Proofs are organized into subproofs. If any statement is followed by a boxed region, then the boxed region is a subproof of the statement.
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.
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is symmetric.
Let so and by definition.
Then by the closure of under multiplication.
Then by definition.
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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 .
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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 ![{\displaystyle f_{2}-f_{1}=0=r_{i}-r_{j}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/38bcbe652fb69c40e99ebe75110da86ad851167a)
Then and therefore since is an enumeration, .
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Define .
Then .
![{\displaystyle (0,1)\subseteq V}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b482fcd089b2672380116b3436eea5d6e3ad01e)
Let .
Let such that and therefore .
Also since both .
Therefore there is some in the enumeration , such that .
Then .
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is monotonic because it is countably additive and monotonic.
Let .
Then .
By countable additivity .
By the nonnegativity of , ![{\displaystyle \mu (B)\geq \mu (A)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f5d63f3d27d2f7d9ed2f04656b29e21b0c3be836)
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By monotonicity and the interval length property, ![{\displaystyle \mu ((0,1))=1\leq \mu (V)\leq \mu ((-1,2))=3}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b43d585e76533b26a24204686288c9ec0fc492bd)
By countable additivity and the disjointness of , ![{\displaystyle \mu (V)=\sum _{i=1}^{\infty }\mu (F+r_{i})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c38fabf9644fcaaa41630901410b6aaf130e198f)
By translation-invariance, there is some constant such that every .
By all of the above, ![{\displaystyle \mu (V)=\sum _{i=1}^{\infty }c}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c094d1457b0c33cfac0f1407295af0f1e241d863)
If then , and if then from the above.
From all of the above, we have established that
![{\displaystyle 1\leq \mu (V)\leq 3}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a2cc98335584dbd1f2037dc1f1daade1d3348fcb)
or .
This is a contradiction.
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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 .
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Therefore the infimum over the set of over-estimates of A exists and is at least zero.
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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
![{\displaystyle {\mathfrak {O}}+c=\{I+c:I\in {\mathfrak {O}}\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fdfdf6636e0505b8f948841f1c17b617dd112978)
Then
is an open interval over-approximation of
.
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
![{\displaystyle \lambda ^{*}(A)=\inf {\mathcal {E}}=\inf {\mathcal {F}}=\lambda ^{*}(A+c)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/60e43a512d6ae40526cc251f5fc64a05be41bb22)
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
![{\displaystyle \lambda ^{*}(A)=\inf {\mathcal {E}}\leq \inf {\mathcal {F}}=\lambda ^{*}(B)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5eece1d0a0d5f5f0c56940c32e477e96376a4ba9)
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
![{\displaystyle I_{1}=(n_{1}-\varepsilon /2^{2},n_{1}+\varepsilon /2^{2}),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f833bb15d89f57cab9aa0f5b53fab92713392e93)
![{\displaystyle I_{2}=(n_{2}-\varepsilon /2^{3},n_{2}+\varepsilon /2^{3}),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/57544e12bde937890c597352ed6672d96ba00dc7)
- ...
![{\displaystyle I_{j}=(n_{j}-\varepsilon /2^{j+1},n_{j}+\varepsilon /2^{j+1})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a6d7cdeb7eca6410166ce127beba977b1c2d832)
- ...
If N is non-finite with final indexed element then for indices set .
Then .
Then as a geometric sum
![{\displaystyle \sum _{j=1}^{\infty }\ell (I_{j})\leq \sum _{j=1}^{\infty }\varepsilon /2^{j}={\frac {1}{2}}\cdot \varepsilon \left({\frac {1}{1-1/2}}\right)=\varepsilon }](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed89463b40148f94ae155ff209dddbf4a8ccf0f9)
From an elementary argument it is clear that is an open interval over-approximation of N.
Therefore by definition as an infimum, .
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Therefore
.
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
![{\displaystyle \sum _{J\in {\mathfrak {O}}}\ell (J)=b-a-\varepsilon }](https://wikimedia.org/api/rest_v1/media/math/render/svg/53727d8f9cd2f495417d924e99571e5b4c69a0f6)
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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,
![{\displaystyle {\begin{aligned}\sum _{j=1}^{n}\ell (I_{j})&=(b_{1}-a_{1})+\cdots +(b_{n}-a_{n})\\&=-a_{1}+\overbrace {(b_{1}-a_{2})} ^{>0}+\overbrace {(b_{2}-a_{3})} ^{>0}+\cdots +\overbrace {(b_{n-1}-a_{n})} ^{>0}+b_{n}\\&>-a+0+\cdots +0+b\\&=b-a\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d3a876fe78c62dbdc44d87518bc5bbd30267935)
where the inequalities indicated by curly braces are due to . The inequalities and are due to earlier inequalities.
