Real-Analysis Lebesgue Measurable-Functions, guaranteed and verified 100% Pass

1


Lebesgue Measurable Functions


We will assume all functions have domains that are a subset of ℝ and take
values in ℝ∪ {±∞}.



Prop. Let 𝑓 have a measurable domain 𝐸. Then the following statements are
equivalent.

1. For each 𝑐 ∈ ℝ, the set {𝑥 ∈ 𝐸 |𝑓 (𝑥 ) > 𝑐 } is measurable.
2. For each 𝑐 ∈ ℝ, the set {𝑥 ∈ 𝐸 |𝑓 (𝑥 ) ≥ 𝑐 } is measurable.
3. For each 𝑐 ∈ ℝ, the set {𝑥 ∈ 𝐸 |𝑓 (𝑥 ) < 𝑐 } is measurable.
4. For each 𝑐 ∈ ℝ, the set {𝑥 ∈ 𝐸 |𝑓 (𝑥 ) ≤ 𝑐 } is measurable.

Each of these properties implies that for each extended real number 𝑐, the set
{𝑥 ∈ 𝐸 |𝑓 (𝑥 ) = 𝑐 } is measurable.


Proof: Sets 1 and 4 , and 2 and 3, are complements . Since complements of
measurable sets are measurable 1 and 4 are equivalent and 2 and 3 are
equivalent.

1⇒2.
1
{𝑥 ∈ 𝐸 |𝑓 (𝑥 ) ≥ 𝑐 } = ⋂∞
𝑛=1 {𝑥 ∈ 𝐸|𝑓 (𝑥 ) > 𝑐 − } . 𝑛

1
By 1, each set {𝑥 ∈ 𝐸|𝑓 (𝑥 ) > 𝑐 − } is measurable.
𝑛


The countable intersection of measurable sets is measurable hence
{𝑥 ∈ 𝐸 |𝑓 (𝑥 ) ≥ 𝑐 } is measurable.

, 2


2⇒1.
1
{𝑥 ∈ 𝐸 |𝑓 (𝑥 ) > 𝑐 } = ⋃∞
𝑛=1 {𝑥 ∈ 𝐸|𝑓 (𝑥 ) ≥ 𝑐 + }. 𝑛

1
By 2, each {𝑥 ∈ 𝐸|𝑓 (𝑥 ) ≥ 𝑐 + } is measurable hence so is
𝑛
{𝑥 ∈ 𝐸 |𝑓 (𝑥 ) > 𝑐 }.

Thus statements 1-4 are equivalent.



Notice that if 𝑐 ∈ ℝ then
{𝑥 ∈ 𝐸 |𝑓 (𝑥 ) = 𝑐 } = {𝑥 ∈ 𝐸 |𝑓 (𝑥 ) ≥ 𝑐 } ∩ {𝑥 ∈ 𝐸 |𝑓(𝑥 ) ≤ 𝑐},
thus {𝑥 ∈ 𝐸 |𝑓 (𝑥 ) = 𝑐 } is measurable because it’s the intersection of two
measurable sets.

Notice that if 𝑐 = ∞ then:
{𝑥 ∈ 𝐸 |𝑓 (𝑥 ) = ∞} = ⋂∞
𝑛=1{𝑥 ∈ 𝐸 |𝑓 (𝑥 ) > 𝑛}.

Thus {𝑥 ∈ 𝐸 |𝑓 (𝑥 ) = ∞} is the countable intersection of measurable sets and
hence measurable.


Def. An extended real valued function defined on 𝐸 is said to be Lebesgue
measurable (or just measurable), provided its domain 𝐸 is measurable and it
satisfies one (and hence all) of the four statements in the previous proposition.



Prop. Let 𝑓: 𝐸 → ℝ ∪ {±∞}, where 𝐸 is measurable. Then 𝑓 is measurable if
and only if for each open set 𝑂, the inverse image of 𝑂,
𝑓 −1 (𝑂) = {𝑥 ∈ 𝐸 | 𝑓(𝑥) ∈ 𝑂}, is a measurable set.