How to apply fundamental theorem of calculus to multiple integrals I have the following problem at hand:
$$\lim_{\epsilon \to 0}\dfrac{1}{4\epsilon^2}\int_{z-\epsilon}^{z+\epsilon}f(e_1,y) \int_{\frac{z+y}{2}-\epsilon}^{\frac{z+y}{2}+\epsilon}g(e_2,x) dx dy$$
$f$ and $g$ are well behaving, continuous functions. I have the following intuitive thought but I can't show it rigorously: I shift $1/2\epsilon$ inside the first integral. Then we have:
$$\lim_{\epsilon \to 0}\dfrac{1}{2\epsilon}\int_{z-\epsilon}^{z+\epsilon}f(e_1,y) \dfrac{1}{2\epsilon}\int_{\frac{z+y}{2}-\epsilon}^{\frac{z+y}{2}+\epsilon}g(e_2,x) dx dy$$
If it would be valid to apply Fundamental theorem of calculus both for inner and outer integrals, we would have obtain $f(e_1,z)g(e_2,z)$ as the result. But I have no idea how to show to in a proper way. What should be done in this case?
 A: I don't know how to apply fundamental theorem of calculus here. But I would use bounding values of the functions.
You can restrict the analysis to $0 \le \epsilon \le 1$ which leads to a compact subset for the integral domain. As the functions are supposed to be continuous, those are also uniformly continuous on that compact domain.
Therefore for $\alpha > 0$, there exists $\epsilon >0$ such that $\vert g(e_2,u) - g(e_2,v) \vert \le \alpha$ for $\vert u - v \vert \le \epsilon$. In particular, you get:
$$\left\vert g(e_2,x) - g(e_2,\frac{z+y}{2}) \right\vert \le \alpha$$ for $x \in [\frac{z+y}{2} - \epsilon, \frac{z+y}{2} + \epsilon]$, hence:
$$\left\vert \int_{\frac{z+y}{2} - \epsilon}^{\frac{z+y}{2} + \epsilon} g(e_2,x) dx - 2 \epsilon g(e_2,\frac{z+y}{2}) \right\vert  \le 2 \alpha \epsilon$$ or
$$- 2 \alpha \epsilon + 2 \epsilon g(e_2,\frac{z+y}{2}) \le \int_{\frac{z+y}{2} - \epsilon}^{\frac{z+y}{2} + \epsilon} g(e_2,x) dx \le  2 \alpha \epsilon + 2 \epsilon g(e_2,\frac{z+y}{2})$$
You can multiply the inequalities by $f(e_1,y)$ and integrate on the domain $[z-\epsilon,z+\epsilon]$. Your double integral is then bounded lower and upper by a single integral. Now you can apply the Fundamental theorem of calculus to the continuous map $y \mapsto f(e_1,y)g(e_2,\frac{z+y}{2})$ to upper and lower bound your double integral by $f(e_1,z)g(e_2,z)$ plus a quantity that can be as small as desired leading to the result you were looking for.
Jean-Pierre (Math Counterexamples).
A: We may as well take $z=0.$ Suppose $h(x,y)$ is continuous in a neighborhood of $(0,0).$ Claim:
$$ (1)\,\,\,\,\frac{1}{4\epsilon^2}\lim_{\epsilon\to 0}\int_{-\epsilon}^\epsilon\int_{y/2-\epsilon}^{y/2+\epsilon}h(x,y)\,dx\,dy = h(0,0).$$
Proof: The linear transformation $T(u,v) = (u+v/2,v)$ maps $[-\epsilon,\epsilon]\times [-\epsilon,\epsilon]$ onto the region of integration in (1). We have $\det J_T = 1,$ where $J_T$ is the Jacobian matrix of $T.$ Thus the expression in (1) equals
$$\frac{1}{4\epsilon^2}\int_{-\epsilon}^{\epsilon} \int_{-\epsilon}^{\epsilon} h(u+v/2,v)\,du\,dv.$$
So we are looking at averages of a continuous function over squares centered at $(0,0).$ As these squares shrink in size, the averages converge to $h(0,0).$ This proves the claim, and also shows the OP's guess at the answer was right.
A: Since $g$ is continuous, by the Mean Value Theorem for integral, we have
$$\frac1{2\epsilon}\int_{\frac{z+y}{2}-\epsilon}^{\frac{z+y}{2}+\epsilon}g(e_2,x)dx=\frac1{2\epsilon}\int_{-\epsilon}^{\epsilon}g(e_2,\frac{z+y}{2}+x)dx=g(e_2,\frac{z+y}{2}+a\epsilon)$$
for $a\in(-1,1)$ independent of $\frac{y+z}{2}$. Then
$$\dfrac{1}{2\epsilon}\int_{z-\epsilon}^{z+\epsilon}f(e_1,y) \dfrac{1}{2\epsilon}\int_{\frac{z+y}{2}-\epsilon}^{\frac{z+y}{2}+\epsilon}g(e_2,x) dx dy=\dfrac{1}{2\epsilon}\int_{z-\epsilon}^{z+\epsilon}f(e_1,y)g(e_2,\frac{z+y}{2}+a\epsilon)dy.$$
Since $f$ and $g$ are continuous, by the Mean Value Theorem for integral again, we have
$$\dfrac{1}{2\epsilon}\int_{z-\epsilon}^{z+\epsilon}f(e_1,y)g(e_2,\frac{z+y}{2}+a\epsilon)dy=f(e_1,z+b\epsilon)g(e_2,\frac{2z+b\epsilon}{2}+a\epsilon)$$
for some $b\in(-1,1)$.
So
$$\lim_{\epsilon \to 0}\dfrac{1}{2\epsilon}\int_{z-\epsilon}^{z+\epsilon}f(e_1,y) \dfrac{1}{2\epsilon}\int_{\frac{z+y}{2}-\epsilon}^{\frac{z+y}{2}+\epsilon}g(e_2,x) dx dy=\lim_{\epsilon \to 0}f(e_1,z+b\epsilon)g(e_2,\frac{2z+b\epsilon}{2}+a\epsilon)=f(e_1,z)g(e_2,z).$$
