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Unit 27: Measurable Functions and Littlewood’s Second Principle
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To understand Point (i), recall from that all continuous functions on are measurable; in Notes
contrast, there are measurable functions that are highly discontinuous (like Dirichlet’s function).
There are more measurable functions than continuous functions because there are a lot more
measurable sets than there are open sets. For example, not only are open sets measurable but so
are points, Cantor-type sets, G sets, F sets, etc. We shall see that, just like continuous functions,
measurable functions are closed under all the usual arithmetic operations such as addition,
multiplication, etc. What exemplifies Point (ii) is that measurable functions are closed under all
limiting operations. For example, a limit of measurable functions is always measurable. This
stands in stark contrast to continuous functions. Indeed, that the characteristic function of a
Cantor set can be expressed as a limit of continuous functions. The reason that measurable
functions are closed under more operations is that measurable sets are closed under operations
(e.g. countable intersections and complements) that open sets are not.
Measurable functions are similar to continuous functions, but there are more of them and they
are more robust. Littlewood’s second principle shows exactly how “similar” measurable
functions are to continuous functions.
27.3 Littlewood’s Second Principle
We now continue our discussion of Littlewood’s Principles where we stated the first principle;
There are three principles, roughly expressible in the following terms: Every [finite Lebesgue]
measurable set is nearly a finite union of intervals; every measurable function is nearly continuous;
every convergent sequence of measurable functions is nearly uniformly convergent.
—Nikolai Luzin
The third principle is contained in Egorov’s theorem, which we’ll get to in the next topic.
The second principle comes from Luzin’s Theorem, named after Nikolai Nikolaevich Luzin
(1883-1950) who proved it in 1912 [70], and this theorem makes precise Littlewood’s comment
that any Lebesgue measurable function is “nearly continuous”.
Theorem 3: Luzin’s Theorem
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Let X be Lebesgue measurable and let f : X be a Lebesgue measurable function. Then
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given any > 0, there exists a closed set C such that C X, m(X\C) < , and f is continuous
on C.
Proof: Here we follow Feldman’s [38] proof that only uses Littlewood’s First Principle. Luzin’s
theorem is commonly proved using Egorov’s theorem and the fact that every measurable function
is the limit of simple functions.
Step 1: We first prove the theorem only requiring that C be measurable; this proof is yet another
example of the “ k –trick.” Let { } be a countable basis of open sets in ; this means that every
2 k
open set in is a union of countably many ’s. (For example, take the ’s as open intervals with
k k
–1
rational end points.) Let > 0. Then, since f ( ) is measurable, by Littlewood’s First Principle
k
there is an open set such that
k
f ( ) and m( \f ( )) < .
–1
–1
k k k k k
2
Now put
1
-
A : = ( k \f ( )).
k
k 1
=
Then A is measurable and
1
-
m(A) £ å m ( k \f ( )) < å k = .
k
k 1 k 1 2
=
=
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