Real-Analysis-The-Lebesgue Integral-f-bounded-E-of-finite-measure, guaranteed and verified 100% Pass

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The Lebesgue Integral ∫𝐸 𝑓: 𝑓 Bounded, 𝑚(𝐸 ) < ∞


Note: From now on integration will mean Lebesgue integration unless
otherwise specified.



Let 𝜓 = ∑𝑛 −1
𝑖=1 𝑎𝑖 𝜒𝐸𝑖 on 𝐸, where 𝐸𝑖 = 𝜓 (𝑎𝑖 ) = {𝑥 ∈ 𝐸 | 𝜓(𝑥 ) = 𝑎𝑖 }

Be a simple function (𝑎𝑖 ′𝑠 are distinct and {𝐸𝑖 } disjoint).



Def. For a simple function 𝜓 defined on a set of finite measure 𝐸, define
.
∫𝑬 𝝍 = ∑𝒏𝒊=𝟏 𝒂𝒊 (𝒎(𝑬𝒊 )).



Notice that this definition of ∫𝐸 𝜓 allows us to evaluate the following integral.


Ex. Let 𝑓 (𝑥 ) = 1 if 𝑥 ∈ ℚ ∩ [0,1] = 𝐸1

= 0 if [0,1]~𝐸1 .

Evaluate ∫[0,1] 𝑓 .



Let 𝐸1 = ℚ ∩ [0,1] and 𝐸2 = [0,1]~𝐸1 , then we can write:

𝑓(𝑥 ) = 1(𝜒𝐸1 ) + 0(𝜒𝐸2 ) = 𝜒𝐸2 .

Thus, ∫[0,1] 𝑓 = 1 (𝑚(𝜒𝐸1 )) = 0.

, 2


Lemma: Let {𝐸𝑖 }𝑛
𝑖=1 be a finite disjoint collection of measurable subsets of a set
of finite measure 𝐸. For 1 ≤ 𝑖 ≤ 𝑛, let 𝑎𝑖 be a real number. If

𝜑 = ∑𝑛𝑖=1 𝑎𝑖 𝜒𝐸𝑖 on 𝐸 then ∫𝐸 𝜑 = ∑𝑛𝑖=1 𝑎𝑖 (𝑚(𝐸𝑖 )).


Proof: The issue here is that {𝑎𝑖 } may not be distinct (i.e., 𝜑 is not written in
canonical form). If we rewrite 𝜑 in canonical form the result readily follows.

𝜑(𝑥)
𝑎2
𝜑(𝑥 ) = 𝑎1 𝜒𝐸1 + 𝑎2 𝜒𝐸2 + 𝑎1 𝜒𝐸3 ;

𝑎1 Let 𝐸 ′ = 𝐸1 ∪ 𝐸3 then we can

Write 𝜑 in canonical form:
𝜑(𝑥 ) = 𝑎1 𝜒𝐸′ + 𝑎2 𝜒𝐸2
𝐸1 𝐸2 𝐸3


Prop. Let 𝜑 and 𝜓 be simple function defined on a set of finite measure 𝐸.
Then

1. for 𝛼, 𝛽 ∈ ℝ ∫𝐸 (𝛼𝜑 + 𝛽𝜓) = 𝛼 ∫𝐸 𝜑 + 𝛽 ∫𝐸 𝜓.
2. if 𝜑 ≤ 𝜓 on 𝐸 then

∫𝐸 𝜑 ≤ ∫𝐸 𝜓.


Proof: Since 𝜑 and 𝜓 are simple we can find a finite disjoint collection of sets
{𝐸𝑖 }𝑛𝑖=1 such that 𝜑 and 𝜓 are constant on each 𝐸𝑖 and 𝐸 = ⋃𝑛𝑖=1 𝐸𝑖 .
𝑦 = 𝜓(𝑥)
𝑦 = 𝜑(𝑥)




𝐸1 𝐸2 𝐸3 𝐸4