Introduction
We wish to define a "measure" for all subsets of \(\mathbb{R}^d\). What are the simplest subsets for which we already have an intuitive notion of what their measure should be? Rectangles.
Define a rectangle as follows. Take two lists of real numbers \(a_1 \dots a_d\) and \(b_1 \dots b_d\), where \(b_i \geq a_i \forall i\). The closed rectangle for those two lists is the cartesian product of all the intervals \([a_i, b_i]\). The open rectangle is similarly \((a_1, b_1) \times \dots \times (a_d, b_d)\). A rectangle in general is any set that contains an open rectangle and is contained in a closed rectangle of the same intervals. (i.e., letting the brackets take on all possible combinations of open/closed).
A rectangle in one dimension is an interval, and a rectangle in two dimensions is the standard rectangle. Notice that we don't allow rectangles to be infinite.
For a rectangle \(Q\) that is the cartesian product of intervals with endpoints \(a_i, b_i\), we define the volume, denoted \(|Q|\) in the obvious manner: \(\prod_{i=1}^d (b_i - a_i)\)
A cube is a rectangle where all the side lengths are the same, i.e., \(b_d - a_d = \dots = b_1 - a_1\)
Then, for any subset \(A \subseteq \mathbb{R}^d\), define its exterior measure \(m_*(A)\) as follows: \[ m_*(A) = \inf\left\{\sum_{k=0}^\infty |Q_k| : A \subseteq \bigcup_{k=0}^\infty Q_k, Q_k \, \text{closed cube} \right\} \] In plain English, it's the infimum of the volume of cubes over all coverings of \(A\) by a countable number of closed cubes. To see that it always exists, consider that the cubes \([-k, k]^d\) cover all of \(\R^d\) and so will cover any subset. Based on this definition, it is worth our time to get some basic facts about rectangle coverings out of the way first.
Coverings by Rectangles
Some definitions: a set of rectangles are almost disjoint if their interiors are disjoint. That means that if you reduce each rectangle to their open version, they become disjoint. So they only "touch" at their boundaries. An almost disjoint union of rectangles is just a union of almost disjoint rectangles.
As soon as possible, we want to prove that the exterior measure of rectangles coincides with the volume we defined earlier. To that end, we'll prove the following lemma:
First, we'll discuss subdivisions of rectangles that are just grids. If our rectangle \(R\) was the product of intervals \([a_1, b_1] \dots [a_d, b_d]\), then we can partition each interval into many subintervals. More precisely, partition \([a_i, b_i]\) into \([a_{i,1}, b_{i,1}] \dots [a_{i, n_i}, b_{i, n_i}]\).
Then, the volume of \(R\) can be written as: \[ \prod_{i=1}^d (b_i - a_i) = \prod_{i=1}^d \sum_{j=1}^{n_i} (b_{i, j} - a_{i, j}) \] Which if you expand out the product becomes: \[ \sum_{j_1, \dots, j_d} \prod_{i=1}^d (b_{i, j_i} - a_{i, j_i}) \] where \(1 \leq j_i \leq n_i\) for all \(i\).
What just happened, in words? First, imagine our rectangle is in 2 dimensions. You can think of partitioning an interval as making cuts with lines at the endpoints (the lines \(x = b_{1, 1}\) or \(y = b_{2, 1}\) for example). Once all cuts are made, we've partitioned our rectangle into a grid. Each subrectangle corresponds to a choice of subintervals, one from each dimension. Then, summing over all subrectangles gives us exactly the expanded product from above. That's because when we expanded the product of sums, each term corresponded to a choice of one term from each sum.
The analogue in higher dimensions is that when we make our cuts at the endpoints of our partitions, we cut with "planes", or \(d-1\) spaces (think of the plane \(z=a\) in \(\mathbb{R}^3\) for example).
