This is problem 10.4 from Munkres' Analysis on Manifolds. If $f,g \colon [0,1] \to \mathbb{R}$ are increasing and non-negative, show that $h(x,y)=f(x)g(y)$ is integrable over $[0,1]^2$.
I have not really come far yet. I know that f and g are increasing, and therefore integrable on $[0,1]$. But I have no idea how to solve this question - any suggestions or answers?
For any partitions $P=\{x_{0},...,x_{n}\}, Q=\{y_{0},...,y_{m}\}$ of $[0,1]$, denote the partition $P\times Q=\{[x_{i-1},x_{i}]\times[y_{j-1},y_{j}]\}_{1\leq i\leq n, 1\leq j\leq m}:=\{R_{i,j}\}_{1\leq i\leq n,1\leq j\leq m}$ of $[0,1]\times[0,1]$, then \begin{align*} &U(h,P\times Q)-L(h,P\times Q)\\ &=\sum_{1\leq i\leq n, 1\leq j\leq m}\left(\sup_{R_{i,j}}h-\inf_{R_{i,j}}h\right)|R_{i,j}|\\ &=\sum_{1\leq i\leq n, 1\leq j\leq m}(f(x_{i})g(y_{j})-f(x_{i-1})g(y_{j-1}))(x_{i}-x_{i-1})(y_{j}-y_{j-1})\\ &=\sum_{1\leq i\leq n, 1\leq j\leq m}(f(x_{i})g(y_{j})-f(x_{i})g(y_{j-1}))(x_{i}-x_{i-1})(y_{j}-y_{j-1})\\ &~~~~~~~~+\sum_{1\leq i\leq n,1\leq j\leq m}(f(x_{i})g(y_{j-1})-f(x_{i-1})g(y_{j-1}))(x_{i}-x_{i-1})(y_{j}-y_{j-1})\\ &\leq\sum_{1\leq i\leq n,1\leq j\leq m}f(1)(g(y_{j})-g(y_{j-1}))(x_{i}-x_{i-1})(y_{j}-y_{j-1})\\ &~~~~~~~~+\sum_{1\leq i\leq n,1\leq j\leq m}g(1)(f(x_{i})-f(x_{i-1}))(x_{i}-x_{i-1})(y_{j}-y_{j-1})\\ &=\sum_{1\leq j\leq m}f(1)(g(y_{j})-g(y_{j-1}))(y_{j}-y_{j-1})+\sum_{1\leq i\leq n}g(1)(f(x_{i})-f(x_{i-1}))(x_{i}-x_{i-1})\\ &=f(1)(U(g,Q)-L(g,Q))+g(1)(U(f,P)-L(f,P)), \end{align*} which can be made arbitrary small.