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
edit
Theorem: There Is No Total Measure of Real Numbers
Let
μ
:
P
(
R
)
→
R
∗
{\displaystyle \mu :{\mathcal {P}}({\mathbb {R}})\to {\mathbb {R}}^{*}}
be any real-valued set function. Then
μ
{\displaystyle \mu }
must not satisfy at least one of the properties: Length measure, nonnegativity, translation-invariance, countable additivity.
Proof :
Assume that there is a function
μ
:
P
(
R
)
→
R
∗
{\displaystyle \mu :{\mathcal {P}}({\mathbb {R}})\to {\mathbb {R}}^{*}}
which has the properties of nonnegativity, interval length, translation-invariant, and countably additive.
This implies a contradiction.
Define the relation
∼
{\displaystyle \sim }
on
(
0
,
1
)
{\displaystyle (0,1)}
by
x
∼
y
{\displaystyle x\sim y}
if
x
−
y
∈
Q
{\displaystyle x-y\in {\mathbb {Q}}}
.
∼
{\displaystyle \sim }
is reflexive.
Let
x
∈
(
0
,
1
)
{\displaystyle x\in (0,1)}
. Then
x
−
x
=
0
∈
Q
{\displaystyle x-x=0\in {\mathbb {Q}}}
and therefore
x
∼
x
{\displaystyle x\sim x}
by definition.
∼
{\displaystyle \sim }
is symmetric.
Let
x
∼
y
{\displaystyle x\sim y}
so
x
,
y
∈
(
0
,
1
)
{\displaystyle x,y\in (0,1)}
and
x
−
y
∈
Q
{\displaystyle x-y\in {\mathbb {Q}}}
by definition.
Then
y
−
x
=
−
(
x
−
y
)
∈
Q
{\displaystyle y-x=-(x-y)\in {\mathbb {Q}}}
by the closure of
Q
{\displaystyle {\mathbb {Q}}}
under multiplication.
Then
y
∼
x
{\displaystyle y\sim x}
by definition.
∼
{\displaystyle \sim }
is transitive.
Let
x
∼
y
{\displaystyle x\sim y}
and
y
∼
z
{\displaystyle y\sim z}
so that by definition
x
−
y
,
y
−
z
∈
Q
{\displaystyle x-y,y-z\in {\mathbb {Q}}}
and all three are in (0,1).
Then
x
−
z
=
(
x
−
y
)
+
(
y
−
z
)
∈
Q
{\displaystyle x-z=(x-y)+(y-z)\in {\mathbb {Q}}}
by the closure of
Q
{\displaystyle {\mathbb {Q}}}
under addition.
Then by definition
x
∼
z
{\displaystyle x\sim z}
.
Because
∼
{\displaystyle \sim }
is reflexive, symmetric, and transitive, therefore it is an equivalence relation.
Because
∼
{\displaystyle \sim }
is an equivalence relation there is a partition P induced by
∼
{\displaystyle \sim }
. If
x
∈
(
0
,
1
)
{\displaystyle x\in (0,1)}
then write the cell of P containing x by
[
x
]
{\displaystyle [x]}
.
Define F to be any arbitrary set with the following property.
x
∈
(
0
,
1
)
{\displaystyle x\in (0,1)}
if and only if there is a unique
a
∈
F
{\displaystyle a\in F}
such that
x
∈
[
a
]
{\displaystyle x\in [a]}
.
Let
r
1
,
r
2
,
…
{\displaystyle r_{1},r_{2},\dots }
be any enumeration of
(
−
1
,
2
)
∩
Q
{\displaystyle (-1,2)\cap {\mathbb {Q}}}
, which must exist because the set is countable.
For any two distinct
1
≤
i
,
j
{\displaystyle 1\leq i,j}
, the sets
F
+
r
i
{\displaystyle F+r_{i}}
and
F
+
r
k
{\displaystyle F+r_{k}}
are disjoint.
Suppose there is some
x
∈
(
F
+
r
i
)
∩
(
F
+
r
j
)
{\displaystyle x\in (F+r_{i})\cap (F+r_{j})}
.
Then
i
=
j
{\displaystyle i=j}
.
Let
x
=
f
1
+
r
i
{\displaystyle x=f_{1}+r_{i}}
and
x
=
f
2
+
r
j
{\displaystyle x=f_{2}+r_{j}}
.
Then
f
2
−
f
1
=
r
i
−
r
j
∈
Q
{\displaystyle f_{2}-f_{1}=r_{i}-r_{j}\in {\mathbb {Q}}}
by the closure of rational numbers under subtraction.
Then
f
1
∼
f
2
{\displaystyle f_{1}\sim f_{2}}
and therefore
[
f
1
]
=
[
f
2
]
{\displaystyle [f_{1}]=[f_{2}]}
.
Because F contains unique representatives of each cell of the partition P , then
f
1
=
f
2
{\displaystyle f_{1}=f_{2}}
.
Returning to an earlier equation, this implies
f
2
−
f
1
=
0
=
r
i
−
r
j
{\displaystyle f_{2}-f_{1}=0=r_{i}-r_{j}}
Then
r
i
=
r
j
{\displaystyle r_{i}=r_{j}}
and therefore since
r
1
,
r
2
,
…
{\displaystyle r_{1},r_{2},\dots }
is an enumeration,
i
=
j
{\displaystyle i=j}
.
Define
V
=
⨆
i
=
1
∞
(
F
+
r
i
)
{\displaystyle V=\bigsqcup _{i=1}^{\infty }(F+r_{i})}
.
Then
(
0
,
1
)
⊆
V
⊆
(
−
1
,
2
)
{\displaystyle (0,1)\subseteq V\subseteq (-1,2)}
.
(
0
,
1
)
⊆
V
{\displaystyle (0,1)\subseteq V}
Let
x
∈
(
0
,
1
)
{\displaystyle x\in (0,1)}
.
Let
a
∈
F
{\displaystyle a\in F}
such that
x
∈
[
a
]
{\displaystyle x\in [a]}
and therefore
x
−
a
∈
Q
{\displaystyle x-a\in {\mathbb {Q}}}
.
Also
x
−
a
∈
(
−
1
,
2
)
{\displaystyle x-a\in (-1,2)}
since both
x
,
a
∈
(
0
,
1
)
{\displaystyle x,a\in (0,1)}
.
Therefore there is some
r
i
{\displaystyle r_{i}}
in the enumeration
r
1
,
r
2
,
…
{\displaystyle r_{1},r_{2},\dots }
, such that
r
i
=
x
−
a
{\displaystyle r_{i}=x-a}
.
Then
x
∈
F
+
r
i
⊆
V
{\displaystyle x\in F+r_{i}\subseteq V}
.
μ
{\displaystyle \mu }
is monotonic because it is countably additive and monotonic.
Let
A
⊆
B
⊆
R
{\displaystyle A\subseteq B\subseteq {\mathbb {R}}}
.
Then
B
=
A
⊔
(
B
∖
A
)
{\displaystyle B=A\sqcup (B\smallsetminus A)}
.
By countable additivity
μ
(
B
)
=
μ
(
A
)
+
μ
(
B
∖
A
)
{\displaystyle \mu (B)=\mu (A)+\mu (B\smallsetminus A)}
.
By the nonnegativity of
μ
{\displaystyle \mu }
,
μ
(
B
)
≥
μ
(
A
)
{\displaystyle \mu (B)\geq \mu (A)}
By monotonicity and the interval length property,
μ
(
(
0
,
1
)
)
=
1
≤
μ
(
V
)
≤
μ
(
(
−
1
,
2
)
)
=
3
{\displaystyle \mu ((0,1))=1\leq \mu (V)\leq \mu ((-1,2))=3}
By countable additivity and the disjointness of
F
+
r
1
,
F
+
r
2
,
…
{\displaystyle F+r_{1},F+r_{2},\dots }
,
μ
(
V
)
=
∑
i
=
1
∞
μ
(
F
+
r
i
)
{\displaystyle \mu (V)=\sum _{i=1}^{\infty }\mu (F+r_{i})}
By translation-invariance, there is some constant
0
≤
c
{\displaystyle 0\leq c}
such that every
F
+
r
i
=
c
{\displaystyle F+r_{i}=c}
.
By all of the above,
μ
(
V
)
=
∑
i
=
1
∞
c
{\displaystyle \mu (V)=\sum _{i=1}^{\infty }c}
If
c
=
0
{\displaystyle c=0}
then
μ
(
V
)
=
0
{\displaystyle \mu (V)=0}
, and if
0
<
c
{\displaystyle 0<c}
then
μ
(
V
)
=
∞
{\displaystyle \mu (V)=\infty }
from the above.