So we've proved the desired result for when our subrectangles form a perfect grid subdivision of our big rectangle. What happens when they don't? In 2 dimensions, we can imagine taking each subrectangle and extending its edges to infinite lines which then subdivide the rectangle into a grid. In 3 dimensions, the analogue is extending each face to a plane that cuts the rectangle.
The general statement is that for each subrectangle \(R_i\), for each of its endpoints \(a_i\) and \(b_i\), taking the hyperplane \(x_i = a_i\) or \(x_i = b_i\) to cut \(R\), forming a grid. (word salad incoming) Because each \(R_i\)'s faces are part of the subdivisions, the grid subrectangles cannot span two of the \(R_i\). Thus it is possible to partition the grid cells into which original subrectangle they belonged to. But of course, the sum over the grid cells is the volume of \(R\), and so we can split up the sum into one over the \(R_i\). But the \(R_i\) have also been subdivided so the sum of the grid cells within them is also just their volume, giving the desired result (TODO: make this last part clearer)
One corollary of the above lemma is that if the subrectangles in general cover our rectangle \(R\), then we get \(|R| \leq \sum_{i=1}^n |R_i| \), as we might expect. I won't put a full formal proof here yet, but the idea is that if a subrectangle intersects with \(R\), then you can just cut it and take the intersection to get a rectangle that's smaller and fits in \(R\). Then you can apply the above Lemma.
Another important class of sets we'll consider are the open sets. For these cases, we'll make use of the following two lemmas:
Consider some open set \(\mathcal{O} \subseteq \mathbb{R}\). For each \(x \in \mathcal{O}\), let \(I_x\) be the largest open interval in \(\mathcal{O}\) that contains \(x\) (this obviously exists by openness)..
Any two intervals \(I_x\) and \(I_y\) are either equal or disjoint. Suppose they had a nonempty intersection. Then \(I_x \cup I_y\) is another open interval that contains \(x\) and so by maximality, \(I_x \cup I_y \subseteq I_x\) which implies \(I_y \subseteq I_x\).
Similarly, \(I_x \cup I_y \subseteq I_y\) which implies \(I_x \subseteq I_y\). The two together imply \(I_x = I_y\).
Then clearly, \(\mathcal{O} = \bigcup_{x \in \mathcal{O}} I_x\). Then by corresponding each interval with a rational in that interval (by density), the number of intervals can't be more than the cardinality of the rationals, so we have countability.
However, this statement is not necessarily true for higher dimensions, we have to weaken the hypotheses slightly (TODO: Put a counterexample here):
The proof is pretty intuitive, so I will give the high level overview here. I want to fill in the details at some point
Consider the grid formed by taking cubes with integer points. Any cubes outside of the open set we discard, any cubes inside of the open set we keep, and any spanning the border we subdivide it by partitioning each dimension into equal halves (so we get \(2^d\) more cubes), taking them to the next step. We keep applying this process to infinity. At step \(k\), the points have coordinates in the lattice \((1/2^k)\mathbb{Z}^d\).
Any point in the open set has an open ball around it, and thus there exists a large enough \(k\) such that there's a cube containing it in the \(k^{th}\) subdivision. This cube is entirely within the open set, and so must've been either accepted at step \(k\), or part of a larger cube that was accepted before.
Thus every point was covered, and we have a countable number of cubes (since we picked a countable number of cubes at each step).
Simple facts
Now we cover some basic facts of the theory of exterior measure.
Fact 1: If \(A \subseteq B\) then \(m_*(A) \leq m_*(B)\) (monotonicity)
This is simply since any covering of \(B\) is a covering of \(A\) and using simple rules of \(\inf\).
Fact 2: \(m_*\left(\bigcup_{i=0}^\infty A_i\right) \leq \sum_{i=0}^\infty m_*(A_i) \) (countable sub-additivity)
The idea is simply that if you take closed cube coverings of each \(A_i\), obviously putting them all together will cover the union of the \(A_i\). And then, you want to be able to say that the sum of volumes of the closed cube coverings is just the sum of the exterior measures of the \(A_i\).