Also
![{\displaystyle {\begin{aligned}e&=\sum _{J\in {\mathfrak {O}}}\ell (J)\\&\geq \sum _{J\in {\mathfrak {O}}'}\ell (J)\\&\geq \sum _{J\in {\mathfrak {O}}''}\ell (J)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6859c7ef87e02f743037805f324a4fcb89e8a247)
because each series is of nonnegative terms, and contains a subsequence of the terms in the series before it.
Combining all of the above,
![{\displaystyle b-a\leq e}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9eb4fbbc59f8ba1504c57830a97d6d76b9ce8fea)
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By definition of the outer measure as an infimum, then
Therefore
![{\displaystyle \lambda ^{*}(I)=b-a}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c66eff6ff1f4c521ff64dfd2f1d52e89d283308)
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If
is a bounded open interval then
.
By monotonicity .
Let be any positive real number less than .
Then by monotonicity,
![{\displaystyle \lambda ^{*}([a+\varepsilon /2,b-\varepsilon /2])=b-a-\varepsilon \leq \lambda ^{*}(I)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/782d6fb6568d0a87a4b6bd587044676db7d442f4)
Letting we have
![{\displaystyle b-a\leq \lambda ^{*}(I)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/146983f83324697e1f41a45262aff045fb6043b6)
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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
![{\displaystyle (a,b)\subseteq I\subseteq [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0448d47adcde162daf615ecb25f5d3107de1e6b7)
so that by monotonicity
![{\displaystyle b-a\leq \lambda ^{*}(I)\leq b-a}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1193d37befa632879508c49aa0f76a9d69d1f9dc)
and so
.
If I is any interval unbounded on the right but finite on the left, then for every
we have
.
By monotonicity
as ![{\displaystyle n\to \infty }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a0d55d9b32f6fa8fab6a84ea444a6b5a24bb45e1)
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,
.
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
![{\displaystyle \lambda ^{*}\left(\bigcup _{k=1}^{\infty }A_{k}\right)\leq \infty =\sum _{k=1}^{\infty }\lambda ^{*}(A_{k})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bafce38827ef1d87ab2b97d08ed70179ed94f2c4)
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
![{\displaystyle \sum _{I\in {\mathfrak {O}}_{k}}\ell (I)<\lambda ^{*}(A_{k})+\varepsilon /2^{k}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d00eb5d486eb4abbd7efda55f376e70bf7afed24)
Then
is an open interval over-approximation of
.
Also,
![{\displaystyle {\begin{aligned}\sum _{I\in {\mathfrak {O}}}\ell (I)&=\sum _{k\in {\mathbb {N}}^{+}}\sum _{I\in {\mathfrak {O}}_{k}}\ell (I)\\&<\sum _{k\in {\mathbb {N}}^{+}}(\lambda ^{*}(A_{k})+\varepsilon /2^{k})\\&=\sum _{k=1}^{\infty }\lambda ^{*}(A_{k})+\varepsilon \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/44702ce1a34938f0d5e9ad6ec2c9696c14acfb5a)
Therefore by definition of the infimum,
![{\displaystyle \lambda ^{*}\left(\bigcup _{k=1}^{\infty }I_{k}\right)<\sum _{k=1}^{\infty }\lambda ^{*}(A_{k})+\varepsilon }](https://wikimedia.org/api/rest_v1/media/math/render/svg/d4069b4464bbba21a5e013bfbb8e45ba34e01a15)
Letting
![{\displaystyle \lambda ^{*}\left(\bigcup _{k=1}^{\infty }I_{k}\right)\leq \sum _{k=1}^{\infty }\lambda ^{*}(A_{k})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/edf51a2371ea5f4157424155807f5f00e0b33fb3)
Theorem: Null Adding and Subtracting.
Let
be two subsets and E a null set. Then
Proof:
By monotonicity, and the fact that
,
![{\displaystyle \lambda ^{*}(A)\leq \lambda ^{*}(A\cup E)\leq \lambda ^{*}(A)+\lambda ^{*}(E)=\lambda ^{*}(A)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e219651c6d3d9810a29762e9da8af12e0746ea46)
Therefore
.
From this result,
![{\displaystyle \lambda ^{*}(A\smallsetminus E)=\lambda ^{*}([A\smallsetminus E]\sqcup E)=\lambda ^{*}(A\cup E)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a95e823896765cb266ce811a56098bcc477009be)
Theorem: Null Sets Are Measurable.
Every null set is measurable.
