Measure, Integration & Real Analysis, 2021a

70 Chapter 2 Measures
The next result states that if we adopt
the philosophy that what happens on a
set of outer measure 0 does not matter
much, then we might as well restrict our
attention to Borel measurable functions.
“He professes to have received no
sinister measure.”
– Measure for Measure,
by William Shakespeare
2.95 every Lebesgue measurable function is almost Borel measurable
Suppose f : R → R is a Lebesgue measurable function. Then there exists a Borel
measurable function g : R → R such that
|{x ∈ R : g(x) ̸= f (x)}| = 0.
Proof There exists a sequence f 1 , f 2 ,...of Lebesgue measurable simple functions
from R to R converging pointwise on R to f (by 2.89). Suppose k ∈ Z + . Then there
exist c 1 ,...,c n ∈ R and disjoint Lebesgue measurable sets A 1 ,...,A n ⊂ R such
that
f k = c 1 χ
A1
+ ···+ c n χ
An
.
For each j ∈{1, . . . , n}, there exists a Borel set B j ⊂ A j such that |A j \ B j | = 0
[by the equivalence of (a) and (d) in 2.71]. Let
g k = c 1 χ
B1
+ ···+ c n χ
Bn
.
Then g k is a Borel measurable function and |{x ∈ R : g k (x) ̸= f k (x)}| = 0.
If x /∈ ⋃ ∞
k=1
{x ∈ R : g k (x) ̸= f k (x)}, then g k (x) = f k (x) for all k ∈ Z + and
hence lim k→∞ g k (x) = f (x). Let
E = {x ∈ R : lim g k (x) exists in R}.
k→∞
Then E is a Borel subset of R [by Exercise 14(b) in Section 2B]. Also,
∞⋃
R \ E ⊂ {x ∈ R : g k (x) ̸= f k (x)}
k=1
and thus |R \ E| = 0. Forx ∈ R, let
2.96 g(x) = lim
k→∞
(χ E
g k )(x).
If x ∈ E, then the limit above exists by the definition of E; ifx ∈ R \ E, then the
limit above exists because (χ E
g k )(x) =0 for all k ∈ Z + .
For each k ∈ Z + , the function χ E
g k is Borel measurable. Thus 2.96 implies that g
is a Borel measurable function (by 2.48). Because
{x ∈ R : g(x) ̸= f (x)} ⊂
∞⋃
{x ∈ R : g k (x) ̸= f k (x)},
k=1
we have |{x ∈ R : g(x) ̸= f (x)}| = 0, completing the proof.
Measure, Integration & Real Analysis, by Sheldon Axler