Real-Analysis Continuity of Measure, guaranteed and verified 100% Pass

1


Continuity of Measure


Def. The restriction of the set function outer measure to the class of measurable
sets is called the Lebesgue measure. We will denote this by 𝑚. Thus if 𝐸 is
measurable 𝑚 (𝐸 ) = 𝑚∗ (𝐸).



Prop. Lebesgue measure is countably additive, that is, if {𝐸𝑘 }∞
𝑘=1 is a countable
disjoint collection of sets then ⋃∞
𝑘=1 𝐸𝑘 is measurable and

𝑚(⋃∞ ∞
𝑘=1 𝐸𝑘 ) = ∑𝑘=1 𝑚(𝐸𝑘 ).



Proof: We already know that ⋃∞
𝑘=1 𝐸𝑘 is measurable and the outer measure is
subadditive thus:

𝑚(⋃∞ ∞
𝑘=1 𝐸𝑘 ) ≤ ∑𝑘=1 𝑚(𝐸𝑘 ).


Now let’s prove the inequality in the other direction.
We know for a finite number of disjoint measurable sets:

𝑚(⋃𝑛𝑘=1 𝐸𝑘 ) = ∑𝑛𝑘=1 𝑚(𝐸𝑘 ).

Since ⋃∞ 𝑛
𝑘=1 𝐸𝑘 ⊇ ⋃𝑘=1 𝐸𝑘 for all 𝑛, we have :

𝑚(⋃∞ 𝑛 𝑛
𝑘=1 𝐸𝑘 ) ≥ 𝑚(⋃𝑘=1 𝐸𝑘 ) = ∑𝑘=1 𝑚(𝐸𝑘 ).



Thus we have: 𝑚(⋃∞ ∞
𝑘=1 𝐸𝑘 ) ≥ ∑𝑘=1 𝑚(𝐸𝑘 ).


Hence: 𝑚 (⋃∞ ∞
𝑘=1 𝐸𝑘 ) = ∑𝑘=1 𝑚(𝐸𝑘 ).

, 2


The Lebesgue measure defined on the 𝜎-algebra of Lebesgue measurable sets
satisfies:

1. 𝑚(𝐼) = 𝑙(𝐼)
2. 𝑚(𝑡 + 𝐸 ) = 𝑚(𝐸)
3. 𝑚(⋃∞ ∞
𝑘=1 𝐸𝑘 ) = ∑𝑘=1 𝑚(𝐸𝑘 ); 𝐸𝑘 are disjoint measurable sets.



Ex. Define 𝐸∆𝐹 = (𝐸~𝐹) ∪ (𝐹~𝐸). Suppose 𝐸 and 𝐹 are measurable sets.
Prove 𝑚(𝐸∆𝐹 ) = 𝑚(𝐸 ∩ 𝐹 𝑐 ) + 𝑚(𝐹 ∩ 𝐸 𝑐 ).

𝐸∆𝐹




𝐸
𝐹


𝐸∆𝐹 = (𝐸~𝐹 ) ∪ (𝐹~𝐸 ) = (𝐸 ∩ 𝐹 𝑐 ) ∪ (𝐹 ∩ 𝐸 𝑐 ).

Since (𝐸 ∩ 𝐹 𝑐 ) and (𝐹 ∩ 𝐸 𝑐 ) are disjoint and measurable we have:

𝑚(𝐸∆𝐹 ) = 𝑚((𝐸 ∩ 𝐹 𝑐 ) ∪ (𝐹 ∩ 𝐸 𝑐 )) = 𝑚((𝐸 ∩ 𝐹 𝑐 )) + 𝑚((𝐹 ∩ 𝐸 𝑐 )) .




Def. {𝐸𝑘 }∞ 𝑘=1 is said to be ascending if for each 𝑘 𝐸𝑘 ⊆ 𝐸𝑘+1 , and descending
if for each 𝑘 𝐸𝑘 ⊇ 𝐸𝑘+1 .