But the closed cube coverings always have some error term and you want to bound the (infinite) sum of the errors. Here we use a classic trick that comes up a lot: take the error terms to be exponentially decreasing, then the sum is finite and can be taken as small as you want it to be
Fact 3: A point has zero exterior measure
We can consider a point as a closed cube where each closed interval is a single point, and thus we get zero exterior measure. If that seems weird, you can instead take a sequence of cubes containing the point that get smaller and limit in exterior measure to 0.
Fact 4: A closed cube has exterior measure equal to its volume
Consider a closed cube \(Q\). Since it covers itself, we have that \(m_*(Q) \leq |Q|\). To show the other direction of the inequality, we want to show that \(|Q|\) is a lower bound for the total volume of any closed cube covering of \(Q\). Consider some arbitrary closed cube covering \((Q_k)\). Then consider slightly expanding each cube to a bigger open cube \(S_k\) such that \(|S_k| \leq (1+\varepsilon)|Q_k|\). By compactness of \(Q\), we can take a finite subcovering from the \(S_k\). By the corollary, we have that \(|Q| \leq \sum_{k=1}^N |S_k|\) giving us that \(|Q| \leq (1+\varepsilon)\sum_{k=1}^\infty |Q_k|\). This is true for arbitrary \(\varepsilon\) and so we get the desired result.
Fact 4.5: Any cube has exterior measure equal to its volume
Take any cube \(Q\). Obviously its closure covers it and has the same volume, so \(m_*(Q) \leq |Q|\). For the reverse inequality, take an \(\varepsilon\)-smaller closed cube \(Q'\). Clearly \(|Q'| \leq m_*(Q)\) but letting \(\varepsilon\) limit to zero we get \(|Q| \leq m_*(Q)\) (this operation is like the reverse of taking an \(\varepsilon\)-bigger open cube that we did above).
Fact 5: Any rectangle has exterior measure equal to its volume
If we do the same expanding + finite subcover from compactness argument from Fact 4, we get \(|R| \leq m_*(R)\) (notice that this is the opposite of the "easy" direction from the previous facts).
For the reverse direction, it suffices to find a closed cube covering with total volume \(\leq |R|\). We do this by allowing for some error and letting the error go to 0.
More precisely, subdivide space into cubes of sidelength \(1/k\). Then let \(Q\) be the set of cubes contained in \(R\), and let \(Q'\) be the set of cubes that intersect both \(R\) and the complement of \(R\). Clearly both sets are finite, and their union is a cover of \(R\). First, note that by doing a similar "extending faces" argument to before, the sum of volumes of cubes in \(Q\) is \(\leq |R|\).
Next, we can estimate the size of \(Q'\). The cubes intersecting one face of the rectangle will be "one cube thick", and will be basically a grid of cubes. The number of cubes in each dimension is \(O((b_i-a_i)k)\) where \(i\) is the dimension in question. Thus, (dropping the constant), the number of cubes on each face is \(O(k^{d-1})\). Then the number of cubes in all of \(Q'\) is still \(O(k^{d-1})\) because we're just summing over some fixed, finite number of faces.
Then since the volume of each cube is \(1/k^d\), the sum of their volumes is \(O(1/k)\).
Thus we get: \begin{align} \sum_{A \in Q \cup Q'} |A| &= \sum_{A \in Q}|A| + \sum_{A \in Q'} |A| \\ &\leq |R| + O(1/k) \\ \end{align} where the error term goes to zero as \(k \to \infty\) giving us the desired inequality.
