Real-Analysis-The-Lebesgue Integral-f-Nonnegative, guaranteed and verified 100% Pass

1



The Lebesgue Integral ∫𝐸 𝑓: 𝑓≥0


Def. A measurable function with domain 𝐸 has finite support if
𝑚({𝑥 ∈ 𝐸 | 𝑓(𝑥 ) ≠ 0}) < ∞.

Let 𝐸0 = {𝑥 ∈ 𝐸 | 𝑓 (𝑥 ) ≠ 0} with 𝑚(𝐸0 ) < ∞. If 𝑓 is bounded and
measurable on 𝐸 we can define:

∫𝑬 𝒇 = ∫𝑬𝟎 𝒇.



So what do we do if 𝑚(𝐸0 ) = ∞ or 𝑚(𝐸0 ) < ∞ but 𝑓 is not bounded?


Def. Let 𝑓 be a nonnegative measurable function on 𝐸. Let

𝐻𝐸 (𝑓) = {bounded measurable functions of finite support|0 ≤ ℎ ≤ 𝑓 on 𝐸}.

we define ∫𝐸 𝑓 by

∫𝑬 𝒇 = 𝐬𝐮𝐩 {∫𝑬 𝒉| 𝒉 ∈ 𝑯𝑬 (𝒇)}.


Notice that ∫𝐸 𝑓 can be finite or ∞.

, 2


Chebychev’s inequality: Let 𝑓 be a nonnegative measurable function on 𝐸. Then
for any 𝑎 > 0,
(𝑎)(𝑚(𝐸)) 𝑦 = 𝑓(𝑥)
1
𝑚({𝑥 ∈ 𝐸 | 𝑓 (𝑥 ) ≥ 𝑎}) ≤ ∫𝐸 𝑓.
𝑎 𝑎


Proof: Let 𝐸𝑎 = {𝑥 ∈ 𝐸 | 𝑓 (𝑥 ) ≥ 𝑎}.


First let’s assume 𝑚(𝐸𝑎 ) = ∞.

Let 𝐸𝑎,𝑛 = 𝐸𝑎 ∩ [−𝑛, 𝑛] and 𝜓𝑛 = 𝑎(𝜒𝐸𝑎,𝑛 ).
𝐸

𝜓𝑛 is a bounded, measurable function of finite support with

𝑎(𝑚(𝐸𝑎,𝑛 )) = ∫𝐸 𝜓𝑛 and 0 ≤ 𝜓𝑛 ≤ 𝑓 on 𝐸 for all 𝑛.

Notice that: 𝐸𝑎 = ⋃∞
𝑛=1 𝐸𝑎,𝑛 and 𝐸𝑎,𝑛+1 ⊇ 𝐸𝑎,𝑛 .

Thus 𝑚(𝐸𝑎 ) = lim 𝑚(⋃∞
𝑛=1 𝐸𝑎,𝑛 ) = lim 𝑚(𝐸𝑎,𝑛 ).
𝑛→∞ 𝑛→∞


Thus ∞ = 𝑎(𝑚(𝐸𝑎 )) = 𝑎 ( lim 𝑚(𝐸𝑎,𝑛 )) = lim ∫𝐸 𝜓𝑛 ≤ ∫𝐸 𝑓 .
𝑛→∞ 𝑛→∞

1
So 𝑚({𝑥 ∈ 𝐸 | 𝑓 (𝑥 ) ≥ 𝑎}) ≤ ∫ 𝑓 because both sides are ∞.
𝑎 𝐸



Now suppose 𝑚(𝐸𝑎 ) < ∞.


Define ℎ = 𝑎(𝜒𝐸𝑎 ).

ℎ is a bounded measurable function with finite support and 0 ≤ ℎ ≤ 𝑓 on 𝐸.