Proof:
Let
be a null set.
Let
be any set.
By monotonicity
therefore
![{\displaystyle \lambda ^{*}(A\cap E)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f24028cc7d5cf9cfaae03d0a3aa9bf37c8c82631)
By a previous result, because E is null,
.
Therefore
![{\displaystyle \lambda ^{*}(A\cap E)+\lambda ^{*}(A\smallsetminus E)=\lambda ^{*}(A)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/34fc29c5420b796a7b7cc4ddcca661bf86d89f02)
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,
![{\displaystyle {\begin{aligned}\lambda ^{*}(A)&=\lambda ^{*}(A\cap E)+\lambda ^{*}(A\smallsetminus E)\\&=\lambda ^{*}(A\smallsetminus E^{c})+\lambda ^{*}(A\cap E^{c})\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/332daf52f41c423046de945ae69e833a1240da38)
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
![{\displaystyle \sum _{J\in {\mathfrak {O}}}\ell (J)<\lambda ^{*}(A)+\varepsilon }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a655f08467c720733d8d6d58f581b8e85e14704e)
Define and .
Then and are open interval over-approximations of and respectively.
Also
![{\displaystyle {\begin{aligned}\lambda ^{*}(A\cap I)+\lambda ^{*}(A\smallsetminus I)&\leq \sum _{J\in {\mathfrak {O}}_{l}}\ell (J)+\sum _{J\in {\mathfrak {O}}_{r}}\ell (J)\\&=\sum _{J\in {\mathfrak {O}}}\ell (J\cap I)+\sum _{J\in {\mathfrak {O}}}\ell (J\cap (-\infty ,a))\\&=\sum _{J\in {\mathfrak {O}}}(\ell (J\cap I)+\ell (J\cap (-\infty ,a)))\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eacea0b3729af87b70745c86630bcd9a02cc59ad)
Also .
Let where the end-points are real numbers.
If then and and then and .
Then
![{\displaystyle \ell (J)=\ell (J\cap I)+\ell (J\cap (-\infty ,a))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/877061bc9a015a98f9ff41e44bfc671543dedf6b)
Mutatis mutandis the same proof handles the case where .
If then and . Then and .
![{\displaystyle \ell (J)=y-x=(y-a)+(a-x)=\ell (J\cap (-\infty ,a))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3e6bab1d4d6db6b57bf4f21a08dfff12895685de)
So we have shown in all cases that .
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Therefore
![{\displaystyle {\begin{aligned}\lambda ^{*}(A\cap I)+\lambda ^{*}(A\smallsetminus I)&\leq \sum _{J\in {\mathfrak {O}}}\ell (J)\\&<\lambda ^{*}(A)+\varepsilon \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/79224699e17cdd2521b581f71efc79ea7d0910ff)
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Letting
![{\displaystyle \lambda ^{*}(A)\geq \lambda ^{*}(A\cap I)+\lambda ^{*}(A\smallsetminus I)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a35e92ffa0d60a3c68c437b59a710e139fad7dc)
Therefore .
Therefore I splits A cleanly.
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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,
![{\displaystyle {\begin{aligned}\lambda ^{*}(A)&=\lambda ^{*}(B)\\&=\lambda ^{*}(B\cap I)+\lambda ^{*}(B\smallsetminus I)\\&=\lambda ^{*}(A\cap I)+\lambda ^{*}(A\smallsetminus I)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a2ab9170a2d11d23d191c491f2a0a1387c8453e)
Since I splits A cleanly in all cases, and A was arbitrary, then I is measurable.
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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
![{\displaystyle E\cap F=E_{11}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a1c709fbfbe14d8d355ab24d02b7752886bb1d81)
![{\displaystyle E\cap F^{c}=E_{10}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/700392c6087b8237d595873c645475eee0d10d03)
![{\displaystyle E^{c}\cap F=E_{01}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c09ded792e83906b9e8aa5f867caa821c0582f7)
![{\displaystyle E^{c}\cap F^{c}=E_{00}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1fbbeacf10a1cd52d434f97aa6872b2d94b82cf)
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,
![{\displaystyle {\begin{aligned}\lambda ^{*}(A)&=\lambda ^{*}(A\cap E)+\lambda ^{*}(A\smallsetminus E)\\&=\lambda ^{*}(A\cap E)+\lambda ^{*}([A\smallsetminus E]\cap F)+\lambda ^{*}([A\smallsetminus E]\smallsetminus F)\\&=\lambda ^{*}(A\cap E)+\lambda ^{*}(A\cap E_{01})+\lambda ^{*}(A\cap E_{00})\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/104912b37659a9bca58edaa71ddc850fb9d9b11d)
By monotonicity and elementary set theory,
![{\displaystyle {\begin{aligned}\lambda ^{*}(A\cap E)+\lambda ^{*}(A\cap E_{01})&\geq \lambda ^{*}([A\cap E]\sqcup [A\cap E_{01}])\\&=\lambda ^{*}(A\cap [E\sqcup E_{01}])\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eaad6bc102045cec0851d3b31ca78c017261117a)
With the further observation by elementary set theory that and , then all of the above now shows
![{\displaystyle {\begin{aligned}\lambda ^{*}(A)&\geq \lambda ^{*}(A\cap [E\sqcup E_{01}])+\lambda ^{*}(A\cap E_{00})\\&=\lambda ^{*}(A\cap [E\cup F])+\lambda ^{*}(A\smallsetminus [E\cup F])\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ebef1d0dba35499773828454852df894f16ab3c7)
|
Therefore .
