Changing the order of integration. Is $F$ continuous on direct product $[0,1]\times [0,1]$?

Question

Let $f :[0,1]\to \mathbb R,g : [0,1]\to \mathbb R$ are $C^1$ functions.

Define $F: [0,1]\times [0,1]\to \mathbb R $ by $F(x,y)=f(x)g(y)e^{f(x)g(y)}$.

Then, I want to show $$\int_0^1 \left( \int_0^1 F(x,y) dx\right)dy=\int_0^1 \left(\int_0^1 F(x,y) dy \right)dx.$$ i.e., I want to find the reason for changing the order of integration.

I think the idea using the boundedness of $F$ works.

If $F$ is continuous $[0,1]\times [0,1]$, the bound of $F$ exists and let the bound $M>0.$

Then, $\int_0^1\left(\int_0^1 |F(x,y)| dx \right)dy\leqq\int_0^1\left(\int_0^1 M dx \right)dy=M<\infty.$

From Fubini, I can change the order of integration.

But I'm not sure whether $F$ is continuous or not.

$f, g$ are continuous on $[0,1]$ respectively, but from this, can I say $F$ is continuous on $[0,1]\times [0,1]$ ? Is there some theorem that confirms the continuity of $F$ ?

Answer 1

From Continuity of cartesian product of functions between topological spaces I deduce that the function $f\times g$ is continuous.

The multiplication function, $M:\Bbb{R}\times\Bbb{R}\rightarrow\Bbb{R}$ is continuous as the preimage of an open interval is an open set which is union of some hyperbolas. Intervals containing zero must be treated seperately.

The function $w:\Bbb{R}\rightarrow\Bbb{R}$ defined by $w(x)=xe^x$ is continuous.

Then $F=w\circ M\circ (f\times g)$ is continuous.

And by Fubini's theorem (https://web.ma.utexas.edu/users/m408m/Display15-2-3.shtml) you are allowed to change the order of the integration.

Answer 2

Yes, $F$ is continuous as composite of continuous functions.

More specifically, multiplication is a continuous map $\Bbb R\times \Bbb R \rightarrow R$, so the map $$\begin{array}{rcccl} \Bbb R\times \Bbb R & \rightarrow &\Bbb R\times \Bbb R & \rightarrow \Bbb R\\ (x,y) & \mapsto & (x,e^y) & \rightarrow & xe^y \end{array}$$ is continuous. Similarly the map $$\begin{array}{rcccl} {[0,1]\times [0,1]} & \rightarrow & \Bbb R \times \Bbb R&\rightarrow &\Bbb R\\ (x,y) & \mapsto & (f(x),g(y)) &\mapsto &(f(x)g(y)) \end{array}$$ is continuous. Combining both maps via the diagonal $\Bbb R \mapsto \Bbb R^2, x\mapsto (x,x)$ gives the function $F$.