Question about the meaning of a "region" in mathemathics. I am currently going over double integral. One concept that seems to appear a lot is "region". I have consulted different resources and it seems like there is no objetive definition of what a "region" is. For example, according to HandWiki,

the word region usually refers to a subset of $\mathbb R^n$
or $\mathbb C^n$
that is open (in the standard Euclidean topology), simply connected and non-empty. A closed region is sometimes defined to be the closure of a region.

This leads me to some confusion. Below follows a well-known property of double integrals.

Given two functions $f$ and $g$ integrable on a region $S$, if $f(x,y) \leqslant g(x,y)$ on $S$ (the inequality is verified for every point of $S$), then the inequality below also holds:
$$ \iint_S f(x,y) \, dA \leqslant \iint_S g(x,y) \, dA.$$

According to this and to the definition of a region that I blockquoted, $S$ must be an open, simply connected and non-empty subset of $\mathbb R^2.$ I have seen some answers on this site that apply this inequality for so-called regions that aren't open.
Take a look at this answer, for example. Here, the user Mark Viola uses the set $S = \{ (x,y) \in \mathbb R^2 \, | \, xy \leqslant 1, x>0, y>0 \}$ and uses the property blockquoted above to assure that
$$ \iint_S \frac{1}{1+x^4 \, y^4} dx \, dy \geqslant \frac{1}{2} \iint_S \, dx \, dy. $$
And well, this makes almost total sense, besides the fact that $S$ is not a region ($S$ isn't an open set).
So I am wondering: I don't doubt the knowledge of Mark, nor the knowledge of other users that I've seen using this same property to non-closed sets. So a natural question comes up to my mind:
For which subsets of $\mathbb R^2$ is the property I blockquoted above about double integrals valid?
 A: The difference between an open, simply connected set,  a closed, simply connected set,  and a simply connected set that is neither open nor closed (All non-empty) just has to do with whether or not it contains its boundary.  In other words,  $x^2+y^2<1$ is open,  $x^2+y^2\leq 1$ is closed, and if it contains only some points on the circle but not others,  it is neither.
The key concept here is that from the point of view of a double integral, the values on the boundary do not matter.   This is the exact same analogue as on a single integral, the points on the boundary do not matter,  you can change the values at for a single integral from $[0,1]$ at $0$ and $1$ without changing the integral,  because we define the integral in terms of limits (Either Riemann sums in terms of smaller and smaller mesh sizes or Darboux integration in terms of upper sums and lower sums, which are exactly equivalent).
The same thing occurs in 2 dimensions,  but now our samples are smaller and smaller rectangles.   Along the boundary edge of a simply connected region,  you can have the values be whatever you want and by taking your mesh size to 0, they have no impact on the total.
That's a tiny bit hand-wavy,  but its the essence of the underlying point:   Integrals don't change on sets of measure 0.   In 2 dimensions, the boundary of a simply connected region is a 1 dimensional manifold,  which has measure 0 (Again, concepts you get to a bit later).
So ultimately, it doesn't matter.  We often like to work with open sets for other reasons, so oftentimes authors will use that terminology even when it isn't required.
