Real Analysis MAA 6616

Lecture 9
Lebesgue Measurable Functions
Definition and Properties of Measurable Functions

All functions considered here will be R-valued, where R is the extended real line
R ∪ {±∞} = [−∞, ∞]. For a function f : E ⊂ Rp −→ R, we will use the following
abbreviation {f < c} for {x ∈ E : f (x) < c} where c ∈ R. Similar abbreviations will be used
for {f > c}, {f ≥ c}, {f ≤ c}, and {f = c}.

Proposition (1)
Let f : E ⊂ Rp −→ R. Then the following properties are equivalent:
1. For every c ∈ R, {f > c} is measurable;
2. For every c ∈ R, {f ≥ c} is measurable;
3. For every c ∈ R, {f < c} is measurable;
4. For every c ∈ R, {f ≤ c} is measurable.
Moreover, each one of these properties implies that {f = c} is measurable.

Proof.
This proposition follows from the fact that the collection M of measurable sets is a σ-algebra: (1) ⇐⇒ (4) and
(2) ⇐⇒ (3) follow from {f ≤ c} = R\{f > c} and {f ≥ c} = R\{f < c}; (1) =⇒ (2) and (2) =⇒ (1) follow
∞ ∞
\ 1 [ 1
from {f ≥ c} = {f > c − } and {f > c} = {f ≥ c + }.
n=1
n n=1
n
Now assume that anyone of the four property holds (and so all four hold). Let c ∈ R. Then {f = c} = {f ≥ c} ∩ {f ≤ c}
\∞ ∞
\
is measurable. Also {f = −∞} = {f < −n} and {f = ∞} = {f > n} are measurable.
n=1 n=1
A function f : E ⊂ Rp −→ R defined on the measurable set E is said to be Lebesgue
measurable if it satisfies any one of the properties of Proposition 1.

Theorem (2)
Let f : E ⊂ Rp −→ R with E measurable. Then f is measurable if and only if for every open
set U ⊂ R, the pre-image f −1 (U) = {x ∈ E : f (x) ∈ U} is a measurable set in Rp .

Proof.
"⇐=" Suppose that for every open set U ⊂ R, f −1 (U) is measurable. Let c ∈ R. Since (c, ∞) ⊂ R is open, then
{f > c} is measurable, then f is measurable.
"=⇒" Suppose that f is measurable (and so it satisfies all four properties of Proposition 1). Let U ⊂ R be an open set. Then
S
there is a countable collection of open and bounded intervals {Ij }j∈N = {(aj , bj )}j∈N such that U = j∈N . Note that
−1 −1
Ij = (aj , ∞) ∩ (−∞, bj ) so that f (Ij ) = {f < bj } ∩ {f > aj } and f (Ij ) is measurable. Therefore
∞
−1 −1
[
f (U) = f (Ij ) is measurable as a countable union of measurable sets.
j=1

Theorem (3)
Let f : E ⊂ Rp −→ R be a continuous function and E measurable. Then f is measurable

Proof.
Let U ⊂ R be an open set. It suffices to show that f −1 (U) is measurable (Theorem 2). Since f is continuous, then there exists
an open set V ⊂ Rp such that f −1 (U) = E ∩ V. Both E and V are measurable and so is their intersection f −1 (U)
Recall that a property (P) is said to hold almost everywhere (abbreviated a.e) in a set E if there
exists a set Z ⊂ E of measure zero such that (P) holds for every point x ∈ E\Z.

Theorem (4)
Let f : E ⊂ Rp −→ R defined on the measurable set E.
1. Suppose that f is measurable Let g : E −→ R. If g = f a.e. on E, then g is measurable
and for every c ∈ R, m ({g > c}) = m ({f > c})
2. The function f is measurable on E if and only if for every measurable set A ⊂ E the
restrictions fA and fE\A of f to A and to E\A are measurable.

Proof.
1. Let Z = {x ∈ E : g(x) ̸= f (x)} then m(Z) = 0. Let c ∈ R We need to show that {g > c} is measurable. We
have

{g > c} = {x ∈ Z : g(x) > c} ∪ {x ∈ E\Z : f (x) > c} = {x ∈ Z : g(x) > c} ∪ ({f > c} ∩ (E\Z)) .

The set {g > c} is measurable as an intersection of three measurable sets. Note that since
{f > c} = {x ∈ E\Z : f (x) > c} ∪ {x ∈ Z : f (x) > c} and m(Z) = 0, then
m({f > c}) = m ({x ∈ E\Z : f (x) > c}). As a consequence we have m ({g > c}) = m ({f > c}).
2. Let A ⊂ E measurable and c ∈ R. We have

{f > c} = {fA > c} ∪ {fE\A > c} = ({f > c} ∩ A) ∪ ({f > c} ∩ (E\A)) .

