let $f$ and $g$ be two continuous functions from $[a,b] \rightarrow \Bbb R$ such that $f(a) = g(a), f(b) = g(b)$ and $f(s) \gt g(s)$ for $s \in (a,b)$.
If $f(x)$ has distribution $\left(\frac{f(x) - g(x)}{\int_a^b f(t) - g(t) dt}\right)$ for $x \in [a,b]$ prove $f(x)$ is a probability density.
I understand we need to show $f(x) \ge 0$ and $\int_a^b f(x) dx = 1$ but i'm struggling to show the latter requirement.
You are confused because you use $f(x)$ in two ways: as one of the functions (along with $g(x)$) and as the name of the distribution function.
You should use a different name like this:
If $h(x)$ has distribution $\left(\frac{f(x) - g(x)}{\int_b^a f(t) - g(t) dt}\right) $, show that $h$ is a distribution function.
This then becomes easy:
$\begin{array}\\ \int_a^b h(x) dx &=\int_a^b \left(\frac{f(x) - g(x)}{\int_b^a (f(t) - g(t)) dt}dx\right)\\ &=\frac1{\int_b^a (f(t) - g(t)) dt}\int_a^b (f(x) - g(x))dx\\ &=1\\ \end{array} $