From all of the above, we have established that
1
≤
μ
(
V
)
≤
3
{\displaystyle 1\leq \mu (V)\leq 3}
μ
(
V
)
=
0
{\displaystyle \mu (V)=0}
or
μ
(
V
)
=
∞
{\displaystyle \mu (V)=\infty }
.
This is a contradiction.
◻
{\displaystyle \Box }
Lesson 2: Outer Measure Properties
edit
Theorem: Outer-measure Is Well-defined and Nonnegative.
For every
A
⊆
R
{\displaystyle A\subseteq {\mathbb {R}}}
the outer-measure takes a unique extended real number value. That is to say
λ
∗
(
A
)
∈
R
∗
{\displaystyle \lambda ^{*}(A)\in {\mathbb {R}}^{*}}
. Also this value is nonnegative,
0
≤
λ
∗
(
A
)
{\displaystyle 0\leq \lambda ^{*}(A)}
.
Proof :
λ
∗
(
A
)
{\displaystyle \lambda ^{*}(A)}
always exists as an extended real number, and
0
≤
λ
∗
(
A
)
{\displaystyle 0\leq \lambda ^{*}(A)}
.
R
{\displaystyle {\mathbb {R}}}
is an open set and
A
⊆
R
{\displaystyle A\subseteq {\mathbb {R}}}
.
Therefore
O
=
{
R
}
{\displaystyle {\mathfrak {O}}=\{{\mathbb {R}}\}}
is an open interval over-approximation of A .
Therefore
∑
I
∈
O
ℓ
(
I
)
=
∞
{\displaystyle \sum _{I\in {\mathfrak {O}}}\ell (I)=\infty }
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
P
{\displaystyle {\mathfrak {P}}}
is any open interval over-approximation of A , and
∑
I
∈
P
ℓ
(
I
)
=
e
{\displaystyle \sum _{I\in {\mathfrak {P}}}\ell (I)=e}
the corresponding over-estimate, then e is a sum of nonnegative numbers.
Therefore
0
≤
e
{\displaystyle 0\leq e}
.
Therefore the infimum over the set of over-estimates of A exists and is at least zero.
◻
{\displaystyle \Box }
Theorem: Outer Measure Is Determined by Countable Sums
For any
A
⊆
R
{\displaystyle A\subseteq {\mathbb {R}}}
,
λ
∗
(
A
)
=
inf
{
∑
I
∈
O
ℓ
(
I
)
:
O
is a disjoint countable open interval over-approximation of
A
}
{\displaystyle \lambda ^{*}(A)=\inf \left\{\sum _{I\in {\mathfrak {O}}}\ell (I):{\mathfrak {O}}{\text{ is a disjoint countable open interval over-approximation of }}A\right\}}
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
O
{\displaystyle {\mathfrak {O}}}
be an uncountable over-approximation of A .
Either
∑
I
∈
O
ℓ
(
I
)
=
∞
{\displaystyle \sum _{I\in {\mathfrak {O}}}\ell (I)=\infty }
or there are only countably many nonzero terms,
ℓ
(
I
)
{\displaystyle \ell (I)}
.
Because the length of an open interval is always strictly positive, then
∑
I
∈
O
ℓ
(
I
)
=
∞
{\displaystyle \sum _{I\in {\mathfrak {O}}}\ell (I)=\infty }
.
But also,
∑
I
∈
{
R
}
ℓ
(
I
)
=
∞
{\displaystyle \sum _{I\in \{{\mathbb {R}}\}}\ell (I)=\infty }
is a countable series and is an over-estimate of A .
Therefore if
E
{\displaystyle {\mathcal {E}}}
is the collection of all over-estimates of A , and if
E
′
⊆
E
{\displaystyle {\mathcal {E}}'\subseteq {\mathcal {E}}}
is the set of over-estimates which are countable series, then
E
′
=
E
{\displaystyle {\mathcal {E}}'={\mathcal {E}}}
.
Therefore
λ
∗
(
A
)
=
inf
E
=
inf
E
′
{\displaystyle \lambda ^{*}(A)=\inf {\mathcal {E}}=\inf {\mathcal {E}}'}
.
◻
{\displaystyle \Box }
Theorem: Outer-measure Is Translation Invariant.
For every subset
A
⊆
R
{\displaystyle A\subseteq {\mathbb {R}}}
and real number
c
∈
R
{\displaystyle c\in {\mathbb {R}}}
the outer-measure of A is invariant under translation by c ,
λ
∗
(
A
)
=
λ
∗
(
A
+
c
)
{\displaystyle \lambda ^{*}(A)=\lambda ^{*}(A+c)}
Proof :
Let
A
⊆
R
{\displaystyle A\subseteq {\mathbb {R}}}
be a set of real numbers, and
c
∈
R
{\displaystyle c\in {\mathbb {R}}}
any real number.
Let
O
{\displaystyle {\mathfrak {O}}}
be any open interval over-approximation of A .
Define
O
+
c
=
{
I
+
c
:
I
∈
O
}
{\displaystyle {\mathfrak {O}}+c=\{I+c:I\in {\mathfrak {O}}\}}
Then
O
+
c
{\displaystyle {\mathfrak {O}}+c}
is an open interval over-approximation of
A
+
c
{\displaystyle A+c}
.
Let
E
=
{
∑
I
∈
O
ℓ
(
I
)
:
O
is an open interval over-approximation of
A
}
{\displaystyle {\mathcal {E}}=\left\{\sum _{I\in {\mathfrak {O}}}\ell (I):{\mathfrak {O}}{\text{ is an open interval over-approximation of }}A\right\}}
be the set of all over-estimates of A .
Likewise define
F
{\displaystyle {\mathcal {F}}}
to be the set of all over-estimates of
A
+
c
{\displaystyle A+c}
.
For every over-estimate
e
=
∑
I
∈
O
ℓ
(
I
)
{\displaystyle e=\sum _{I\in {\mathfrak {O}}}\ell (I)}
of A , and for every
ℓ
(
I
)
{\displaystyle \ell (I)}
, we have that
ℓ
(
I
)
=
ℓ
(
I
+
c
)
{\displaystyle \ell (I)=\ell (I+c)}
.
Therefore
∑
I
∈
O
ℓ
(
I
)
=
∑
I
∈
O
ℓ
(
I
+
c
)
=
∑
I
+
c
∈
O
+
c
ℓ
(
I
+
c
)
{\displaystyle \sum _{I\in {\mathfrak {O}}}\ell (I)=\sum _{I\in {\mathfrak {O}}}\ell (I+c)=\sum _{I+c\in {\mathfrak {O}}+c}\ell (I+c)}
.
Therefore
e
∈
O
+
c
{\displaystyle e\in {\mathfrak {O}}+c}
and so
E
⊆
F
{\displaystyle {\mathcal {E}}\subseteq {\mathcal {F}}}
.
Mutatis mutandis the same proof shows
F
⊆
E
{\displaystyle {\mathcal {F}}\subseteq {\mathcal {E}}}
.
Therefore
E
=
F
{\displaystyle {\mathcal {E}}={\mathcal {F}}}
and therefore
λ
∗
(
A
)
=
inf
E
=
inf
F
=
λ
∗
(
A
+
c
)
{\displaystyle \lambda ^{*}(A)=\inf {\mathcal {E}}=\inf {\mathcal {F}}=\lambda ^{*}(A+c)}
◻
{\displaystyle \Box }
Theorem: Outer-measure Is Monotonic.
If
A
⊆
B
⊆
R
{\displaystyle A\subseteq B\subseteq {\mathbb {R}}}
then
λ
∗
(
A
)
≤
λ
∗
(
B
)
{\displaystyle \lambda ^{*}(A)\leq \lambda ^{*}(B)}
.
Proof :
Let
A
⊆
B
⊆
R
{\displaystyle A\subseteq B\subseteq {\mathbb {R}}}
.
Any open interval over-approximation of B is immediately also an open interval over-approximation of A , which follows trivially by definitions.
Set
E
{\displaystyle {\mathcal {E}}}
as the set of over-estimates of A and
F
{\displaystyle {\mathcal {F}}}
the over-estimates of B .
Then
F
⊆
E
{\displaystyle {\mathcal {F}}\subseteq {\mathcal {E}}}
.
