Claiming that $\int_c^\infty I(y)dy = \int_a^\infty J(x)dx$ 
Claim: Let a real-valued function $f(x,y)$ be definied on $D = [a,+\infty)\times [c,+\infty)$.
If $I(y) = \int_a^\infty f(x,y)dx$ and $J(x) = \int_c^\infty f(x,y)dy$ are continuous and at least one of the following
$$\int_c^\infty I(y)dy = \int_c^\infty dy \int_a^\infty f(x,y)dx \\ \int_a^\infty J(x)dx = \int_a^\infty dx \int_c^\infty f(x,y)dy$$ improper integrals conveverges, then so does the other and
$$\int_c^\infty I(y)dy = \int_a^\infty J(x)dx$$

I haven't found anything very related to this theorem in my textbooks. I was thinking of taking two arbitrary sequences $(a_n)$ and $(c_n)$ such that $a_n \to +\infty$ and $c_n \to +\infty$ and defining
$$I_n(y) = \int_a^{a_n} f(x,y) dy$$ and $$J_n(x) = \int_c^{c_n} f(x,y)dy$$ Then, obviously $I_n(y)_\rightarrow^\rightarrow I(y)$ and $J_n(x)_\rightarrow^\rightarrow J(x)$. However, it seems I cannot go any further. I wonder if after the definition of the functional sequence above, one could use the ready results from that section.
 A: This is false under the given hypotheses.  Take $D = [1,\infty)\times [1,\infty)$ and $f(x,y) = \frac{x-y}{(x+y)^3}$.
Noting that
$$\frac{x-y}{(x+y)^3} = \frac{\partial}{\partial x}\frac{-x}{(x+y)^2} = \frac{\partial}{\partial y}\frac{y}{(x+y)^2},$$
we have
$$I(y) = \int_1^\infty f(x,y) \, dx = \int_x^\infty\frac{\partial}{\partial x}\frac{-x}{(x+y)^2}\, dx = \frac{1}{(1+y)^2},\\  J(x) = \int_1^\infty f(x,y) \, dy = \int_x^\infty\frac{\partial}{\partial y}\frac{y}{(x+y)^2}\, dy = \frac{-1}{(1+x)^2}$$
Here, $I$ and $J$ are continuous and both iterated integrals exist, with
$$\int_1^\infty I(y) \, dy = \int_1^\infty \frac{dy}{(1+y)^2} = \frac{1}{2}, \\\int_1^\infty J(x) \, dx = -\int_1^\infty \frac{dx}{(1+x)^2} = -\frac{1}{2} $$
However, $\int_1^\infty I(y) \, dy \neq \int_1^\infty J(x) \, dx$.

The theorem is true if we add the hypothesis that $f$ is nonnegative.  Here is an elementary  proof for improper Riemann integrals without reference to Lebesgue integration and Tonelli's theoem.
Since $f$ is nonnegative, we have  $J_n(x) = \int_c^{c+n}f(x,y) \, dy \nearrow \int_c^\infty f(x,y) \, dy =J(x)$. Since $J$ is continuous, it is Riemann integrable and by the monotone convergence theorem it follows that for all $A > a$,
$$\int_a^A\left(\int_c^\infty f(x,y) \, dy \right) \, dx =\int_c^\infty\left(\int_a^A f(x,y) \, dx \right) \, dy$$
Suppose that $\int_c^\infty I(y) \, dy$ exists. Then we have
$$\int_a^A\left(\int_c^\infty f(x,y) \, dy \right) \, dx= \int_c^\infty\left(\int_a^A f(x,y) \, dx \right) \, dy \leqslant \int_c^\infty\left(\int_a^\infty f(x,y) \, dx \right) \, dy = \int_c^\infty I(y) \, dy$$
Thus,
$$\int_a^\infty J(x) \, dx = \lim_{A \to \infty}\int_a^A\left(\int_c^\infty f(x,y) \, dy \right) \, dx \leqslant \int_c^\infty I(y) \, dy$$
The argument can be repeated when the other iterated integral exists to obtain the reverse inequality
$$\int_c^\infty I(y) \, dy \leqslant \int_a^\infty J(x) \, dx, $$
and, thus,
$$\int_c^\infty I(y) \, dy = \int_a^\infty J(x) \, dx$$
