Measure Theory/Section 1 Proofs, Measure

Section 1 Proofs, Measure

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Proofs are organized into subproofs. If any statement is followed by a boxed region, then the boxed region is a subproof of the statement.

Lesson 1: No Total Measure of Real Numbers

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Theorem: There Is No Total Measure of Real Numbers

Let   be any real-valued set function.  Then   must not satisfy at least one of the properties: Length measure, nonnegativity, translation-invariance, countable additivity.

Proof:

Assume that there is a function   which has the properties of nonnegativity, interval length, translation-invariant, and countably additive.

This implies a contradiction.

Define the relation   on   by   if  .

  is reflexive.

Let  . Then   and therefore   by definition.

  is symmetric.

Let   so   and   by definition.

Then   by the closure of   under multiplication.

Then   by definition.

  is transitive.

Let   and   so that by definition   and all three are in (0,1).

Then   by the closure of   under addition.

Then by definition  .

Because   is reflexive, symmetric, and transitive, therefore it is an equivalence relation.

Because   is an equivalence relation there is a partition P induced by  . If   then write the cell of P containing x by  .

Define F to be any arbitrary set with the following property.   if and only if there is a unique   such that  .

Let   be any enumeration of  , which must exist because the set is countable.

For any two distinct  , the sets   and   are disjoint.

Suppose there is some  .

Then  .

Let   and  .

Then   by the closure of rational numbers under subtraction.

Then   and therefore  .

Because F contains unique representatives of each cell of the partition P, then  .

Returning to an earlier equation, this implies  

Then   and therefore since   is an enumeration,  .

Define  .

Then  .

 

Let  .

Let   such that   and therefore  .

Also   since both  .

Therefore there is some   in the enumeration  , such that  .

Then  .

 

Let  , and therefore there is some   in the enumeration  , and then there is some   such that  .

Since both   and   then  .

  is monotonic because it is countably additive and monotonic.

Let  .

Then  .

By countable additivity  .

By the nonnegativity of  ,  

By monotonicity and the interval length property,  

By countable additivity and the disjointness of  ,  

By translation-invariance, there is some constant   such that every  .

By all of the above,  

If   then  , and if   then   from the above.

From all of the above, we have established that

  •  
  •   or  .

This is a contradiction.

 

Lesson 2: Outer Measure Properties

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Theorem: Outer-measure Is Well-defined and Nonnegative.

For every   the outer-measure takes a unique extended real number value.  That is to say  .  Also this value is nonnegative,  .  

Proof:

  always exists as an extended real number, and  .

  is an open set and  .

Therefore   is an open interval over-approximation of A.

Therefore   is an over-estimate of A.

Therefore the set of over-estimates of A is non-empty.

Also the set of over-estimates of A is bounded below by zero.

If   is any open interval over-approximation of A, and   the corresponding over-estimate, then e is a sum of nonnegative numbers.

Therefore  .

Therefore the infimum over the set of over-estimates of A exists and is at least zero.

 

Theorem: Outer Measure Is Determined by Countable Sums

For any  , 

   

Proof:

For any uncountable series of nonnegative numbers, either the sum is infinity or at most countably many terms are nonzero. The proof of this fact is deferred to Terence Tao's book Analysis II.

Let   be an uncountable over-approximation of A.

Either   or there are only countably many nonzero terms,  .

Because the length of an open interval is always strictly positive, then  .

But also,   is a countable series and is an over-estimate of A.

Therefore if   is the collection of all over-estimates of A, and if   is the set of over-estimates which are countable series, then  .

Therefore  .

 

Theorem: Outer-measure Is Translation Invariant.

For every subset   and real number   the outer-measure of A is invariant under translation by c, 
   

Proof:

Let   be a set of real numbers, and   any real number.

Let   be any open interval over-approximation of A.

Define

 

Then   is an open interval over-approximation of  .

If   then   is an open interval. Hence all the elements of   are open intervals.

If   then there is some   such that  .

Then there is some   such that  .

Then   and  .

So   covers A.

Let   be the set of all over-estimates of A.

Likewise define   to be the set of all over-estimates of  .

For every over-estimate   of A, and for every  , we have that  .

Therefore  .

Therefore   and so  .

Mutatis mutandis the same proof shows  .

Therefore   and therefore

 

 

Theorem: Outer-measure Is Monotonic.

If   then  .

Proof:

Let  .

Any open interval over-approximation of B is immediately also an open interval over-approximation of A, which follows trivially by definitions.

Set   as the set of over-estimates of A and   the over-estimates of B.

Then  .

From elementary analysis, therefore

 

 

Theorem: Countable Sets Are Null.

If   is a countable set, then  .

If   then we can set the open interval over-approximation  .

Then

 .

Therefore, for the rest of the proof, assume  .

Let   be a countable set, with enumeration  .

Let  .

