In our analysis course studied a theorem (it will be stated below). After studying its analogue for triple integrals, it was noted that generelization of this theorem to multiple integrals is called Fubini's theorem. But when I found it on the Internet it was hard to me to understand how the theorem stated below is a partial case of Fubini's theorem. Is it that I don't understand something or the theorem we learnt is not a partial case of a Fubini's theorem ? In the latter case, is there any common name for the theorem we learnt ?
Theorem: $A\subset \mathbb{R}^{2}$ is a compact set which can be expressed as union of sets of points between graphs of two continuous functions on some segment $[a,b]$. And $f\in C(A,\mathbb{R})$. Then the following equality holds and is properly defined: $$\iint_{A} f(x,y) dx dy=\int_{pr_{x}A}\Big(\int_{A_x}f(x,y) dy\Big)dx$$ Where $pr_{x}A=\{x\in\mathbb{R}: \exists y\in\mathbb{R}, (x,y)\in A\}$ and $A_{x}=\{y\in\mathbb{R}:(x,y)\in A\}$
This theorem looks like a partial case of Fubini’s theorem. To see this, first extend $f$ to a function on the whole $\mathbb R^2$ by being equal to $0$ outside $A$. Then, the left hand side of the equality above becomes
$$\iint_{\mathbb R^2} f(x,y) dx dy,$$
since $f$ is $0$ in the complement of $A$, and the right hand side becomes
$$\int_{\mathbb R}\Big(\int_{\mathbb R}f(x,y) dy\Big)dx$$
for essentially the same reason (i.e., $f(x,y)=0$ if $y\notin A_x$, and $\int_{\mathbb R}f(x,y) dy = 0$ if $x\notin pr_x A$, this last point being true because $f(x,y)=0\quad\forall y$ if $x\notin pr_x A$).
The above expressions are equal because of Fubini’s theorem. You can apply it because $f$ is bounded (it is continuous on a compact set) and has a compact support, thus it has finite integral.