From elementary analysis, therefore
λ
∗
(
A
)
=
inf
E
≤
inf
F
=
λ
∗
(
B
)
{\displaystyle \lambda ^{*}(A)=\inf {\mathcal {E}}\leq \inf {\mathcal {F}}=\lambda ^{*}(B)}
◻
{\displaystyle \Box }
Theorem: Countable Sets Are Null.
If
A
⊆
R
{\displaystyle A\subseteq {\mathbb {R}}}
is a countable set, then
λ
∗
(
A
)
=
0
{\displaystyle \lambda ^{*}(A)=0}
.
If
N
=
∅
{\displaystyle N=\emptyset }
then we can set the open interval over-approximation
O
=
{
∅
}
{\displaystyle {\mathfrak {O}}=\{\emptyset \}}
.
Then
λ
∗
(
N
)
=
∑
I
∈
{
∅
}
ℓ
(
I
)
=
ℓ
(
∅
)
=
0
{\displaystyle \lambda ^{*}(N)=\sum _{I\in \{\emptyset \}}\ell (I)=\ell (\emptyset )=0}
.
Therefore, for the rest of the proof, assume
N
≠
∅
{\displaystyle N\neq \emptyset }
.
Let
N
⊆
R
{\displaystyle N\subseteq {\mathbb {R}}}
be a countable set, with enumeration
n
1
,
n
2
,
…
{\displaystyle n_{1},n_{2},\dots }
.
Let
ε
∈
R
+
{\displaystyle \varepsilon \in {\mathbb {R}}^{+}}
.
Then
λ
∗
(
N
)
≤
ε
{\displaystyle \lambda ^{*}(N)\leq \varepsilon }
.
Define the sequence of open intervals by
I
1
=
(
n
1
−
ε
/
2
2
,
n
1
+
ε
/
2
2
)
,
{\displaystyle I_{1}=(n_{1}-\varepsilon /2^{2},n_{1}+\varepsilon /2^{2}),}
I
2
=
(
n
2
−
ε
/
2
3
,
n
2
+
ε
/
2
3
)
,
{\displaystyle I_{2}=(n_{2}-\varepsilon /2^{3},n_{2}+\varepsilon /2^{3}),}
...
I
j
=
(
n
j
−
ε
/
2
j
+
1
,
n
j
+
ε
/
2
j
+
1
)
{\displaystyle I_{j}=(n_{j}-\varepsilon /2^{j+1},n_{j}+\varepsilon /2^{j+1})}
...
If N is non-finite with final indexed element
n
J
{\displaystyle n_{J}}
then for indices
J
<
k
{\displaystyle J<k}
set
I
k
=
∅
{\displaystyle I_{k}=\emptyset }
.
Then
ℓ
(
I
j
)
≤
ε
/
2
j
{\displaystyle \ell (I_{j})\leq \varepsilon /2^{j}}
.
Then as a geometric sum
∑
j
=
1
∞
ℓ
(
I
j
)
≤
∑
j
=
1
∞
ε
/
2
j
=
1
2
⋅
ε
(
1
1
−
1
/
2
)
=
ε
{\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 }
From an elementary argument it is clear that
{
I
j
}
j
=
1
∞
{\displaystyle \{I_{j}\}_{j=1}^{\infty }}
is an open interval over-approximation of N .
Therefore by definition as an infimum,
λ
∗
(
N
)
≤
∑
j
=
1
∞
ℓ
(
I
j
)
=
ε
{\displaystyle \lambda ^{*}(N)\leq \sum _{j=1}^{\infty }\ell (I_{j})=\varepsilon }
.
Therefore
λ
∗
(
A
)
=
0
{\displaystyle \lambda ^{*}(A)=0}
.
◻
{\displaystyle \Box }
Lesson 3: Outer Measure Interval Length
edit
Theorem: Outer Measure Interval Length.
For every interval
I
⊆
R
{\displaystyle I\subseteq {\mathbb {R}}}
, its outer-measure is its length,
λ
∗
(
I
)
=
ℓ
(
I
)
{\displaystyle \lambda ^{*}(I)=\ell (I)}
.
Proof :
Let
I
⊆
R
{\displaystyle I\subseteq {\mathbb {R}}}
be any interval with end-points a and b . Note that the bounds are extended real numbers, so a may be
−
∞
{\displaystyle -\infty }
and b may be
∞
{\displaystyle \infty }
.
Note that if
a
=
b
{\displaystyle a=b}
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
a
<
b
{\displaystyle a<b}
.
If
I
=⊆
R
{\displaystyle I=\subseteq {\mathbb {R}}}
is any closed, bounded interval of real numbers, then
λ
∗
(
I
)
=
ℓ
(
I
)
{\displaystyle \lambda ^{*}(I)=\ell (I)}
.
Let
ε
∈
R
+
{\displaystyle \varepsilon \in {\mathbb {R}}^{+}}
be a positive real number less than
b
−
a
2
{\displaystyle {\frac {b-a}{2}}}
.
Then
λ
∗
(
I
)
≤
b
−
a
−
ε
{\displaystyle \lambda ^{*}(I)\leq b-a-\varepsilon }
.
Define the open interval over-approximation
O
=
{
(
a
−
ε
/
2
,
b
+
ε
/
2
)
}
{\displaystyle {\mathfrak {O}}=\{(a-\varepsilon /2,b+\varepsilon /2)\}}
.
Then
∑
J
∈
O
ℓ
(
J
)
=
b
−
a
−
ε
{\displaystyle \sum _{J\in {\mathfrak {O}}}\ell (J)=b-a-\varepsilon }
Letting
ε
→
0
{\displaystyle \varepsilon \to 0}
we have
λ
∗
(
I
)
≤
b
−
a
{\displaystyle \lambda ^{*}(I)\leq b-a}
.
Also
b
−
a
{\displaystyle b-a}
is a lower bound on the set of all over-estimates of I .
Let
e
=
∑
J
∈
O
ℓ
(
J
)
{\displaystyle e=\sum _{J\in {\mathfrak {O}}}\ell (J)}
be any over-estimate of I .
Let
O
′
⊆
O
{\displaystyle {\mathfrak {O}}'\subseteq {\mathfrak {O}}}
be a finite subcover of I , which must exist because [a ,b ] is a compact set and
O
{\displaystyle {\mathfrak {O}}}
an open cover.
Let
O
″
⊆
O
′
{\displaystyle {\mathfrak {O}}''\subseteq {\mathfrak {O}}'}
be the subset which results from successively removing from
O
′
{\displaystyle {\mathfrak {O}}'}
any interval which is a subset of some other interval.
O
″
{\displaystyle {\mathfrak {O}}''}
is still a cover of I .
List the elements
O
″
=
{
(
a
1
,
b
1
)
,
(
a
2
,
b
2
)
,
…
,
(
a
n
,
b
n
)
}
{\displaystyle {\mathfrak {O}}''=\{(a_{1},b_{1}),(a_{2},b_{2}),\dots ,(a_{n},b_{n})\}}
in increasing order of the left end-point.
a
i
≤
a
i
+
1
{\displaystyle a_{i}\leq a_{i+1}}
for
1
≤
i
≤
n
{\displaystyle 1\leq i\leq n}
.
For any
1
≤
i
≤
n
{\displaystyle 1\leq i\leq n}
, if
a
i
=
a
i
+
1
{\displaystyle a_{i}=a_{i+1}}
then either
(
a
i
,
b
i
)
⊆
(
a
i
+
1
,
b
i
+
1
)
{\displaystyle (a_{i},b_{i})\subseteq (a_{i+1},b_{i+1})}
or
(
a
i
+
1
,
b
i
+
1
)
⊆
(
a
i
,
b
i
)
{\displaystyle (a_{i+1},b_{i+1})\subseteq (a_{i},b_{i})}
.
Either case is impossible because no interval can be a subset of any other, hence we must have the strict inequality
a
i
<
a
i
+
1
{\displaystyle a_{i}<a_{i+1}}
.
Because a is covered by the set, then it is in one of the intervals. Whichever one it is, we must then have
a
1
<
a
{\displaystyle a_{1}<a}
.
Mutatis mutandis , the same argument shows
b
<
b
n
{\displaystyle b<b_{n}}
.
For each
1
≤
i
≤
n
{\displaystyle 1\leq i\leq n}
, if
b
i
≤
a
i
+
1
{\displaystyle b_{i}\leq a_{i+1}}
then
b
i
{\displaystyle b_{i}}
is not covered by any interval in
O
″
{\displaystyle {\mathfrak {O}}''}
. Therefore
a
i
+
1
<
b
i
{\displaystyle a_{i+1}<b_{i}}
.