Then  .

Define the sequence of open intervals by

 
 
...
 
...

If N is non-finite with final indexed element   then for indices   set  .

Then  .

Then as a geometric sum

 

From an elementary argument it is clear that   is an open interval over-approximation of N.

Therefore by definition as an infimum,  .

Therefore  .

 

Lesson 3: Outer Measure Interval Length

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Theorem: Outer Measure Interval Length.

For every interval  , its outer-measure is its length,  .

Proof:

Let   be any interval with end-points a and b. Note that the bounds are extended real numbers, so a may be   and b may be  .

Note that if   then I is a countable set, and by a previous result, has measure zero which is equal to its length. Therefore, throughout the rest of the proof, we assume  .

If   is any closed, bounded interval of real numbers, then  .

Let   be a positive real number less than  .

Then  .

Define the open interval over-approximation  .

Then

 

Letting   we have  .

Also   is a lower bound on the set of all over-estimates of I.

Let   be any over-estimate of I.

Let   be a finite subcover of I, which must exist because [a,b] is a compact set and   an open cover.

Let   be the subset which results from successively removing from   any interval which is a subset of some other interval.

  is still a cover of I.

List the elements   in increasing order of the left end-point.   for  .

For any  , if   then either   or  .

Either case is impossible because no interval can be a subset of any other, hence we must have the strict inequality  .

Because a is covered by the set, then it is in one of the intervals. Whichever one it is, we must then have  .

Mutatis mutandis, the same argument shows  .

For each  , if   then   is not covered by any interval in  . Therefore  .

Then by merely shifting parentheses for regrouping terms,

 

where the inequalities indicated by curly braces are due to  . The inequalities   and   are due to earlier inequalities.

Also

 

because each series is of nonnegative terms, and contains a subsequence of the terms in the series before it.

Combining all of the above,

 

By definition of the outer measure as an infimum, then  

Therefore

 

If   is a bounded open interval then  .

By monotonicity  .

Let   be any positive real number less than  .

Then by monotonicity,

 

Letting   we have

 

Up to this point we now have shown that every bounded open interval, and every bounded closed interval, has outer measure equal to its length.

If I is any other bounded interval, then

 

so that by monotonicity

 

and so  .

If I is any interval unbounded on the right but finite on the left, then for every   we have  .

By monotonicity

  as  

Mutatis mutandis the same argument shows that if I is finite on the right and unbounded on the left, or if  , then  .

In every case, therefore,  .

 

Lesson 4: Outer Measure Subadditivity

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Theorem: Outer-measure not Additive

  is not countably additive.

Proof:

In earlier theorems we have proved that   satisfies the properties Nonnegativity, Interval length, and Translation invariance.

We also know from an earlier theorem that no function, defined on  , has all four properties, Nonnegativity, Interval length, Translation invariance, and Countable additivity.

Since   is defined on   then it cannot be countably additive.

 

Theorem: Outer-measure Subadditivity.

Let   be any countable sequence of subsets of real numbers.  Then 
   

Proof:

Suppose that for some   we have  .

The sum of   with any nonnegative number is again  .

Therefore, regardless of the value of the left-hand side, we have

 

Therefore throughout the rest of the proof, assume that each   has finite outer measure.

Let   be any positive real number.

For each   there is an open interval over-approximation of  , call it  , such that

 

Then   is an open interval over-approximation of  .

Also,

 

Therefore by definition of the infimum,

 

Letting  

 

 

Theorem: Null Adding and Subtracting.

Let   be two subsets and E a null set.  Then 
   

Proof:

By monotonicity, and the fact that  ,

 

Therefore  .

From this result,

 

 

Lesson 5: Measurable Sets

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Theorem: Null Sets Are Measurable.

Every null set is measurable.

Proof:

Let   be a null set.

Let   be any set.

By monotonicity   therefore

 

By a previous result, because E is null,  .

Therefore

 

so E splits A cleanly.

Since A was arbitrary, therefore E is measurable.

 

Theorem: Measurable Sets Closed Under Complement.

If   is a measurable subset then   is measurable.

Proof:

Let   be a measurable set.

Let   be any set of real numbers.

Then by definition of the set difference,

 

So   splits A cleanly, and since A was arbitrary, then   is measurable.

 

Theorem: Open Rays Are Measurable.

For every   the open ray   is measurable.

Proof:

Let  , and let   be any set of real numbers. Let   be the open ray to the right of a.

If   then I splits A cleanly.

By monotonicity,  .

If   then regardless of the right-hand side we must have  .

Therefore for the rest of this proof, assume   is finite.

Let   be any positive real number.

Then  .

By definition as a finite infimum, there is an open interval over-approximation   of A such that

 

Define   and  .

Then   and   are open interval over-approximations of   and   respectively.

Also

 

Also  .

Let   where the end-points   are real numbers.

