Let $h:[0,1] \times [0,1] \rightarrow \mathbb{R}$ be the function $h(x,y)=f(x)g(y)$. Show h is integrable.

Hint 1: The set of all points of discontinuity of a monotone function is countable and a countable set has measure $0$.

Hint 2: A function $f$ is integrable if and only if the set of points of discontinuity of $f$ has measure $0$

We have $|f(x)| \leqslant M_f$ and $|g(y)| \leqslant M_g$ for all $x,y \in [0,1]$. Assume $M_f, M_g > 0$. (Otherwise we have the trivial case $h = 0$).

Since $f$ and $g$ are each bounded and monotone and, hence, Riemann integrable on $[0,1]$, there exist partitions $P' = (x_0,x_1,\ldots, x_n)$ and $P''= (y_0,y_1,\ldots, y_m)$ such that

$$U(P',f) - L(P',f) < \frac{\epsilon}{2M_g}, \quad U(P'',g) - L(P'',g) < \frac{\epsilon}{2M_f}$$

Leaving some steps for you to complete ...

(1) Forming the partition $P = (P',P,'')$ of $[0,1]^2$ show that

$$U(P,h) - L(P,h)= [U(P',f)- L(P',f)] U(P'',g) + [U(P'',g)- L(P'',g)] L(P',f)$$

(2) Show that $|U(P'',g)| \leqslant M_g$ and $|L(P',f)| \leqslant M_f$, and

$$U(P,h) - L(P,h) < \epsilon$$