Then by merely shifting parentheses for regrouping terms,
∑
j
=
1
n
ℓ
(
I
j
)
=
(
b
1
−
a
1
)
+
⋯
+
(
b
n
−
a
n
)
=
−
a
1
+
(
b
1
−
a
2
)
⏞
>
0
+
(
b
2
−
a
3
)
⏞
>
0
+
⋯
+
(
b
n
−
1
−
a
n
)
⏞
>
0
+
b
n
>
−
a
+
0
+
⋯
+
0
+
b
=
b
−
a
{\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}}}
where the inequalities indicated by curly braces are due to
a
i
+
1
<
b
i
{\displaystyle a_{i+1}<b_{i}}
. The inequalities
−
a
1
>
−
a
{\displaystyle -a_{1}>-a}
and
b
n
>
b
{\displaystyle b_{n}>b}
are due to earlier inequalities.
Also
e
=
∑
J
∈
O
ℓ
(
J
)
≥
∑
J
∈
O
′
ℓ
(
J
)
≥
∑
J
∈
O
″
ℓ
(
J
)
{\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}}}
because each series is of nonnegative terms, and contains a subsequence of the terms in the series before it.
Combining all of the above,
b
−
a
≤
e
{\displaystyle b-a\leq e}
By definition of the outer measure as an infimum, then
b
−
a
≤
λ
∗
(
I
)
{\displaystyle b-a\leq \lambda ^{*}(I)}
Therefore
λ
∗
(
I
)
=
b
−
a
{\displaystyle \lambda ^{*}(I)=b-a}
If
I
=
(
a
,
b
)
{\displaystyle I=(a,b)}
is a bounded open interval then
λ
∗
(
I
)
=
b
−
a
{\displaystyle \lambda ^{*}(I)=b-a}
.
By monotonicity
λ
∗
(
I
)
≤
λ
(
[
a
,
b
]
)
=
b
−
a
{\displaystyle \lambda ^{*}(I)\leq \lambda ([a,b])=b-a}
.
Let
ε
∈
R
+
{\displaystyle \varepsilon \in {\mathbb {R}}^{+}}
be any positive real number less than
b
−
a
2
{\displaystyle {\frac {b-a}{2}}}
.
Then by monotonicity,
λ
∗
(
[
a
+
ε
/
2
,
b
−
ε
/
2
]
)
=
b
−
a
−
ε
≤
λ
∗
(
I
)
{\displaystyle \lambda ^{*}([a+\varepsilon /2,b-\varepsilon /2])=b-a-\varepsilon \leq \lambda ^{*}(I)}
Letting
ε
→
0
{\displaystyle \varepsilon \to 0}
we have
b
−
a
≤
λ
∗
(
I
)
{\displaystyle b-a\leq \lambda ^{*}(I)}
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
(
a
,
b
)
⊆
I
⊆
[
a
,
b
]
{\displaystyle (a,b)\subseteq I\subseteq [a,b]}
so that by monotonicity
b
−
a
≤
λ
∗
(
I
)
≤
b
−
a
{\displaystyle b-a\leq \lambda ^{*}(I)\leq b-a}
and so
λ
∗
(
I
)
=
b
−
a
{\displaystyle \lambda ^{*}(I)=b-a}
.
If I is any interval unbounded on the right but finite on the left, then for every
n
∈
N
+
{\displaystyle n\in {\mathbb {N}}^{+}}
we have
(
a
,
n
)
⊆
I
{\displaystyle (a,n)\subseteq I}
.
By monotonicity
λ
∗
(
(
a
,
n
)
)
=
n
−
a
≤
λ
∗
(
I
)
→
∞
{\displaystyle \lambda ^{*}((a,n))=n-a\leq \lambda ^{*}(I)\to \infty }
as
n
→
∞
{\displaystyle n\to \infty }
Mutatis mutandis the same argument shows that if I is finite on the right and unbounded on the left, or if
I
=
R
{\displaystyle I={\mathbb {R}}}
, then
λ
∗
(
I
)
=
∞
{\displaystyle \lambda ^{*}(I)=\infty }
.
In every case, therefore,
λ
∗
(
I
)
=
ℓ
(
I
)
{\displaystyle \lambda ^{*}(I)=\ell (I)}
.
◻
{\displaystyle \Box }
Lesson 4: Outer Measure Subadditivity
edit
Theorem: Outer-measure not Additive
λ
∗
{\displaystyle \lambda ^{*}}
is not countably additive.
Proof :
In earlier theorems we have proved that
λ
∗
{\displaystyle \lambda ^{*}}
satisfies the properties Nonnegativity , Interval length , and Translation invariance .
We also know from an earlier theorem that no function, defined on
P
(
R
)
{\displaystyle {\mathcal {P}}({\mathbb {R}})}
, has all four properties, Nonnegativity , Interval length , Translation invariance , and Countable additivity .
Since
λ
∗
{\displaystyle \lambda ^{*}}
is defined on
P
(
R
)
{\displaystyle {\mathcal {P}}({\mathbb {R}})}
then it cannot be countably additive.
◻
{\displaystyle \Box }
Theorem: Outer-measure Subadditivity.
Let
A
1
,
A
2
,
⋯
⊆
R
{\displaystyle A_{1},A_{2},\dots \subseteq {\mathbb {R}}}
be any countable sequence of subsets of real numbers. Then
λ
∗
(
⋃
k
=
1
∞
A
k
)
≤
∑
k
=
1
∞
λ
∗
(
A
k
)
{\displaystyle \lambda ^{*}\left(\bigcup _{k=1}^{\infty }A_{k}\right)\leq \sum _{k=1}^{\infty }\lambda ^{*}(A_{k})}
Proof :
Suppose that for some
i
∈
N
+
{\displaystyle i\in {\mathbb {N}}^{+}}
we have
λ
∗
(
A
i
)
=
∞
{\displaystyle \lambda ^{*}(A_{i})=\infty }
.
The sum of
∞
{\displaystyle \infty }
with any nonnegative number is again
∞
{\displaystyle \infty }
.
Therefore, regardless of the value of the left-hand side, we have
λ
∗
(
⋃
k
=
1
∞
A
k
)
≤
∞
=
∑
k
=
1
∞
λ
∗
(
A
k
)
{\displaystyle \lambda ^{*}\left(\bigcup _{k=1}^{\infty }A_{k}\right)\leq \infty =\sum _{k=1}^{\infty }\lambda ^{*}(A_{k})}
Therefore throughout the rest of the proof, assume that each
A
i
{\displaystyle A_{i}}
has finite outer measure.
Let
ε
∈
R
+
{\displaystyle \varepsilon \in {\mathbb {R}}^{+}}
be any positive real number.
For each
1
≤
k
{\displaystyle 1\leq k}
there is an open interval over-approximation of
A
k
{\displaystyle A_{k}}
, call it
O
k
{\displaystyle {\mathfrak {O}}_{k}}
, such that
∑
I
∈
O
k
ℓ
(
I
)
<
λ
∗
(
A
k
)
+
ε
/
2
k
{\displaystyle \sum _{I\in {\mathfrak {O}}_{k}}\ell (I)<\lambda ^{*}(A_{k})+\varepsilon /2^{k}}
Then
O
=
⋃
k
=
1
∞
O
k
{\displaystyle {\mathfrak {O}}=\bigcup _{k=1}^{\infty }{\mathfrak {O}}_{k}}
is an open interval over-approximation of
⋃
k
=
1
∞
A
k
{\displaystyle \bigcup _{k=1}^{\infty }A_{k}}
.
Also,
∑
I
∈
O
ℓ
(
I
)
=
∑
k
∈
N
+
∑
I
∈
O
k
ℓ
(
I
)
<
∑
k
∈
N
+
(
λ
∗
(
A
k
)
+
ε
/
2
k
)
=
∑
k
=
1
∞
λ
∗
(
A
k
)
+
ε
{\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}}}
Therefore by definition of the infimum,
λ
∗
(
⋃
k
=
1
∞
I
k
)
<
∑
k
=
1
∞
λ
∗
(
A
k
)
+
ε
{\displaystyle \lambda ^{*}\left(\bigcup _{k=1}^{\infty }I_{k}\right)<\sum _{k=1}^{\infty }\lambda ^{*}(A_{k})+\varepsilon }
Letting
ε
→
0
{\displaystyle \varepsilon \to 0}
λ
∗
(
⋃
k
=
1
∞
I
k
)
≤
∑
k
=
1
∞
λ
∗
(
A
k
)
{\displaystyle \lambda ^{*}\left(\bigcup _{k=1}^{\infty }I_{k}\right)\leq \sum _{k=1}^{\infty }\lambda ^{*}(A_{k})}
◻
{\displaystyle \Box }
Theorem: Null Adding and Subtracting.
