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

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Approximating Lebesgue Measurable Sets


So far we know the following about measurable sets:
1. They form a 𝜎-algebra (so closed under countable unions and complements
hence also closed under countable intersections)
2. They contain all Borel sets (the smallest 𝜎-algebra containing all open
subsets of ℝ)
3. All sets of measure 0 are measurable.



Measurable sets contain the following excision property. If 𝐴 is measurable of
finite outer measure and 𝐴 ⊆ 𝐸, then

𝑚∗ (𝐸~𝐴) = 𝑚∗ (𝐸 ) − 𝑚∗ (𝐴)
since:

𝑚∗ (𝐸 ) = 𝑚∗ (𝐸 ∩ 𝐴) + 𝑚∗ (𝐸 ∩ 𝐴𝑐 ) = 𝑚∗ (𝐴) + 𝑚∗ (𝐸~𝐴).

And since 𝑚∗ (𝐴) is finite we have:

𝑚∗ (𝐸~𝐴) = 𝑚∗ (𝐸 ) − 𝑚∗ (𝐴).


Theorem: Let 𝐸 be any set of real numbers. Then each of the following is
equivalent to the measurability of 𝐸:

1. For each 𝜖 > 0, there’s an open set 𝑂 ⊇ 𝐸 where 𝑚 ∗ (𝑂~𝐸 ) < 𝜖
2. There is a 𝐺𝛿 set 𝐺 ⊇ 𝐸 for which 𝑚∗ (𝐺~𝐸 ) = 0
3. For each 𝜖 > 0, there’s an closed set 𝐹 ⊆ 𝐸 where 𝑚∗ (𝐸~𝐹 ) < 𝜖
4. There is an 𝐹𝜎 set 𝐹 ⊆ 𝐸 where 𝑚∗ (𝐸~𝐹 ) = 0.

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Proof: Assume 𝐸 is measurable and let 𝜖 > 0 be given.

If 𝑚∗ (𝐸 ) < ∞ then by the definition of an outer measure there is a collection
{𝐼𝑘 }∞
𝑘=1 of open intervals which covers 𝐸 and has

∑∞ ∗
𝑘=1 𝑙( 𝐼𝑘 ) < 𝑚 (𝐸 ) + 𝜖.


Let 𝑂 = ⋃∞
𝑘=1 𝐼𝑘 , then 𝑂 is an open set containing 𝐸. In addition:

𝑚∗ (𝑂) ≤ ∑∞ ∗
𝑘=1 𝑙( 𝐼𝑘 ) < 𝑚 (𝐸 ) + 𝜖,

So 𝑚∗ (𝑂) − 𝑚∗ (𝐸 ) < 𝜖 .

Since 𝐸 is measurable and has finite measure, by the excision property:

𝑚∗ (𝑂~𝐸 ) = 𝑚∗ (𝑂) − 𝑚∗ (𝐸 ) < 𝜖.


If 𝑚∗ (𝐸 ) = ∞, let 𝐸𝑘 = 𝐸 ∩ [𝑘, 𝑘 + 1] and 𝐸 = ⋃𝑘∈ℤ 𝐸𝑘 , a countable
union measurable sets, each with finite measure.

By the first part we know there is an open set 𝑂𝑘 ⊇ 𝐸𝑘 with
𝜖
𝑚∗ (𝑂𝑘 ~𝐸𝑘 ) < .
2|𝑘|+2


Now let 𝑂 = ⋃𝑘∈ℤ 𝑂𝑘 .


𝑂 is open, 𝑂 ⊇ 𝐸, and 𝑂~𝐸 = ⋃𝑘∈ℤ 𝑂𝑘 ~𝐸 ⊆ ⋃𝑘∈ℤ( 𝑂𝑘 ~𝐸𝑘 ).
(Draw a picture with 2 pairs of sets, 𝑂1 , 𝑂2 , and 𝐸1 , 𝐸2 to see this is true.)

Therefore:
𝜖
𝑚∗ (𝑂~𝐸 ) ≤ ∑𝑘∈ℤ 𝑚∗ (𝑂𝑘 ~𝐸𝑘 ) < ∑𝑘∈ℤ <𝜖
2|𝑘|+2