Fact 6: \(m_*(A) = \inf m_*(\fO)\) where the infimum runs over all open sets containing \(A\)
One direction inequality is easy thanks to monotonicity. For the other direction, take a closed cube covering of \(A\) with error \(\e\). Then, expand each cube to a slighly bigger open cube, and do the trick of making the errors exponentially decreasing. Then by subadditivity the exterior measure of the union of the open cubes is at most the sum of the exterior measures, which will be \(m_*(A)\) plus some negligible error term. But the union of the open cubes is an open set containing \(A\), giving the desired result.
Fact 7: If \(d(A_1, A_2) > 0\), \(m_*(A_1 \cup A_2) = m_*(A_1) + m_*(A_2)\)
This statement is a special case of "additivity". In general, we might want and expect that the measure of the disjoint union of two sets is just the sum of the measures. However, this is not generally true, and we'll see this later. However, we can prove additivity right now for sets that have positive distance.
Onto the proof. We have one direction already by subadditivity. For the opposite direction, the high-level idea is as follows: take a closed cube covering \((Q_k)_{k=1}^\infty \) of \(A_1 \cup A_2\) with an error of \(\e\). Because there is a positive distance between \(A_1\) and \(A_2\), we can subdivide the cubes so that no cube spans both. Thus (after throwing out subcubes that ended up in the gap), we can partition our cover into disjoint covers \( ({Q_1}_k)_{k=1}^\infty, ({Q_2}_k)_{k=1}^\infty \) of \(A_1\) and \(A_2\), respectively.
Then, \begin{align} m_*(A_1) + m_*(A_2) &\leq \sum_{k=1}^\infty |{Q_1}_k| + \sum_{k=1}^\infty |{Q_2}_k| \\ & \leq \sum_{k=1}^\infty |Q_k| \\ & \leq m_*(A_1 \cup A_2) + \e \end{align}
And we get the desired result because \(\e\) was arbitrary.Here's some details on how to subdivide the cubes. In an open ball, the largest distance between any two points is the diameter. Thus, if you make your cubes small enough to fit inside an open ball of radius \(\frac{d(A_1, A_2)}{2}\), then you know there can't be two points from different sets simultaneously in the cube, as desired.
Fact 8: If a set \(A\) is the almost disjoint union of countably many cubes, i.e. \(A = \bigcup_{k=1}^\infty\), then \(m_*(A) = \sum_{k=1}^\infty m_*(Q_k) \)
One direction is obvious by monotonicity. For the opposite direction, take the first \(N\) cubes, with \(N\) arbitrary. Shrink them with exponentially decaying error terms to get \((Q'_k)_{k=1}^N \). These cubes now have positive distance, so you can apply Fact 7 repeatedly: \begin{align} m_*(A) &\geq m_*(\bigcup_{k=1}^\infty Q_k) \\ & \geq m_*(\bigcup_{k=1}^N Q_k) \\ & \geq m_*(\bigcup_{k=1}^N Q'_k) \\ & \geq \sum_{k=1}^N m_*(Q_k) - \frac{\e}{2^k} \\ & \geq -\e + \sum_{k=1}^N m_*(Q_k) \end{align} Thus \(\sum_{k=1}^N m_*(Q_k) \leq m_*(A) + \e\) for arbitrary \(\e\) so we can drop it. Then, this is true for arbitrary \(N\) and so true in the limit, giving the desired result.
One corollary is that open sets are "nice", since they are the almost disjoint union of cubes.
Recap
We've introduced the definition of exterior measure and proved a bunch of basic facts about it. Thankfully, it satisfies a lot of intuitive results that we might expect/hope it to.
The important thing that is missing is real, honest additivity. To tackle this, we'll introduce a class of sets that satisfy an additional property. These will be our "measurable" sets, and will satisfy additivity. Later, when we discuss pathologies, we'll see why the extra property was necessary and counterexamples to additivity of the exterior measure.
The most important results here are the ones about coverings of open sets, monotonicity/subadditivity, and the alternative definition of exterior measure in terms of open sets (Fact 6). It was a bit of a lengthy/grindy topic, but it gets a little better (sometimes not by very much though).