Let
A
,
E
⊆
R
{\displaystyle A,E\subseteq {\mathbb {R}}}
be two subsets and E a null set. Then
λ
∗
(
A
)
=
λ
∗
(
A
∪
E
)
=
λ
∗
(
A
∖
E
)
{\displaystyle \lambda ^{*}(A)=\lambda ^{*}(A\cup E)=\lambda ^{*}(A\smallsetminus E)}
Proof :
By monotonicity, and the fact that
λ
∗
(
E
)
=
0
{\displaystyle \lambda ^{*}(E)=0}
,
λ
∗
(
A
)
≤
λ
∗
(
A
∪
E
)
≤
λ
∗
(
A
)
+
λ
∗
(
E
)
=
λ
∗
(
A
)
{\displaystyle \lambda ^{*}(A)\leq \lambda ^{*}(A\cup E)\leq \lambda ^{*}(A)+\lambda ^{*}(E)=\lambda ^{*}(A)}
Therefore
λ
∗
(
A
)
=
λ
∗
(
A
∪
E
)
{\displaystyle \lambda ^{*}(A)=\lambda ^{*}(A\cup E)}
.
From this result,
λ
∗
(
A
∖
E
)
=
λ
∗
(
[
A
∖
E
]
⊔
E
)
=
λ
∗
(
A
∪
E
)
{\displaystyle \lambda ^{*}(A\smallsetminus E)=\lambda ^{*}([A\smallsetminus E]\sqcup E)=\lambda ^{*}(A\cup E)}
◻
{\displaystyle \Box }
Lesson 5: Measurable Sets
edit
Theorem: Null Sets Are Measurable.
Every null set is measurable.
Proof :
Let
E
⊆
R
{\displaystyle E\subseteq {\mathbb {R}}}
be a null set.
Let
A
⊆
R
{\displaystyle A\subseteq {\mathbb {R}}}
be any set.
By monotonicity
λ
∗
(
A
∩
E
)
≤
λ
∗
(
E
)
=
0
{\displaystyle \lambda ^{*}(A\cap E)\leq \lambda ^{*}(E)=0}
therefore
λ
∗
(
A
∩
E
)
=
0
{\displaystyle \lambda ^{*}(A\cap E)=0}
By a previous result, because E is null,
λ
∗
(
A
∖
E
)
=
λ
∗
(
A
)
{\displaystyle \lambda ^{*}(A\smallsetminus E)=\lambda ^{*}(A)}
.
Therefore
λ
∗
(
A
∩
E
)
+
λ
∗
(
A
∖
E
)
=
λ
∗
(
A
)
{\displaystyle \lambda ^{*}(A\cap E)+\lambda ^{*}(A\smallsetminus E)=\lambda ^{*}(A)}
so E splits A cleanly.
Since A was arbitrary, therefore E is measurable.
◻
{\displaystyle \Box }
Theorem: Measurable Sets Closed Under Complement.
If
E
⊆
R
{\displaystyle E\subseteq {\mathbb {R}}}
is a measurable subset then
E
c
{\displaystyle E^{c}}
is measurable.
Proof :
Let
E
⊆
R
{\displaystyle E\subseteq {\mathbb {R}}}
be a measurable set.
Let
A
⊆
R
{\displaystyle A\subseteq {\mathbb {R}}}
be any set of real numbers.
Then by definition of the set difference,
λ
∗
(
A
)
=
λ
∗
(
A
∩
E
)
+
λ
∗
(
A
∖
E
)
=
λ
∗
(
A
∖
E
c
)
+
λ
∗
(
A
∩
E
c
)
{\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}}}
So
E
c
{\displaystyle E^{c}}
splits A cleanly, and since A was arbitrary, then
E
c
{\displaystyle E^{c}}
is measurable.
◻
{\displaystyle \Box }
Theorem: Open Rays Are Measurable.
For every
a
∈
R
{\displaystyle a\in {\mathbb {R}}}
the open ray
(
a
,
∞
)
{\displaystyle (a,\infty )}
is measurable.
Proof :
Let
a
∈
R
{\displaystyle a\in {\mathbb {R}}}
, and let
A
⊆
R
{\displaystyle A\subseteq {\mathbb {R}}}
be any set of real numbers. Let
I
=
(
a
,
∞
)
{\displaystyle I=(a,\infty )}
be the open ray to the right of a .
If
a
∉
A
{\displaystyle a\notin A}
then I splits A cleanly.
By monotonicity,
λ
∗
(
A
)
≤
λ
∗
(
A
∩
I
)
+
λ
∗
(
A
∖
I
)
{\displaystyle \lambda ^{*}(A)\leq \lambda ^{*}(A\cap I)+\lambda ^{*}(A\smallsetminus I)}
.
If
λ
∗
(
A
)
=
∞
{\displaystyle \lambda ^{*}(A)=\infty }
then regardless of the right-hand side we must have
λ
∗
(
A
)
≥
λ
∗
(
A
∩
I
)
+
λ
∗
(
A
∖
I
)
{\displaystyle \lambda ^{*}(A)\geq \lambda ^{*}(A\cap I)+\lambda ^{*}(A\smallsetminus I)}
.
Therefore for the rest of this proof, assume
λ
∗
(
A
)
{\displaystyle \lambda ^{*}(A)}
is finite.
Let
ε
∈
R
+
{\displaystyle \varepsilon \in {\mathbb {R}}^{+}}
be any positive real number.
Then
λ
∗
(
A
)
+
ε
>
λ
∗
(
A
∩
I
)
+
λ
∗
(
A
∖
I
)
{\displaystyle \lambda ^{*}(A)+\varepsilon >\lambda ^{*}(A\cap I)+\lambda ^{*}(A\smallsetminus I)}
.
By definition as a finite infimum, there is an open interval over-approximation
O
{\displaystyle {\mathfrak {O}}}
of A such that
∑
J
∈
O
ℓ
(
J
)
<
λ
∗
(
A
)
+
ε
{\displaystyle \sum _{J\in {\mathfrak {O}}}\ell (J)<\lambda ^{*}(A)+\varepsilon }
Define
O
l
=
{
J
∩
(
−
∞
,
a
)
:
J
∈
O
}
{\displaystyle {\mathfrak {O}}_{l}=\{J\cap (-\infty ,a):J\in {\mathfrak {O}}\}}
and
O
r
=
{
J
∩
(
a
,
∞
)
:
J
∈
O
}
{\displaystyle {\mathfrak {O}}_{r}=\{J\cap (a,\infty ):J\in {\mathfrak {O}}\}}
.
Then
O
l
{\displaystyle {\mathfrak {O}}_{l}}
and
O
r
{\displaystyle {\mathfrak {O}}_{r}}
are open interval over-approximations of
A
∖
I
{\displaystyle A\smallsetminus I}
and
A
∩
I
{\displaystyle A\cap I}
respectively.
Also
λ
∗
(
A
∩
I
)
+
λ
∗
(
A
∖
I
)
≤
∑
J
∈
O
l
ℓ
(
J
)
+
∑
J
∈
O
r
ℓ
(
J
)
=
∑
J
∈
O
ℓ
(
J
∩
I
)
+
∑
J
∈
O
ℓ
(
J
∩
(
−
∞
,
a
)
)
=
∑
J
∈
O
(
ℓ
(
J
∩
I
)
+
ℓ
(
J
∩
(
−
∞
,
a
)
)
)
{\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}}}
Also
ℓ
(
J
∩
I
)
+
ℓ
(
J
∩
(
−
∞
,
a
)
)
=
ℓ
(
J
)
{\displaystyle \ell (J\cap I)+\ell (J\cap (-\infty ,a))=\ell (J)}
.
Let
J
=
(
x
,
y
)
{\displaystyle J=(x,y)}
where the end-points
x
≤
y
{\displaystyle x\leq y}
are real numbers.
If
a
<
x
{\displaystyle a<x}
then
J
∩
I
=
J
{\displaystyle J\cap I=J}
and
J
∩
(
−
∞
,
a
)
=
∅
{\displaystyle J\cap (-\infty ,a)=\emptyset }
and then
ℓ
(
J
∩
I
)
=
y
−
x
{\displaystyle \ell (J\cap I)=y-x}
and
ℓ
(
J
∩
(
−
∞
,
a
)
)
=
0
{\displaystyle \ell (J\cap (-\infty ,a))=0}
.
