Clarification on step in proof of IBP

Question

I was reading proof for the integration by parts formula for absolutely continuous functions $f$ and $g$,

$$ \int_a^b fg'dx = f(b)g(b) - f(a)g(a) - \int_a^bf'gdx. $$

One step was, after showing $fg$ is absolutely continuous and that $(fg)' = f'g + fg'$ a.e., the LHS and RHS of the product rule identity above are integrated over $[a, b]$. But to do this implies $f'g$ and $fg'$ to be integrable... why is this true? I know

• $f$ and $g$ are absolutely continuous
• $f$ and $g$ are differentiable a.e. on the interval
• $f'$ and $g'$ are in $L^1$

But I don't understand why $f'g$ and $fg'$ are then necessarily integrable. I think I am missing easy Theorem here...

Answer 1

Absolutely continuous functions are, in particular, continuous, so $$ \int_a^b|f'g|\leq\max_{x\in [a,b]}|g(x)|\int_a^b|f'|<\infty $$