f(a)g(a).$$ But $(fg)^{'} = f^{'}g + fg^{'}$ using the product rule, then distribute the integral sign over the previous 2 terms and then by linearity of integration for integrable functions we can get the result.

My question is:

Why $f^{'}g$ and $fg^{'}$ are integrable?

Answer 1

Absolute continuity implies that $f'$ is integrable. Since $g$ is bounded it follows that $f'g$ is integrable.