Then
ℓ
(
J
)
=
ℓ
(
J
∩
I
)
+
ℓ
(
J
∩
(
−
∞
,
a
)
)
{\displaystyle \ell (J)=\ell (J\cap I)+\ell (J\cap (-\infty ,a))}
Mutatis mutandis the same proof handles the case where
y
<
a
{\displaystyle y<a}
.
If
x
<
a
<
y
{\displaystyle x<a<y}
then
J
∩
I
=
(
a
,
y
)
{\displaystyle J\cap I=(a,y)}
and
J
∩
(
−
∞
,
a
)
=
(
x
,
a
)
{\displaystyle J\cap (-\infty ,a)=(x,a)}
. Then
ℓ
(
J
∩
I
)
=
y
−
a
{\displaystyle \ell (J\cap I)=y-a}
and
ℓ
(
J
∩
(
−
∞
,
a
)
)
=
a
−
x
{\displaystyle \ell (J\cap (-\infty ,a))=a-x}
.
ℓ
(
J
)
=
y
−
x
=
(
y
−
a
)
+
(
a
−
x
)
=
ℓ
(
J
∩
(
−
∞
,
a
)
)
{\displaystyle \ell (J)=y-x=(y-a)+(a-x)=\ell (J\cap (-\infty ,a))}
So we have shown in all cases that
ℓ
(
J
)
=
ℓ
(
J
∩
I
)
+
ℓ
(
J
∩
(
−
∞
,
a
)
)
{\displaystyle \ell (J)=\ell (J\cap I)+\ell (J\cap (-\infty ,a))}
.
Therefore
λ
∗
(
A
∩
I
)
+
λ
∗
(
A
∖
I
)
≤
∑
J
∈
O
ℓ
(
J
)
<
λ
∗
(
A
)
+
ε
{\displaystyle {\begin{aligned}\lambda ^{*}(A\cap I)+\lambda ^{*}(A\smallsetminus I)&\leq \sum _{J\in {\mathfrak {O}}}\ell (J)\\&<\lambda ^{*}(A)+\varepsilon \end{aligned}}}
Letting
ε
→
0
{\displaystyle \varepsilon \to 0}
λ
∗
(
A
)
≥
λ
∗
(
A
∩
I
)
+
λ
∗
(
A
∖
I
)
{\displaystyle \lambda ^{*}(A)\geq \lambda ^{*}(A\cap I)+\lambda ^{*}(A\smallsetminus I)}
Therefore
λ
∗
(
A
)
=
λ
∗
(
A
∩
I
)
+
λ
∗
(
A
∖
I
)
{\displaystyle \lambda ^{*}(A)=\lambda ^{*}(A\cap I)+\lambda ^{*}(A\smallsetminus I)}
.
Therefore I splits A cleanly.
If
a
∈
A
{\displaystyle a\in A}
then I splits A cleanly.
Because
{
a
}
{\displaystyle \{a\}}
is countable, therefore it is a null set, due to a theorem from an earlier lesson.
Therefore we may union and subtract
{
a
}
{\displaystyle \{a\}}
from a set without changing its outer measure.
Denote
A
∖
{
a
}
=
B
{\displaystyle A\smallsetminus \{a\}=B}
.
Therefore, by all of the above,
λ
∗
(
A
)
=
λ
∗
(
B
)
=
λ
∗
(
B
∩
I
)
+
λ
∗
(
B
∖
I
)
=
λ
∗
(
A
∩
I
)
+
λ
∗
(
A
∖
I
)
{\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}}}
Since I splits A cleanly in all cases, and A was arbitrary, then I is measurable.
◻
{\displaystyle \Box }
Lesson 6: Measurable Sets Are a Sigma-algebra
edit
Theorem: The Measurable Sets Form a
σ
{\displaystyle \sigma }
-algebra.
The collection
M
{\displaystyle {\mathcal {M}}}
is a
σ
{\displaystyle \sigma }
-algebra.
Proof :
From two earlier theorems,
∅
∈
M
{\displaystyle \emptyset \in {\mathcal {M}}}
because countable sets are null and null sets are measurable.
Also from an earlier theorem,
M
{\displaystyle {\mathcal {M}}}
is closed under taking complements.
If
E
,
F
∈
M
{\displaystyle E,F\in {\mathcal {M}}}
are two measurable sets then their union is measurable,
E
∪
F
∈
M
{\displaystyle E\cup F\in {\mathcal {M}}}
.
Let
E
∩
F
=
E
11
{\displaystyle E\cap F=E_{11}}
E
∩
F
c
=
E
10
{\displaystyle E\cap F^{c}=E_{10}}
E
c
∩
F
=
E
01
{\displaystyle E^{c}\cap F=E_{01}}
E
c
∩
F
c
=
E
00
{\displaystyle E^{c}\cap F^{c}=E_{00}}
Let
A
⊆
R
{\displaystyle A\subseteq {\mathbb {R}}}
be any set of real numbers.
Then
E
∪
F
{\displaystyle E\cup F}
splits A cleanly.
By the measurability of E followed by the measurability of F , followed by elementary set theory,
λ
∗
(
A
)
=
λ
∗
(
A
∩
E
)
+
λ
∗
(
A
∖
E
)
=
λ
∗
(
A
∩
E
)
+
λ
∗
(
[
A
∖
E
]
∩
F
)
+
λ
∗
(
[
A
∖
E
]
∖
F
)
=
λ
∗
(
A
∩
E
)
+
λ
∗
(
A
∩
E
01
)
+
λ
∗
(
A
∩
E
00
)
{\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}}}
By monotonicity and elementary set theory,
λ
∗
(
A
∩
E
)
+
λ
∗
(
A
∩
E
01
)
≥
λ
∗
(
[
A
∩
E
]
⊔
[
A
∩
E
01
]
)
=
λ
∗
(
A
∩
[
E
⊔
E
01
]
)
{\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}}}
With the further observation by elementary set theory that
A
∩
E
00
=
A
∖
(
E
∪
F
)
{\displaystyle A\cap E_{00}=A\smallsetminus (E\cup F)}
and
E
∪
E
01
=
E
∪
F
{\displaystyle E\cup E_{01}=E\cup F}
, then all of the above now shows
λ
∗
(
A
)
≥
λ
∗
(
A
∩
[
E
⊔
E
01
]
)
+
λ
∗
(
A
∩
E
00
)
=
λ
∗
(
A
∩
[
E
∪
F
]
)
+
λ
∗
(
A
∖
[
E
∪
F
]
)
{\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}}}
Therefore
E
∪
F
∈
M
{\displaystyle E\cup F\in {\mathcal {M}}}
.
If
E
1
,
E
2
,
…
,
E
n
∈
M
{\displaystyle E_{1},E_{2},\dots ,E_{n}\in {\mathcal {M}}}
is any finite sequence of measurable sets then their union is measurable,
⋃
k
=
1
n
E
k
∈
M
{\displaystyle \bigcup _{k=1}^{n}E_{k}\in {\mathcal {M}}}
If
n
=
1
{\displaystyle n=1}
the case is trivial.
Suppose the theorem is true for some
n
∈
N
+
{\displaystyle n\in \mathbb {N} ^{+}}
, and consider the case for
n
+
1
{\displaystyle n+1}
sets.
Then
⋃
k
=
1
n
E
k
∈
M
{\displaystyle \bigcup _{k=1}^{n}E_{k}\in {\mathcal {M}}}
.
Then by the result which we proved above for two sets,
⋃
k
=
1
n
+
1
E
k
=
⋃
k
=
1
n
E
k
∪
E
n
+
1
∈
M
{\displaystyle \bigcup _{k=1}^{n+1}E_{k}=\bigcup _{k=1}^{n}E_{k}\cup E_{n+1}\in {\mathcal {M}}}
Thus the proof by induction is complete.
If
E
1
,
E
2
,
⋯
∈
M
{\displaystyle E_{1},E_{2},\dots \in {\mathcal {M}}}
is any countable sequence of measurable sets then
⋃
k
=
1
∞
E
k
∈
M
{\displaystyle \bigcup _{k=1}^{\infty }E_{k}\in {\mathcal {M}}}
.
Let
E
=
⋃
k
=
1
∞
E
k
{\displaystyle E=\bigcup _{k=1}^{\infty }E_{k}}
.
