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

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Lebesgue Measurable Sets


The outer measure 𝑚∗ has 4 important properties:
1. It’s defined for all sets of real numbers.
2. 𝑚∗ (𝐼) = 𝑙(𝐼) = 𝑏 − 𝑎, for any interval (𝑎, 𝑏), [𝑎, 𝑏), (𝑎, 𝑏], [𝑎, 𝑏].
3. 𝑚∗ (⋃∞ ∞ ∗ ∗
𝑘=1 𝐸𝑘 ) ≤ ∑𝑘=1 𝑚 (𝐸𝑘 ) (i.e. 𝑚 is countably subadditive).
4. 𝑚∗ (𝑡 + 𝐸 ) = 𝑚∗ (𝐸) for any 𝑡 ∈ ℝ (i.e. 𝑚∗ is translation invariant).
The problem is there are disjoint sets 𝐴, 𝐵 such that:

𝑚∗ (𝐴 ∪ 𝐵) < 𝑚∗ (𝐴) + 𝑚∗ (𝐵).
This does not correspond well to one’s intuition about how a measure should
work. To solve this problem we will simply remove these “bad” sets.



For any set 𝐸 ⊆ ℝ, Notice that we can always write a set 𝐴 as:

𝐴 = (𝐴 ∩ 𝐸) ∪ (𝐴 ∩ 𝐸 𝑐 ).

Def. A set 𝐸 ⊆ ℝ is said to be measurable provided for any set 𝐴:

𝑚∗ (𝐴) = 𝑚∗ (𝐴 ∩ 𝐸 ) + 𝑚∗ (𝐴 ∩ 𝐸 𝑐 ).


Notice that if 𝐸 is a measurable set and 𝐷 is any set with 𝐸 ∩ 𝐷 = 𝜙,

and we take the set 𝐴 = 𝐸 ∪ 𝐷, we get:

𝑚∗ (𝐸 ∪ 𝐷) = 𝑚∗ ([𝐸 ∪ 𝐷] ∩ 𝐸 ) + 𝑚∗ ([𝐸 ∪ 𝐷] ∩ 𝐸 𝑐 )
= 𝑚∗ (𝐸 ) + 𝑚∗ (𝐷) i.e. there’s no inequality.

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If we write 𝐴 = [𝐴 ∩ 𝐸 ] ∪ [𝐴 ∩ 𝐸 𝑐 ] and 𝐴 is any set (possibly not measurable)
we know from the subadditivity of 𝑚∗ :

𝑚∗ (𝐴) ≤ 𝑚∗ (𝐴 ∩ 𝐸 ) + 𝑚∗ (𝐴 ∩ 𝐸 𝑐 ).


Thus 𝐸 is measurable if, and only if:

𝑚∗ (𝐴) ≥ 𝑚∗ (𝐴 ∩ 𝐸 ) + 𝑚∗ (𝐴 ∩ 𝐸 𝑐 ).


This last inequality will always hold if 𝑚∗ (𝐴) = ∞. So it’s enough to show 𝐸 is
measurable by showing this inequality holds for sets 𝐴 where 𝑚∗ (𝐴) is finite.



The definition of a set 𝐸 being measurable is symmetric in 𝐸 and 𝐸 𝑐 .



Thus 𝐸 is measurable if, and only if, 𝐸 𝑐 is measurable.



Prop. If 𝑚∗ (𝐸 ) = 0 then 𝐸 is measurable.



Proof: Let 𝐴 be any set.

𝐴 ∩ 𝐸 ⊆ 𝐸 and 𝐴 ∩ 𝐸 𝑐 ⊆ 𝐴 thus,
0 ≤ 𝑚∗ (𝐴 ∩ 𝐸 ) ≤ 𝑚∗ (𝐸 ) = 0 and 𝑚∗ (𝐴 ∩ 𝐸 𝑐 ) ≤ 𝑚∗ (𝐴).

Thus we have:

𝑚∗ (𝐴) ≥ 𝑚∗ (𝐴 ∩ 𝐸 𝑐 ) = 0 + 𝑚∗ (𝐴 ∩ 𝐸 𝑐 ) = 𝑚∗ (𝐴 ∩ 𝐸 ) + 𝑚∗ (𝐴 ∩ 𝐸 𝑐 )
Hence 𝐸 is measurable.