It follows immediately that f is measurable if both fA and fE\A are measurable.

Operations on Measurable Functions

Theorem (5)
Let f : E ⊂ Rp −→ R be finite a.e. in E and let α : R −→ R be a continuous function.
Assume that f is measurable. Then the composition α ◦ f is measurable.

Proof.
Let Z = {f = ±∞}. Then Z has measure zero. Set h = α ◦ f so that h is well defined on E\Z by h(x) = α(f (x)). Let
U ⊂ R be an open set. We need to verify that h−1 (U) = f −1 (α−1 (U)) ∩ (E\Z) is measurable. Since α is continuous,
then α−1 (U) is open. Therefore, f −1 (α−1 (U)) is measurable and so is h−1 (U).

Remark (1)
It follows from Theorem 5 that if f is measurable, then so are the functions |f |, f 2 , |f |q for
q > 0, eλf etc.

Let E ⊂ Rp . The characteristic function of E is the function


1 if x ∈ E
χE : Rp −→ R ; χE (x) = .
0 if x ∈
/E

It follows directly from the definition of measurable functions that χE is measurable if and only
if E is measurable.
Remark (2)
Composition of measurable functions is not necessarily measurable. Consider the
ψ(x) = x + ϕ(x) where ϕ is the Cantor-Lebesgue function. We know that ψ is a
homeomorphism between [0, 1] and [0, 2] and that there exists a set S ⊂ C such ψ(S) is not
measurable (where C is the Cantor set). We can extend ψ as a global homeomorphism
Ψ : R −→ R (for example Ψ(x) = 2x for x < 0 and for x > 1). Hence Ψ−1 is continuous.
Then the composition h = χS ◦ Ψ−1 is not measurable. Indeed, {h = 1} = Ψ(χS (1)) = ψ(S)
is not measurable.

Remark (3)
Given a function h, its measurability is not affected if the values are changed on a set of
measure zero. With this understanding, the sum f + g of two functions valued on [−∞, ∞] is
well defined provided that the set of indeterminacy where f (x) = ∞ and g(x) = −∞ has
measure zero. Similarly for the product fg when the product takes the form 0 · ∞.

Theorem (6)
Let f , g : E −→ R be measurable functions that are finite a.e. in E. Then
1. The sum f + g and difference f − g are measurable.
2. The product fg is measurable.
3. The quotient f /g is measurable provided that g ̸= 0 a.e. in E.
Proof.
1. First observe that the set {f > g} = {x ∈ E : f (x) > g(x)} is measurable. Indeed it can be written as a countable
union of measurable sets: {f > g} = ∪r∈Q ({f > r} ∩ {g < r}). Next, for any λ ∈ R, the function f + λ or
−f + λ is measurable since {f + λ > c} = {f > c − λ} is measurable. Now to prove that f + g is measurable,
let c ∈ R, then {f + g > c} = {f > c − g} is measurable by the above observations.
2. We have 4fg = (f + g)2 − (f − g)2 . It follows from part 1 that f ± g are measurable and so (f ± g)2 are also
measurable (Theorem 4). Consequently, fg is measurable.
3. Since up to a set of measure zero, we have {(1/g) > c} = {g < (1/c)}, the measurability of g implies that of
1/g and so of the quotient f /g (part 2)

Let f1 , · · · , fn be functions defined on the same domain E ⊂ Rp . Define the functions

max{f1 , · · · , fn } and min{f1 , · · · , fn } in E by

max{f1 , · · · , fn }(x) = max{f1 (x), · · · , fn (x)} and

min{f1 , · · · , fn }(x) = min{f1 (x), · · · , fn (x)} .

Theorem (7)
Let f1 , · · · , fn : E −→ R be measurable functions. Then the functions max{f1 , · · · , fn } and
min{f1 , · · · , fn } are also measurable.
Proof.
Let c ∈ R. We have

n
[ n
\
{max{f1 , · · · , fn } > c} = {fj > c} and {min{f1 , · · · , fn } > c} = {fj > c} .
j=1 j=1

Thus {max{f1 , · · · , fn } > c} is measurable as a finite union of measurable sets and {min{f1 , · · · , fn } > c} is
measurable as a finite intersection of measurable sets.

For a function f : E −→ R we associate the functions:

|f | = max{f , −f } , f + = max{f , 0} , and f − = max{−f , 0}.

Note that
f = f+ − f−
a difference of two nonnegative functions. Also |f | = f + + f − .