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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,
![{\displaystyle \bigcup _{k=1}^{n+1}E_{k}=\bigcup _{k=1}^{n}E_{k}\cup E_{n+1}\in {\mathcal {M}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b36817c862cea62648599dfd36c6001758d0bc2c)
Thus the proof by induction is complete.
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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,
![{\displaystyle \lambda ^{*}(A)\leq \lambda ^{*}(A\cap E)+\lambda ^{*}(A\smallsetminus E)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc992d8bbc00d870e33d1aa33f4db586f54bd119)
For each
![{\displaystyle {\begin{aligned}\lambda ^{*}(A)&=\lambda ^{*}\left(A\cap \bigsqcup _{k=1}^{n}F_{k}\right)+\lambda ^{*}\left(A\smallsetminus \bigsqcup _{k=1}^{n}F_{k}\right)\quad {\text{ by the earlier result}}\\&\geq \lambda ^{*}\left(A\cap \bigsqcup _{k=1}^{n}F_{k}\right)+\lambda ^{*}\left(A\smallsetminus \bigsqcup _{k=1}^{\infty }F_{k}\right)\quad {\text{ by monotonicity}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/61e300fd9ce8b851bd54f87b2159f55a59b71abe)
Moreover, .
Lemma: Finite Split by Measurables
If are any finite sequence of disjoint measurable sets and is any set of real numbers, then
![{\displaystyle \lambda ^{*}\left(B\cap \bigsqcup _{k=1}^{n}G_{k}\right)=\sum _{k=1}^{n}\lambda ^{*}(B\cap G_{k})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5e251bbd4a6cda84212e9d2a4e88a72c8b0a8bf6)
Proof:
If are any two disjoint measurable sets and is any set of real numbers, then .
Because H is measurable then
![{\displaystyle {\begin{aligned}\lambda ^{*}(B\cap [G\sqcup H])&=\lambda ^{*}(B\cap [G\sqcup H]\cap H)+\lambda ^{*}(B\cap [G\sqcup H]\smallsetminus H)\\&=\lambda ^{*}(B\cap H)+\lambda ^{*}(B\cap G)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/abab7caf03ddf6aa1f4a9657251ae62936b38049)
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If is any finite sequence of disjoint measurable sets, and is any set of real numbers, then
![{\displaystyle \lambda ^{*}\left(B\cap \bigsqcup _{k=1}^{n}G_{k}\right)=\sum _{k=1}^{n}\lambda ^{*}(B\cap G_{k})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5e251bbd4a6cda84212e9d2a4e88a72c8b0a8bf6)
The case for is trivial.
Suppose the claim holds for some and now consider the claim for sets.
