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:
A set

is measurable if for every

, there is some open set

containing

such that

, where

is the Lebesgue measure.
A function

is measurable if for every open interval

, the preimage of

under

,

, is a measurable set.
We will use these facts about the Lebesgue measure throughout.
Let

be the Lebesgue measure on

. Let

be a collection of measurable sets.
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

.
First Principle
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.
We say that

is a closed cube of length

left-aligned at

if
![Rendered by QuickLaTeX.com Q=\Pi_{i=1}^{d}[a_{i},a_{i}+\ell]](https://www.victorchen.org/wp-content/ql-cache/quicklatex.com-d860438fcfa91c2264e5ea0629d646d9_l3.png)
.
The structure of measurable sets actually inherit from the structure of open sets, as stated in the following proposition:
Let

be open. Then

is a countable union of closed cubes, and every pair of these cubes is disjoint except at the boundary.
We first state and prove the First Principle before proving the proposition.
Let

be a measurable set with finite measure. For every

, there exist

disjoint cubes

such that

, where

.
(of theorem) Fix

. It suffices to show that there exist

disjoint cubes

such that

and

. Since

is measurable, there is an open set

containing

such that

. In particular,

since

. From the above proposition,

, where

is a closed cube and pairwise disjoint except at the boundary. For each closed cube

, there exists a cube

such that

, and

is pairwise disjoint. Now, by the upward monotone convergence of measure, there is some

such that

. Let

. Then

by the monotonicity of

. For the other case,

where the lines follow from the monotonicity of
, additivity of
, disjointness of
, construction of
, sub-additivity of
, and construction of
, respectively.
(of proposition) Note that

can be divided into cubes of length

:
![Rendered by QuickLaTeX.com \[ G_{n}=\{ \Pi_{i=1}^{d}[a_{i}2^{-n},(a_{i}+1)2^{-n}]:a_{1},\ldots,a_{d}\in\mathbb{Z}\}, \]](https://www.victorchen.org/wp-content/ql-cache/quicklatex.com-f7ab6b9b4cd9688c9e871eb007c275b1_l3.png)
with
. Let
be an open set in
, and define
![Rendered by QuickLaTeX.com \[ Q=\bigcup_{n\in\mathbb{N}}\{B\in G_{n}:B\subset V\}. \]](https://www.victorchen.org/wp-content/ql-cache/quicklatex.com-c8bb50427dc195756ec24c87e68246d7_l3.png)
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,
Let

be a measure space with

. Suppose

is an algebra. Then for every

,

, where

is the intersection of all

-algebra containing

, there exists

such that

.
Fix

and let

. Since

contains

, it suffices to show that

is itself a

-algebra. It is easy to see that

is closed under complement since

is. For countable union, for every

, let

,

. By definition, for every

, there is some

such that

, where

will be chosen later. By upward monotone convergence, there exists some

such that

. Write

and

, and note that

is in

. Then

where the inequalities follow from the monotocity of
, sub-additivity
of
, and our choice of
, respectively. Similarly,

Thus,
3
and setting
finishes the proof.
Second Principle
(Lusin’s Theorem) Let

be a finite, measurable function. For every

, there exists

such that

and

, the restriction of

to

, is continuous.
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.
Let

be a measurable function. Suppose

. Then for every

, there exists a continuous function

such that

.
We proceed in stages, where in each stage

is restricted to an elementary, but more general class of functions than the previous stage. Note that in all cases, the hypothesis that

is measurable and

always applies.
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.
(Lusin’s Theorem) We proceed in stages by proving the statement for progressively more general classes of functions. In all casees, the hypothesis that

is measurable and finite always applies.
Case 1:
.
Fix
. By the previous proposition, for every positive integer
,
(to be specified), there is a continuous function
such that
. Define
![Rendered by QuickLaTeX.com \[ E_{n}=\{x\in\mathbb{R}^{d}:|f(x)-g_{n}(x)|\geq1/n\}. \]](https://www.victorchen.org/wp-content/ql-cache/quicklatex.com-8750dfe827825124c11f7ff2c93cc468_l3.png)
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.
Third Principle
(Egorov’s Theorem)

, with

. Let

be a sequence of measurable functions with

converging to

pointwise as

. Then for every

, there exists a measurable subset

with

, where

converging to

uniformly within

.
By definition,

converges to

uniformly on

iff for every

, for every positive integer

, there is some positive integer

with

. Let
![Rendered by QuickLaTeX.com \[ A_{m,N}=\{x\in X:\forall n\geq N\big|f_{n}(x)-f(x)\big|<1/m\}. \]](https://www.victorchen.org/wp-content/ql-cache/quicklatex.com-62e630bb751f67de4c8d989f40f33a15_l3.png)
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
.
Then

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
:
(Abstract Egorov’s Theorem) Let

be a measure space with

. Let

be a sequence of measurable functions, with

converging to

pointwise as

. Then for every

, there exists a measurable subset

with

, where

converges to

uniformly within

.
Lastly, we can lift the restriction that the domain of
has finite measure, but uniform convergence must still be local.
(Egorov’s Theorem II) Let

be a sequence of measurable functions with

converging to

pointwise as

. Then for every

, there exists a measurable subset

with

, where

converges to

locally uniform within

, i.e., for every bounded subset

,

converges to

uniformly on

.