How is the area of a set of points in $\Bbb R^2$ defined? Let $S$ be a subset of $\Bbb R^2$. If no vertical slice of $S$ contains gaps, we could define the area of $S$ through the following.
$$A(S) = \int_{-\infty}^\infty\left(\sup\{y\in\Bbb R\mid (x,y)\in S\} - \inf\{y\in\Bbb R\mid (x,y)\in S\}\right)dx$$
The same basic idea could be used to determine the area of any set of points $S$, by turning the set into a function to be integrated, but removing the gaps in each slice.

Question: Is this how, or equivalent to how, the area of a set of points in $\Bbb R^2$ is commonly defined?

 A: There are different competing definitions. What you describe is closer to the Riemann integral definition (which is often the first taught), which has a few problems when dealing with arbitrary sets. Instead the most commonly accepted definition is http://en.wikipedia.org/wiki/Lebesgue_integration The gist of how it works is we try to fill in the space as best we can with small rectangles or other "simple" sets of various shapes and sizes that we know how to calculate the area of. We take the largest sum of those simple areas to be the "area" of the larger picture.
The concept of Lebesgue integration covers calculating area as well, but is more general as it allows for varying density.  You may be interested as well to read about Lebesgue Measure (which is necessary to fully understand/appreciate the Lebesgue integral) http://en.wikipedia.org/wiki/Lebesgue_measure
One of the particularly nice things, is that due to the way measure is defined, we may in effect "skip over" sets during our integration where the measure of the set is identically zero.
For example, in the normal Riemann integration, $\int\limits_0^1 \chi_{\mathbb{Q}}(x)dx$ where $\chi_\mathbb{Q}(x) = 1$ if $x$ is a rational number and is $0$ if $x$ is irrational, is not defined since in every partition the interval will contain both rational and irrational numbers.  However for Lebesgue integration, since $m(\mathbb{Q})=0$ you have $\int_{[0,1]} \chi_\mathbb{Q} dm = \int_{[0,1]\backslash \mathbb{Q}} \chi_\mathbb{Q} dm = \int_{[0,1]\backslash \mathbb{Q}} 0 dm= 0$
You do have that if $f$ is Riemann integrable then it is also Lebesgue integrable and the results are the same, however there are examples (such as the one above) where a function is Lebesgue integrable and is not Riemann integrable.
