Measure, Integration & Real Analysis, 2021a

66 Chapter 2 Measures
Luzin’s Theorem
Our next result is surprising. It says that
an arbitrary Borel measurable function is
almost continuous, in the sense that its
restriction to a large closed set is continuous.
Here, the phrase large closed set
means that we can take the complement
of the closed set to have arbitrarily small
measure.
Be careful about the interpretation of
Nikolai Luzin (1883–1950) proved
the theorem below in 1912. Most
mathematics literature in English
refers to the result below as Lusin’s
Theorem. However, Luzin is the
correct transliteration from Russian
into English; Lusin is the
transliteration into German.
the conclusion of Luzin’s Theorem that f | B is a continuous function on B. This is
not the same as saying that f (on its original domain) is continuous at each point
of B. For example, χ Q
is discontinuous at every point of R. However, χ Q
| R\Q is a
continuous function on R \ Q (because this function is identically 0 on its domain).
2.91 Luzin’s Theorem
Suppose g : R → R is a Borel measurable function. Then for every ε > 0, there
exists a closed set F ⊂ R such that |R \ F| < ε and g| F is a continuous function
on F.
Proof First consider the special case where g = d 1 χ
D1
+ ···+ d n χ
Dn
for some
distinct nonzero d 1 ,...,d n ∈ R and some disjoint Borel sets D 1 ,...,D n ⊂ R.
Suppose ε > 0. For each k ∈{1, . . . , n}, there exist (by 2.71) a closed set F k ⊂ D k
and an open set G k ⊃ D k such that
|G k \ D k | < ε
2n
and
|D k \ F k | < ε
2n .
Because G k \ F k =(G k \ D k ) ∪ (D k \ F k ),wehave|G k \ F k | <
n ε for each k ∈
{1,...,n}.
Let
( ⋃ n ) n⋂
F = F k ∪ (R \ G k ).
k=1
Then F is a closed subset of R and R \ F ⊂ ⋃ n
k=1
(G k \ F k ). Thus |R \ F| < ε.
Because F k ⊂ D k , we see that g is identically d k on F k . Thus g| Fk is continuous
for each k ∈{1, . . . , n}. Because
n⋂
(R \ G k ) ⊂
k=1
k=1
n⋂
(R \ D k ),
we see that g is identically 0 on ⋂ n
k=1
(R \ G k ). Thus g| ⋂ n
k=1 (R\G k ) is continuous.
Putting all this together, we conclude that g| F is continuous (use Exercise 9 in this
section), completing the proof in this special case.
Now consider an arbitrary Borel measurable function g : R → R. By2.89, there
exists a sequence g 1 , g 2 ,...of functions from R to R that converges pointwise on R
to g, where each g k is a simple Borel measurable function.
k=1
Measure, Integration & Real Analysis, by Sheldon Axler