Let
A
⊆
R
{\displaystyle A\subseteq {\mathbb {R}}}
.
Also let
F
1
=
E
1
{\displaystyle F_{1}=E_{1}}
and for
1
<
k
{\displaystyle 1<k}
define
F
k
=
E
k
∖
⋃
j
=
1
k
−
1
E
j
{\displaystyle F_{k}=E_{k}\smallsetminus \bigcup _{j=1}^{k-1}E_{j}}
so that
⟨
F
k
⟩
{\displaystyle \langle F_{k}\rangle }
forms a sequence of disjoint sets.
Each
F
k
{\displaystyle F_{k}}
is measurable by the result that complements and finite unions of measurable sets are measurable.
Also
E
=
⋃
k
=
1
∞
F
k
{\displaystyle E=\bigcup _{k=1}^{\infty }F_{k}}
.
By subadditivity,
λ
∗
(
A
)
≤
λ
∗
(
A
∩
E
)
+
λ
∗
(
A
∖
E
)
{\displaystyle \lambda ^{*}(A)\leq \lambda ^{*}(A\cap E)+\lambda ^{*}(A\smallsetminus E)}
For each
n
∈
N
+
{\displaystyle n\in {\mathbb {N}}^{+}}
λ
∗
(
A
)
=
λ
∗
(
A
∩
⨆
k
=
1
n
F
k
)
+
λ
∗
(
A
∖
⨆
k
=
1
n
F
k
)
by the earlier result
≥
λ
∗
(
A
∩
⨆
k
=
1
n
F
k
)
+
λ
∗
(
A
∖
⨆
k
=
1
∞
F
k
)
by monotonicity
{\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}}}
Moreover,
λ
∗
(
A
∩
⨆
k
=
1
n
F
k
)
=
∑
k
=
1
n
λ
∗
(
A
∩
F
k
)
{\displaystyle \lambda ^{*}\left(A\cap \bigsqcup _{k=1}^{n}F_{k}\right)=\sum _{k=1}^{n}\lambda ^{*}(A\cap F_{k})}
.
Lemma: Finite Split by Measurables
If
G
1
,
G
2
,
…
,
G
n
∈
M
{\displaystyle G_{1},G_{2},\dots ,G_{n}\in {\mathcal {M}}}
are any finite sequence of disjoint measurable sets and
B
⊆
R
{\displaystyle B\subseteq {\mathbb {R}}}
is any set of real numbers, then
λ
∗
(
B
∩
⨆
k
=
1
n
G
k
)
=
∑
k
=
1
n
λ
∗
(
B
∩
G
k
)
{\displaystyle \lambda ^{*}\left(B\cap \bigsqcup _{k=1}^{n}G_{k}\right)=\sum _{k=1}^{n}\lambda ^{*}(B\cap G_{k})}
Proof :
If
G
,
H
∈
M
{\displaystyle G,H\in {\mathcal {M}}}
are any two disjoint measurable sets and
B
⊆
R
{\displaystyle B\subseteq {\mathbb {R}}}
is any set of real numbers, then
λ
∗
(
B
∩
(
G
⊔
H
)
)
=
λ
∗
(
B
∩
G
)
+
λ
∗
(
B
∩
H
)
{\displaystyle \lambda ^{*}(B\cap (G\sqcup H))=\lambda ^{*}(B\cap G)+\lambda ^{*}(B\cap H)}
.
Because H is measurable then
λ
∗
(
B
∩
[
G
⊔
H
]
)
=
λ
∗
(
B
∩
[
G
⊔
H
]
∩
H
)
+
λ
∗
(
B
∩
[
G
⊔
H
]
∖
H
)
=
λ
∗
(
B
∩
H
)
+
λ
∗
(
B
∩
G
)
{\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}}}
If
G
1
,
G
2
,
…
,
G
n
∈
M
{\displaystyle G_{1},G_{2},\dots ,G_{n}\in {\mathcal {M}}}
is any finite sequence of disjoint measurable sets, and
B
⊆
R
{\displaystyle B\subseteq {\mathbb {R}}}
is any set of real numbers, then
λ
∗
(
B
∩
⨆
k
=
1
n
G
k
)
=
∑
k
=
1
n
λ
∗
(
B
∩
G
k
)
{\displaystyle \lambda ^{*}\left(B\cap \bigsqcup _{k=1}^{n}G_{k}\right)=\sum _{k=1}^{n}\lambda ^{*}(B\cap G_{k})}
The case for
n
=
1
{\displaystyle n=1}
is trivial.
Suppose the claim holds for some
1
≤
n
{\displaystyle 1\leq n}
and now consider the claim for
n
+
1
{\displaystyle n+1}
sets.
λ
∗
(
B
∩
⨆
k
=
1
n
+
1
G
k
)
=
λ
∗
(
[
B
∩
⨆
k
=
1
n
G
k
]
⊔
[
B
∩
G
n
+
1
]
)
=
λ
∗
(
B
∩
⨆
k
=
1
n
G
k
)
+
λ
∗
(
B
∩
G
n
+
1
)
by the result for pairs
=
∑
k
=
1
n
λ
∗
(
B
∩
G
k
)
+
λ
∗
(
B
∩
G
n
+
1
)
=
∑
k
=
1
n
+
1
λ
∗
(
B
∩
G
k
)
{\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}}}
Therefore
λ
∗
(
A
)
≥
∑
k
=
1
n
λ
∗
(
A
∩
F
k
)
+
λ
∗
(
A
∖
E
)
{\displaystyle \lambda ^{*}(A)\geq \sum _{k=1}^{n}\lambda ^{*}(A\cap F_{k})+\lambda ^{*}(A\smallsetminus E)}
Now taking the limit
n
→
∞
{\displaystyle n\to \infty }
on each side of this last inequality,
λ
∗
(
A
)
≥
∑
k
=
1
∞
λ
∗
(
A
∩
F
k
)
+
λ
∗
(
A
∖
E
)
{\displaystyle \lambda ^{*}(A)\geq \sum _{k=1}^{\infty }\lambda ^{*}(A\cap F_{k})+\lambda ^{*}(A\smallsetminus E)}
By subadditivity,
λ
∗
(
A
∩
E
)
=
λ
∗
(
A
∩
⨆
k
=
1
∞
F
k
)
=
λ
∗
(
⨆
k
=
1
∞
(
A
∩
F
k
)
)
≤
∑
k
=
1
∞
λ
∗
(
A
∩
F
k
)
{\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}}}
Putting together results from above,
λ
∗
(
A
)
≥
λ
∗
(
A
∩
E
)
+
λ
∗
(
A
∖
E
)
{\displaystyle \lambda ^{*}(A)\geq \lambda ^{*}(A\cap E)+\lambda ^{*}(A\smallsetminus E)}
Since A was arbitrary, this shows that E splits every set cleanly, and therefore E is measurable.
The above shows that
∅
∈
M
{\displaystyle \emptyset \in {\mathcal {M}}}
.
M
{\displaystyle {\mathcal {M}}}
is closed under complement and countable unions.
Therefore
M
{\displaystyle {\mathcal {M}}}
is a
σ
{\displaystyle \sigma }
-algebra.
◻
{\displaystyle \square }
Theorem:
σ
{\displaystyle \sigma }
-algebra Space and Intersection
Let X be any set and
S
{\displaystyle {\mathcal {S}}}
a
σ
{\displaystyle \sigma }
-algebra on X . Then
X
∈
S
{\displaystyle X\in {\mathcal {S}}}
and
S
{\displaystyle {\mathcal {S}}}
is closed under countable intersections.
Proof :
By closure under complements,
X
=
∅
c
∈
S
{\displaystyle X=\emptyset ^{c}\in {\mathcal {S}}}
.
By de Morgan's law, if
E
1
,
E
2
,
⋯
∈
S
{\displaystyle E_{1},E_{2},\dots \in {\mathcal {S}}}
then
⋂
k
=
1
∞
E
k
=
(
⋃
k
=
1
∞
E
k
c
)
c
{\displaystyle {\begin{aligned}\bigcap _{k=1}^{\infty }E_{k}&=\left(\bigcup _{k=1}^{\infty }E_{k}^{c}\right)^{c}\end{aligned}}}
By the closure under complements, each
E
k
c
{\displaystyle E_{k}^{c}}
is measurable.
By closure under unions,
⋃
k
=
1
∞
E
k
c
{\displaystyle \bigcup _{k=1}^{\infty }E_{k}^{c}}
is measurable.
