Why do we say that $A \propto P^2$ for mathematical objects of $\mathbb R ^2$?
I consider $P$ the perimeter and $A$ the area of the object.
I have some difficulty in understanding this relationship.
What do we consider by the perimeter and the surface of a mathematical object? And how this relationship of proportionality can be proved ?
The informal business:
This relation is in fact very intuitive for many people, and is used / taught long before the formal concept of measure is introduced. Area is measured in units-squared - it is a "two dimensional quantity". Perimeter is measured in length, in units-(1) - it is a "one dimensional quantity". The area of a shape can be found by 'counting squares', and when the shape is scaled, the area increases in the same manner that the squares do. That is, the scale factor is... squared, for the area. For the perimeter - built from so many straight lines, you might say - scales exactly with the scale factor. So the perimeter squared scales with the scale factor squared, which is exactly how the area scales.
Volume grows with the scale factor cubed, because volume is a "three dimensional quantity" - like a cube. A nice video exposition of this is here, which introduces the notion of a fractal dimension via the relative rates at which a quantity grows, relative to the scale factor. He implicitly references the fact that $H_s(r\cdot A)=r^s\cdot H_s(A)$ for any $s$-Hausdorff-dimensional measurable set $A$ and scale factor $r$.
The formal business:
Caveat: My answer is completely wrong if you decide to define perimeter and area differently. However, the use of Hausdorff measure to formally describe these informal notions is widespread.
Suppose you have a non-null Lebesgue measurable subset $A\subseteq\Bbb R^2$ with Hausdorff-1 measurable (and dimensional) boundary $P:=\partial A$. This is not a trivial condition: there are many fractal-like subsets of $\Bbb R^2$ for which this is false, for instance the Sierpinski's triangle or two-dimensional Cantor dust.
Then, $H_2(rA)=r^2\cdot H_2(A)$ is well known, wherever $r\ge0$. Now $H_1(rP)=r\cdot H_1(P)$ is also known, where $r\ge0$. So if we interpret the "perimeter" of this set to be the Hausdorff-1 measure $H_1(P)$ and the "area" of it to be $H_2(A)$, then: $$\frac{[H_1(rP)]^2}{H_2(rA)}=\frac{r^2\cdot[H_1(P)]^2}{r^2\cdot H_2(A)}=\frac{[H_1(P)]^2}{H_2(A)}:=c$$For any $r\ge0$.
So the perimeter squared is equal to $c$ times the area, for any 'scaled up or down' version of $A$, $rA$, that you can find.
As Andrew notes, this constant $c$ evidently depends on the set $A$ - you can think of this being the fundamental shape. This could be $c=4\pi$, in the case of $A$ being a circle. This could also be $c=16$, in the case $A$ is a square, and so on.
But we also require that the set $A$ is measurable in two dimensions, and that the boundary $\partial A$ is one-dimensional and Hausdorff-one measurable (you can think of this as: the perimeter is a 'nice' line, i.e. essentially anything you might be able to draw yourself with pen and paper).
You can see a proof of those relations in proposition $3.8$ of this exposition.