Area enclosed by $[|x|].[|y|]=2$, where $[.]$ is the greatest integer function

Question

The graph doesn't look like a closed figure, can we determine the area?

Answer 1

Drawing is a little wrong , Area will have Boundary ...
Area is $8$

We can see that the Gray Area will have the Criteria given ....
Each Gray Point has either $1 \times 2 = 2$ or $2 \times 1 = 2$

No other Point will have that Criteria

You are then going to get Area $8$

Answer 2

Let $$f(x,y) = \lfloor |x| \rfloor \lfloor |y| \rfloor.$$

Clearly, $f(y,x) = f(x,y)$ for all $x, y$. We also have $f(-x,y) = f(x,y) = f(x,-y)$. So with all these symmetries, it suffices to consider the region $0 \le y \le x$. In this region, we can do away with the absolute value and consider the simplified equation $$g(x,y) = \lfloor x \rfloor \lfloor y \rfloor.$$ Then the only way that $g = 2$ is if $\lfloor x \rfloor = 2$ and $\lfloor y \rfloor = 1$, since $x \ge y \ge 0$ and $\lfloor \cdot \rfloor$ always returns an integer. So if $\lfloor x \rfloor = 2$, then $x \in [2, 3)$, and if $\lfloor y \rfloor = 1$, then $y \in [1, 2)$. This implies that $g = 2$ if and only if $$(x,y) \in [2, 3) \times [1, 2),$$ which is a square that includes the boundary on the left and bottom sides only, and the area enclosed by this region is $1$. So the total area for which $f = 2$ is simply $8$ times this, since there is a symmetry on the diagonal $y = x$, and symmetries across the axes.

More generally, for a positive integer $n$, the set of points $$S_n = \{(x,y) \in \mathbb R^2 : f(x,y) = n\}$$ has area equal to $4\sigma_0(n)$, where $\sigma_0(n)$ is the number of integer divisors of $n$. Moreover, $S_n \cap S_m = \varnothing$ for all $n \ne m \ge 0$, and $$\bigcup_{n=0}^\infty S_n = \mathbb R^2;$$ that is to say, the sequence $\{S_n\}_{n \ge 0}$ is a partition of the real plane.