Real-Analysis Littlewoods -Three Principles Egoroffs Theorem and Lusins Theorem-1, guaranteed and verified 100% Pass

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Littlewood’s Three Principles


Littlewood’s three principles:
1. Every measurable set is nearly a finite union of intervals
2. Every measurable function is nearly continuous
3. Every pointwise convergent sequence of measurable functions is nearly
uniformly convergent.
We have seen Littlewood’s first principle already. It takes the form of the
theorem:

Theorem: Let 𝐸 be a measurable set of finite outer measure. Then for each
𝜖 > 0, there is a finite disjoint collection of open intervals {𝐼𝑘 }𝑛𝑘=1 for which if
𝑂 = ⋃𝑛𝑘=1 𝐼𝑘 , then
𝑚∗ (𝐸~𝑂) + 𝑚∗ (𝑂~𝐸 ) < 𝜖.


A precise statement of Littlewood’s third principle is:

Theorem (Egoroff): Assume 𝐸 has finite measure. Let {𝑓𝑛 } be a sequence of
measurable functions on 𝐸 that converge pointwise on 𝐸 to the real valued
function 𝑓. Then for each 𝜖 > 0, there is a closed set 𝐹 ⊆ 𝐸 for which
{𝑓𝑛 } → 𝑓 uniformly on 𝐹 and 𝑚(𝐸~𝐹 ) < 𝜖.




To prove Egoroff’s theorem we use the following:

Lemma: Under the assumptions of Egoroff’s theorem, for each 𝛼 > 0 and
𝛿 > 0, there is a measurable subset 𝐵 ⊆ 𝐸 and an 𝑁 ∈ ℤ+ such that if 𝑛 ≥ 𝑁
then
|𝑓𝑛 − 𝑓| < 𝛼 on 𝐵 and 𝑚(𝐸~𝐵) < 𝛿.

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Proof: Since {𝑓𝑛 } → 𝑓 pointwise, 𝑓 is measurable.

Hence the set {𝑥 ∈ 𝐸 | |𝑓 (𝑥 ) − 𝑓𝑘 (𝑥 )| < 𝛼} is measurable.

Let 𝐸𝑛 = {𝑥 ∈ 𝐸 | |𝑓 (𝑥 ) − 𝑓𝑘 (𝑥 )| < 𝛼 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘 ≥ 𝑛}.

Then 𝐸𝑛 = ⋂∞
𝑘=𝑛{𝑥 ∈ 𝐸 | |𝑓 (𝑥 ) − 𝑓𝑘 (𝑥 )| < 𝛼}

is measurable because it’s the countable intersection of measurable sets.


Notice that 𝐸𝑛 ⊆ 𝐸𝑛+1 ⊆ ⋯ is an ascending collection of sets with:

𝐸 = ⋃∞
𝑛=1 𝐸𝑛 since {𝑓𝑛 } → 𝑓 pointwise on 𝐸.


By the continuity of measure we know that: 𝑚(𝐸 ) = lim 𝑚(𝐸𝑛 ).
𝑛→∞


Since 𝑚(𝐸 ) < ∞, we can choose an 𝑁 for which 𝑚(𝐸 ) − 𝑚(𝐸𝑁 ) < 𝛿.

Define 𝐵 = 𝐸𝑁 , then by the excision property:

𝑚(𝐸~𝐵) = 𝑚(𝐸 ) − 𝑚(𝐵) = 𝑚(𝐸 ) − 𝑚(𝐸𝑁 ) < 𝛿.


Proof of Egoroff’s theorem.
1 𝜖
Using the previous lemma, with 𝛼 = and 𝛿= , Let 𝐵𝑛 be the
𝑛 2𝑛+1
measurable subset of 𝐸 and 𝑁(𝑛) which satisfies the conclusion of the lemma.
𝜖 1
Thus: 𝑚(𝐸~𝐵𝑛 ) < and |𝑓𝑘 − 𝑓 | < on 𝐵𝑛 for all 𝑘 ≥ 𝑁(𝑛).
2𝑛+1 𝑛


Define: 𝐵 = ⋂∞
𝑛=1 𝐵𝑛 .

Then 𝑚(𝐸~𝐵) = 𝑚(⋃∞
𝑛=1(𝐸~𝐵𝑛 ))
𝜖 𝜖
≤ ∑∞
𝑛=1 𝑚 (𝐸~𝐵𝑛 ) < ∑∞
𝑛=1 𝑛+1 = .
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