![{\displaystyle {\begin{aligned}\lambda ^{*}\left(B\cap \bigsqcup _{k=1}^{n+1}G_{k}\right)&=\lambda ^{*}\left(\left[B\cap \bigsqcup _{k=1}^{n}G_{k}\right]\sqcup \left[B\cap G_{n+1}\right]\right)\\&=\lambda ^{*}\left(B\cap \bigsqcup _{k=1}^{n}G_{k}\right)+\lambda ^{*}(B\cap G_{n+1})\quad {\text{ by the result for pairs}}\\&=\sum _{k=1}^{n}\lambda ^{*}(B\cap G_{k})+\lambda ^{*}(B\cap G_{n+1})\\&=\sum _{k=1}^{n+1}\lambda ^{*}(B\cap G_{k})\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ddfe80ff294e181fd707627f66f419972d76a39c)
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Therefore
![{\displaystyle \lambda ^{*}(A)\geq \sum _{k=1}^{n}\lambda ^{*}(A\cap F_{k})+\lambda ^{*}(A\smallsetminus E)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1fc6b083c9f29bbdcacb6420bc46f820ac6701a)
Now taking the limit on each side of this last inequality,
![{\displaystyle \lambda ^{*}(A)\geq \sum _{k=1}^{\infty }\lambda ^{*}(A\cap F_{k})+\lambda ^{*}(A\smallsetminus E)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3cf45a512650c85a8d6e2617e9a1a3b88eb5fca4)
By subadditivity,
![{\displaystyle {\begin{aligned}\lambda ^{*}(A\cap E)&=\lambda ^{*}\left(A\cap \bigsqcup _{k=1}^{\infty }F_{k}\right)\\&=\lambda ^{*}\left(\bigsqcup _{k=1}^{\infty }(A\cap F_{k})\right)\\&\leq \sum _{k=1}^{\infty }\lambda ^{*}(A\cap F_{k})\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/439087285743623b66435f00c91cebecf2fdcf9e)
Putting together results from above,
![{\displaystyle \lambda ^{*}(A)\geq \lambda ^{*}(A\cap E)+\lambda ^{*}(A\smallsetminus E)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8456d76b7044d837c7e075a28beb17530efa4768)
Since A was arbitrary, this shows that E splits every set cleanly, and therefore E is measurable.
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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
![{\displaystyle {\begin{aligned}\bigcap _{k=1}^{\infty }E_{k}&=\left(\bigcup _{k=1}^{\infty }E_{k}^{c}\right)^{c}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53466b017d840b50bf0b0b6dca1d1d3804436c4e)
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.
Theorem: Length-measure Is Countably Additive.
Let
be a disjoint sequence of measurable sets. Then
Proof:
By subadditivity,
![{\displaystyle \lambda \left(\bigsqcup _{k=1}^{\infty }E_{k}\right)\leq \sum _{k=1}^{\infty }\lambda (E_{k})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/62169ec9d7675f3750b78021dae9ef7befa21054)
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
.
![{\displaystyle {\begin{aligned}\lambda \left(\bigsqcup _{k=1}^{\infty }E_{k}\right)&\geq \lambda \left(\bigsqcup _{k=1}^{n}E_{k}\right)\\&=\sum _{k=1}^{n}\lambda (E_{k})\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8609aec48f9270e5e9638df12e645fdbe9562a2a)
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,
![{\displaystyle {\begin{aligned}\lambda (F)&=\lambda (F\cap E)+\lambda (F\smallsetminus E)\\&=\lambda (E)+\lambda (F\smallsetminus E)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7324e0cb45183b5cd15efaa888dd18ef8a68d025)
Whence
![{\displaystyle \lambda (F\smallsetminus E)=\lambda (F)-\lambda (E)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d5ec7e852ea8849465a5b3e6e904d9b00751534)
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,
![{\displaystyle {\begin{aligned}\lambda \left(\bigcup _{k=1}^{\infty }E_{k}\right)&=\lambda \left(\bigsqcup _{k=1}F_{k}\right)\\&=\sum _{k=1}^{\infty }\lambda (F_{k})\\&=\lim _{n\to \infty }\sum _{k=1}^{n}\lambda (F_{k})\\&=\lim _{n\to \infty }\sum _{k=1}^{n}(\lambda (E_{k})-\lambda (E_{k-1}))\quad {\text{ by excision}}\\&=\lim _{n\to \infty }\lambda (E_{n})\quad {\text{ by telescoping sum}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3cb1e3fb0e0d14cc86a501a23fd7173e9d2beeb)
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,
![{\displaystyle {\begin{aligned}\lim _{n\to \infty }\lambda (F_{n})&=\lim _{n\to \infty }(\lambda (E_{1})-\lambda (E_{n}))\\&=\lambda (E_{1})-\lim _{n\to \infty }\lambda (E_{n})\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1d7853796403824dd9ed88a39a904af153552e6)
and
![{\displaystyle {\begin{aligned}\lambda \left(\bigcup _{k=1}^{\infty }F_{n}\right)&=\lambda \left(E_{1}\smallsetminus \bigcap _{k=1}^{\infty }E_{k}\right)\\&=\lambda (E_{1})-\lambda \left(\bigcap _{k=1}^{\infty }E_{k}\right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2fde2fa6acf417dd50c5b5e89fed04c7742c868b)
By upward continuity of measure, therefore the two objects above are equal and therefore
.
Whence
![{\displaystyle \lim _{n\to \infty }\lambda (E_{n})=\lambda \left(\bigcap _{k=1}^{\infty }E_{k}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dde9f29f43cf223586fe168e9b68b64d16553178)