In measure theory, what is an example of a set of not well-defined area in unit square in $\mathbb{R}^2$? I am trying to understand the need of sigma-algebras in probability spaces. In https://stats.stackexchange.com/questions/199280/why-do-we-need-sigma-algebras-to-define-probability-spaces the following example is given

Consider this example. Suppose you have a unit square in
$\mathbb{R}^2$, and you're interested in the probability of randomly
selecting a point that is a member of a specific set in the unit
square. In lots of circumstances, this can be readily answered based
on a comparison of areas of the different sets. For example, we can
draw some circles, measure their areas, and then take the probability
as the fraction of the square falling in the circle. Very simple.
But what if the area of the set of interest is not well-defined?
If the area is not well-defined, then we can reason to two different
but completely valid (in some sense) conclusions about what the area
is. So we could have $P(A)=1$ on the one hand and $P(A)=0$ on the
other hand, which implies $0=1$. This breaks all of math beyond
repair. You can now prove $5<0$ and a number of other preposterous
things. Clearly this isn't too useful.

What does it mean for a set $A$ to be not well-defined in this example?
Also, How could we have both $P(A)=0$ and $P(A)=1$ at the same time?
 A: We cannot really have $P(A) = 0$ and $P(A) = 1$ at the same time (of course not, because $0 \neq 1$). It simply is not possible to assign $P(A)$ a value in a reasonable way.
What the post talks about is the impossibility of having a well-defined Lebesgue-measure for all (maybe extremely complicated) subsets of the unit square.
There are set which have "inner measure $<$ outer measure" which can be seen as the contradiction you wanted to understand. See for example How does the Vitali set violate the definition of measurable sets?.
A: "The measure of a set not being well-defined" means (per def.) that the set is not in the $\sigma$-algebra over which the measure is defined and the set cannot be added to the $\sigma$-algebra and the measure be extended to the new $\sigma$-algebra without violating some requirement of "being a measure".
I.e. these are sets, such that no matter which value you assign to them, you arrive at a contradiction, if you assume your measure satisfies the definition of "being a measure".
That's probably what was meant by "having $P(A)=0$ and $P(A)=1$ simultaneously".
Note that for different measures, different sets may be non-measurable.
There are two resolutions: adapt the properties a measure needs to have, or allow some sets to be non-measurable. Because we often need the properties of a usual measure, we allow non-measurable sets.
This is no problem, because (informally) such sets are "impossible to write down explicitly" so they won't appear in any real-world numeric problem.
Examples of such non-measurable sets are the Vitali sets or those involved in the Banach-Tarski paradox. The former is easier to understand.
Edit to answer comment:
As @unwissen said, if $N$ is non-measurable (wrt. some measure on $\mathbb{R}$), then $N \times [a,b]$ is non-measurable (wrt. the corresponding product measure on $\mathbb{R}^2$).
In many situations you can take the cartesian product with a simple enough set, to lift a counterexample from a lower dimension to a higher one.
