I found an interesting exercise in Terry Tao’s Measure Theory book:
Proposition: for any \(N \in \mathbb{R}\) and interval \(I = [a, b]\) over \(\mathbb{R}\), \(|I| = \lim_{N \to \infty} \frac{|I ~ \cap ~ \frac{\mathbb{Z}}{N} |}{N},\) where \(\frac{\mathbb{Z}}{N} := \{ \frac{n}{N} \mid n \in \mathbb{Z} \}\).
The interesting thing about this proposition is that it expresses the length of an interval \(|I| = (b - a)\) (which is a continuous measure) in terms of the cardinality of a discrete set, \(I \cap \frac{\mathbb{Z}}{N}\). To develop an intuition for how this proposition could work, first notice that \(I \cap \frac{\mathbb{Z}}{N}\) is the set of points in \(\frac{\mathbb{Z}}{N}\) which lie within the interval \(I\). For example, if \(a = 3\), \(b = 8\), and \(N = 25\), then \(I \cap \frac{\mathbb{Z}}{N} = \{ {75}/{25}, {76}/{25}, \ldots, {200}/{25} \}\), which means its cardinality is \((200 - 75) + 1 = 126\).
Plugging this back into the right-hand side of the proposition, we see that:
\[\frac{|I ~ \cap ~ \frac{\mathbb{Z}}{N} |}{N} = \frac{126}{25} = 5 + 0.04.\]As we can see, this is pretty close to \(|I| = 5\). So, if \(| I \cap \frac{\mathbb{Z}}{N} |\) can be expressed as \(N(b - a) + \epsilon\) where \(\epsilon\) is some constant, then the proposition pretty much holds at the limit.
To formalize this intuition, consider \(\frac{l}{N}\) and \(\frac{h}{N}\), the smallest and largest elements of \(I \cap \frac{\mathbb{Z}}{N}\), where \(l, h \in \mathbb{Z}\). From our prior example, it is clear that the cardinality of that set is:
\[\vert I \cap \frac{\mathbb{Z}}{N} \vert = h - l + 1.\]Next, note that by definition, \(\frac{l}{N} \ge a\) and \(\frac{(l - 1)}{N} \lt a\), so we get \(l \lt Na + 1\). Similarly, we get \(h \gt Nb - 1\). Plugging these inequalities into the expression for the interval size, we get:
\[\vert I \cap \frac{\mathbb{Z}}{N} \vert = h - l + 1 \gt Nb - Na - 1.\]which we can rewrite as:
\[\vert I \cap \frac{\mathbb{Z}}{N} \vert = N(b - a) - 1 + \epsilon\]where \(\epsilon > 0\) is a constant. Plugging this back into the right hand side of the original assertion, we get:
\[\begin{align*} \lim_{N \to \infty} \frac{|I \cap \frac{\mathbb{Z}}{N} |}{N} & = \lim_{N \to \infty} \frac{N(b - a) - 1 + \epsilon}{N} \\ & = (b - a) + \lim_{N \to \infty} \frac{\epsilon - 1}{N} \\ & = b - a. \end{align*}\]