In measure theory, the notion of measurability restricts sets and functions so that limit operations are sensible. With non-measurabilty manifesting only through the Axiom of Choice, measurability is a semantically rich class, and in particular, Littlewood’s Three Principles specify precisely how measurability relates back to more elementary building blocks as follows:
- Every measurable set is nearly open or closed.
- Every measurable function is nearly continuous.
- Every convergent sequence of measurable functions is nearly uniformly convergent.
Here in this post, we flesh out these principles in detail, with an emphasis on how these various concepts approximate one another. Each section corresponds to one principle and is independent from another. We primarily focus on the Lebesgue measure on Euclidean space before discussing the abstract generalization. While there are many equivalent definitions of measurability, for simplicity we will only use these definitions in this discussion:
We will use these facts about the Lebesgue measure throughout.
1. (Monotonicity) If , then .
2. (Sub-additivity) .
3. (Additivity) If are disjoint, .
4. (Upward monotone convergence) If , .
5. (Downward monotone convergence) If with for some , then .
By definition, every measurable set is already arbitrary close in “volume” to an open set. Nonetheless, we can say a lot more about its structure: in , a measurable set can be approximated by a disjoint collection of cubes. In the visual setting where , a measurable set can be approximated by a “pixelation” of squares. We first define the notion of cubes.
The structure of measurable sets actually inherit from the structure of open sets, as stated in the following proposition:
We first state and prove the First Principle before proving the proposition.
where the lines follow from the monotonicity of , additivity of , disjointness of , construction of , sub-additivity of , and construction of , respectively.
with . Let be an open set in , and define
In other words, contains all cubes of length for some integer that reside entirely within . Since is countable and is a subset of countable union of , is countable. By construction . For the other direction, since is open, for every , there is an open ball centered at of radius , , that is contained inside . By Pythagorean Theorem, there exists a cube with length centered at that is contained inside . For every , let . By maximality, , with and at a distance less than from . Thus, is inside the cube of length left-aligned at , implying . Lastly, to see that cubes are almost disjoint except at the boundary, note we may modify so that if a cube is selected from , then we can remove its sub-cubes from for every . By this modification, a selected cube in is never contained in another.
The First Principle can also be extended into any abstract measure space , where is the underlying set, is the set of measurable subsets of , and is the measure for . In the concrete setting, let be a bounded subset of , the set of all measurable sets contained in , and the Lebesgue measure. The First Principle effectively states that every measurable set is equivalent (up to a difference of measure zero) to a countable union of closed sets, and a direct generalization of what we have just proved is that every bounded element of the -algebra “generated” from closed sets can be approximated by closed sets themselves. More precisely,
where the inequalities follow from the monotocity of , sub-additivity
of , and our choice of , respectively. Similarly,
Thus, 3 and setting finishes the proof.
As an example, consider the indicator function where is the set of rationals. When restricted to the domain of , the function becomes identically zero. However, the function remains discontinuous at the points in . Thus, the continuity asserted in Lusin’s Theorem applies only to the restriction of the original function.
Before proving Lusin’s Theorem, we first prove that the continuous functions are dense in absolutely integrable functions.
Case 1: , where is a cube in (as defined in the First Principle).
When , is an interval, and has two points of discontinuity at the endpoints of this interval. A continuous function can be defined so that it agrees with everywhere, except when near each endpoint, is a nearly vertical line dropping from to . The case for larger can be similarly constructed.
Case 2: , where is a finite collection of cubes.
By Case 1, each has a continuous function such that . is continuous, and by triangle inequality, .
Case 3: , where is a measurable set.
From the First Principle, can be approximated by a finite union of disjoint cubes, and the result follows by Case 2 and applying triangle inequality.
Case 4: is simple, i.e., , where is a finite collection of measurable sets.
Follows from a similar argument as in Case 2.
Case 5: .
By definition, . Since , there exists some simple with . Result follows from Case 4 and triangle inequality.
Case 6: General case.
Write , where and . Then Case 5 applies to and , and the general case follows from Case 5 and triangle inequality.
Case 1: .
Fix . By the previous proposition, for every positive integer , (to be specified), there is a continuous function such that . Define
By Markov’s Inequality, . Then , which is at most if we specify . Let . Then by construction, converges to uniformly inside . The restriction of to remains continuous, and since continuity is perserved under uniform limit, is continuous as well.
Case 2: is a bounded function.
Consider the “vertical” truncation of : . Since is bounded, is bounded and . By Case 1, there exists some such that , with continuous when restricted to . Let , then , thus every vertical truncation is continuous when restricted to . Since continuity is a local property, for every , must agree with some on a neighborhood of , so is continuous on as well.
Case 3: General case.
Consider the “horizontal” truncation of : . Since is bounded, by Case 2, there is some such that , with continuous when restricted to . Similarly as in Case 2, we define , with , and every truncation is continuous when restricted to . Since is finite, for every , must agree for some on a neighborhood of and is continuous on as well.
We see that converges to uniformly on , where will be chosen later. First note that is measurable: the limit of measurable functions is measurable, so and thus is meaurable, implying that , the pre-image of the open interval under , is measurable. Since is a countable intersection of , is also measurable. Second, note that is an increasing sequence, and its complement is a decreasing sequence. Since converges to , as . Since , by the downward monotone convergence of measure, as . Thus, there exists some such that .
where the lines follow by the definition of , sub-additivity of measure, construction of , and geometric series, respectively.
It is easy to extend the above setting directly to an abstract one, with the proof following verbatim as before, except replacing the Lebesgue measure by an arbitrary measure :
Lastly, we can lift the restriction that the domain of has finite measure, but uniform convergence must still be local.
so any bounded must be inside for some , so , i.e., converges to uniformly on .