If   then   and   and then   and  .

Then

 

Mutatis mutandis the same proof handles the case where  .

If   then   and  . Then   and  .

 

So we have shown in all cases that  .

Therefore

 

Letting  

 

Therefore  .

Therefore I splits A cleanly.

If   then I splits A cleanly.

Because   is countable, therefore it is a null set, due to a theorem from an earlier lesson.

Therefore we may union and subtract   from a set without changing its outer measure.

Denote  .

Therefore, by all of the above,

 

Since I splits A cleanly in all cases, and A was arbitrary, then I is measurable.

 

Lesson 6: Measurable Sets Are a Sigma-algebra

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Theorem: The Measurable Sets Form a  -algebra.

The collection   is a  -algebra.

Proof:

From two earlier theorems,   because countable sets are null and null sets are measurable.

Also from an earlier theorem,   is closed under taking complements.

If   are two measurable sets then their union is measurable,  .

Let

 
 
 
 

Let   be any set of real numbers.

Then   splits A cleanly.

By the measurability of E followed by the measurability of F, followed by elementary set theory,

 

By monotonicity and elementary set theory,

 

With the further observation by elementary set theory that   and  , then all of the above now shows

 

Therefore  .

If   is any finite sequence of measurable sets then their union is measurable,  

If   the case is trivial.

Suppose the theorem is true for some  , and consider the case for   sets.

Then  .

Then by the result which we proved above for two sets,

 

Thus the proof by induction is complete.

If   is any countable sequence of measurable sets then  .

Let  .

Let  .

Also let   and for   define   so that   forms a sequence of disjoint sets.

Each   is measurable by the result that complements and finite unions of measurable sets are measurable.

Also  .

By subadditivity,

 

For each  

 

Moreover,  .

Lemma: Finite Split by Measurables

If   are any finite sequence of disjoint measurable sets and   is any set of real numbers, then

 

Proof:

If   are any two disjoint measurable sets and   is any set of real numbers, then  .

Because H is measurable then

 

If   is any finite sequence of disjoint measurable sets, and   is any set of real numbers, then

 

The case for   is trivial.

Suppose the claim holds for some   and now consider the claim for   sets.

 

Therefore

 

Now taking the limit   on each side of this last inequality,

 

By subadditivity,

 

Putting together results from above,

 

Since A was arbitrary, this shows that E splits every set cleanly, and therefore E is measurable.

The above shows that

  •  .
  •   is closed under complement and countable unions.

Therefore   is a  -algebra.

 

Theorem:  -algebra Space and Intersection

Let X be any set and   a  -algebra on X.  Then   and   is closed under countable intersections.

Proof:

By closure under complements,  .

By de Morgan's law, if   then

 

By the closure under complements, each   is measurable.

By closure under unions,   is measurable.

By closure under complements,   is measurable.

 

Theorem: Intervals Are Measurable.

If   is an interval then  .

Proof:

From an earlier result, all open rays to the right,  , are measurable.

Let   be any closed ray to the left.

Then   is the complement of an open ray to the right.

By the closure of measurable sets under complements, therefore I is measurable.

Now let   a bounded interval open on the left, for   two real numbers.

Then  , and by closure of measurable sets under intersections, then J is measurable.

Now let   be any bounded open interval.

Then define the sequence of intervals open on the left,   for  .

Then   and each   is measurable.

Since the measurable sets are closed under countable unions, then K is measurable.

Mutatis mutandis the same basic proof demonstrates the measurability of all other remaining kinds of intervals.

 

Lesson 8: Properties of Length Measure

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Theorem: Length-measure Is Countably Additive.

Let   be a disjoint sequence of measurable sets.  Then 
   

Proof:

By subadditivity,

 

Fix any  .

In the following, the first inequality is by monotonicity. The second is due to the lemma from the previous section, Finite Split by Measurables, setting B in that lemma to  .

 

Taking   in the above shows  .

Therefore  .

 

Theorem: Length-measure Excision.

Let   be two measurable sets, and  , and assume that  .  Then  .

Proof:

By monotonicity   and hence   is a finite real number. Therefore arithmetic operations are well-defined for   and  .

By additivity,

 

Whence

 

 

Theorem: Upward Continuity of Measure.

Let   be an ascending sequence of measurable sets.  Then 

   

Proof:

Define   and for  , then  .

Then   is a sequence of disjoint measurable sets, and  .

Therefore by additivity,

 

 

Theorem: Downward Continuity of Measure.

Let   be a descending sequence of measurable sets.  If   is finite then 
   

Proof:

By monotonicity, every   and therefore arithmetic operations on these are well-defined.

Define   for each  .

Then   is an increasing sequence of measurable sets, and hence the upward continuity of measure applies to it.

By excision,

 

and

 

By upward continuity of measure, therefore the two objects above are equal and therefore  .

Whence