Prove $F(x, y) = (f(x))^{g(y)}$ is integrable over $A = [0, 1] \times [0, 1].$ I'm writing to ask for your help in the last step to finish the following question:
Let $f, g: [0, 1] \to \mathbb{R}$ be integrables functions with $f(x) \geq m > 0,$ for all $x \in [0, 1]$. Prove that $F(x, y) = (f(x))^{g(y)}$ is integrable over $A = [0, 1] \times [0, 1].$
My attempt:
First, note that: $F(x, y) = (f(x))^{g(y)} = \exp(g(y)\cdot \ln(f(x)))$. 
Then since $f([0, 1]) \subset [m, +\infty)$ and the logarithm is Lipschitz on $[m, +\infty)$, so $\ln \circ f$ is integrable over $[0, 1]$. 
Second, Defining $\phi : [0, 1] \times [0, 1] \to \mathbb{R}$ such that $\phi(x, y) = g(y)\cdot\ln(f(x))$, it is possible to prove that $\phi$ is integrable due to the fact that $g$ and $\ln \circ f$ are. 
Finally, since $F = \exp \circ \phi,$ what is needed to finish the proof is to be show that $\exp \circ \phi$ is also integrable. But here is where I met my Waterloo.
Any kind of help would be really appreciated. Thanks in advance.
 A: Note: Every use of the word "integrable" here is in the sense of Riemann, in the standard/proper sense.
You also need to show that $F$ is bounded on $A$, but I'll leave that to you. A general fact is that continuous composed with integrable is integrable. A simple proof of this is obtained by using Lebesgue's criterion for Riemann integrability, which states a function is Riemann integrable if and only if the set of discontinuities has measure zero.
Let $\xi$ be a point where $F = \exp \circ \phi$ is discontinuous. Then either $\exp$ is discontinuous at $\phi(\xi)$, or $\phi$ is discontinuous at $\xi$. It clearly can't be the former since $\exp$ is $C^{\infty}$. What this shows is the set of discontinuities $\mathcal{D}_F$ of $F$ is contained in that of $\phi$; i.e
\begin{equation}
\mathcal{D}_F \subseteq \mathcal{D}_{\phi}.
\end{equation}
Since $\phi$ is integrable, the RHS has measure zero; hence the LHS being a subset  also has measure zero, thereby proving $F$ is Riemann integrable on $A$.