By closure under complements,
(
⋃
k
=
1
∞
E
k
c
)
c
{\displaystyle \left(\bigcup _{k=1}^{\infty }E_{k}^{c}\right)^{c}}
is measurable.
◻
{\displaystyle \Box }
Theorem: Intervals Are Measurable.
If
I
⊆
R
{\displaystyle I\subseteq {\mathbb {R}}}
is an interval then
I
∈
M
{\displaystyle I\in {\mathcal {M}}}
.
Proof :
From an earlier result, all open rays to the right,
(
a
,
∞
)
⊆
R
{\displaystyle (a,\infty )\subseteq {\mathbb {R}}}
, are measurable.
Let
I
=
(
−
∞
,
a
]
⊆
R
{\displaystyle I=(-\infty ,a]\subseteq {\mathbb {R}}}
be any closed ray to the left.
Then
I
=
(
a
,
∞
)
c
{\displaystyle I=(a,\infty )^{c}}
is the complement of an open ray to the right.
By the closure of measurable sets under complements, therefore I is measurable.
Now let
J
=
(
a
,
b
]
⊆
R
{\displaystyle J=(a,b]\subseteq {\mathbb {R}}}
a bounded interval open on the left, for
a
<
b
{\displaystyle a<b}
two real numbers.
Then
J
=
(
−
∞
,
b
]
∩
(
a
,
∞
)
{\displaystyle J=(-\infty ,b]\cap (a,\infty )}
, and by closure of measurable sets under intersections, then J is measurable.
Now let
K
=
(
a
,
b
)
{\displaystyle K=(a,b)}
be any bounded open interval.
Then define the sequence of intervals open on the left,
(
a
,
b
−
1
/
n
]
{\displaystyle (a,b-1/n]}
for
n
∈
N
+
{\displaystyle n\in {\mathbb {N}}^{+}}
.
Then
K
=
⋃
n
=
1
∞
(
a
,
b
−
1
/
n
]
{\displaystyle K=\bigcup _{n=1}^{\infty }(a,b-1/n]}
and each
(
a
,
b
−
1
/
n
]
{\displaystyle (a,b-1/n]}
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.
◻
{\displaystyle \Box }
Lesson 8: Properties of Length Measure
edit
Theorem: Length-measure Is Countably Additive.
Let
E
1
,
E
2
,
⋯
∈
M
{\displaystyle E_{1},E_{2},\dots \in {\mathcal {M}}}
be a disjoint sequence of measurable sets. Then
λ
(
⨆
k
=
1
∞
E
k
)
=
∑
k
=
1
∞
λ
(
E
k
)
{\displaystyle \lambda \left(\bigsqcup _{k=1}^{\infty }E_{k}\right)=\sum _{k=1}^{\infty }\lambda (E_{k})}
Proof :
By subadditivity,
λ
(
⨆
k
=
1
∞
E
k
)
≤
∑
k
=
1
∞
λ
(
E
k
)
{\displaystyle \lambda \left(\bigsqcup _{k=1}^{\infty }E_{k}\right)\leq \sum _{k=1}^{\infty }\lambda (E_{k})}
Fix any
n
∈
N
+
{\displaystyle n\in {\mathbb {N}}^{+}}
.
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 \emptyset }
.
λ
(
⨆
k
=
1
∞
E
k
)
≥
λ
(
⨆
k
=
1
n
E
k
)
=
∑
k
=
1
n
λ
(
E
k
)
{\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}}}
Taking
n
→
∞
{\displaystyle n\to \infty }
in the above shows
λ
(
⨆
k
=
1
∞
E
k
)
≥
∑
k
=
1
∞
λ
(
E
k
)
{\displaystyle \lambda \left(\bigsqcup _{k=1}^{\infty }E_{k}\right)\geq \sum _{k=1}^{\infty }\lambda (E_{k})}
.
Therefore
λ
(
⨆
k
=
1
∞
E
k
)
=
∑
k
=
1
∞
λ
(
E
k
)
{\displaystyle \lambda \left(\bigsqcup _{k=1}^{\infty }E_{k}\right)=\sum _{k=1}^{\infty }\lambda (E_{k})}
.
◻
{\displaystyle \Box }
Theorem: Length-measure Excision.
Let
E
,
F
∈
M
{\displaystyle E,F\in {\mathcal {M}}}
be two measurable sets, and
E
⊆
F
{\displaystyle E\subseteq F}
, and assume that
λ
(
F
)
<
∞
{\displaystyle \lambda (F)<\infty }
. Then
λ
(
F
∖
E
)
=
λ
(
F
)
−
λ
(
E
)
{\displaystyle \lambda (F\smallsetminus E)=\lambda (F)-\lambda (E)}
.
Proof :
By monotonicity
λ
(
E
)
≤
λ
(
F
)
<
∞
{\displaystyle \lambda (E)\leq \lambda (F)<\infty }
and hence
λ
(
E
)
{\displaystyle \lambda (E)}
is a finite real number. Therefore arithmetic operations are well-defined for
λ
(
E
)
{\displaystyle \lambda (E)}
and
λ
(
F
)
{\displaystyle \lambda (F)}
.
By additivity,
λ
(
F
)
=
λ
(
F
∩
E
)
+
λ
(
F
∖
E
)
=
λ
(
E
)
+
λ
(
F
∖
E
)
{\displaystyle {\begin{aligned}\lambda (F)&=\lambda (F\cap E)+\lambda (F\smallsetminus E)\\&=\lambda (E)+\lambda (F\smallsetminus E)\end{aligned}}}
Whence
λ
(
F
∖
E
)
=
λ
(
F
)
−
λ
(
E
)
{\displaystyle \lambda (F\smallsetminus E)=\lambda (F)-\lambda (E)}
◻
{\displaystyle \Box }
Theorem: Upward Continuity of Measure.
Let
E
1
,
E
2
,
⋯
∈
M
{\displaystyle E_{1},E_{2},\dots \in {\mathcal {M}}}
be an ascending sequence of measurable sets. Then
lim
n
→
∞
λ
(
E
n
)
=
λ
(
⋃
k
=
1
∞
E
k
)
{\displaystyle \lim _{n\to \infty }\lambda (E_{n})=\lambda \left(\bigcup _{k=1}^{\infty }E_{k}\right)}
Proof :
Define
F
1
=
E
1
{\displaystyle F_{1}=E_{1}}
and for
2
≤
n
{\displaystyle 2\leq n}
, then
F
n
=
E
n
∖
E
n
−
1
{\displaystyle F_{n}=E_{n}\smallsetminus E_{n-1}}
.
Then
⟨
F
n
⟩
{\displaystyle \langle F_{n}\rangle }
is a sequence of disjoint measurable sets, and
⋃
k
=
1
∞
E
k
=
⨆
k
=
1
∞
F
k
{\displaystyle \bigcup _{k=1}^{\infty }E_{k}=\bigsqcup _{k=1}^{\infty }F_{k}}
.
Therefore by additivity,
λ
(
⋃
k
=
1
∞
E
k
)
=
λ
(
⨆
k
=
1
F
k
)
=
∑
k
=
1
∞
λ
(
F
k
)
=
lim
n
→
∞
∑
k
=
1
n
λ
(
F
k
)
=
lim
n
→
∞
∑
k
=
1
n
(
λ
(
E
k
)
−
λ
(
E
k
−
1
)
)
by excision
=
lim
n
→
∞
λ
(
E
n
)
by telescoping sum
{\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}}}
◻
{\displaystyle \Box }
Theorem: Downward Continuity of Measure.
Let
E
1
,
E
2
,
⋯
∈
M
{\displaystyle E_{1},E_{2},\dots \in {\mathcal {M}}}
be a descending sequence of measurable sets. If
λ
(
E
1
)
{\displaystyle \lambda (E_{1})}
is finite then
lim
n
→
∞
λ
(
E
m
)
=
λ
(
⋂
k
=
1
∞
E
k
)
{\displaystyle \lim _{n\to \infty }\lambda (E_{m})=\lambda \left(\bigcap _{k=1}^{\infty }E_{k}\right)}
Proof :
By monotonicity, every
λ
(
E
n
)
<
∞
{\displaystyle \lambda (E_{n})<\infty }
and therefore arithmetic operations on these are well-defined.
Define
F
n
=
E
1
∖
E
n
{\displaystyle F_{n}=E_{1}\smallsetminus E_{n}}
for each
n
∈
N
+
{\displaystyle n\in {\mathbb {N}}^{+}}
.
Then
⟨
F
n
⟩
{\displaystyle \langle F_{n}\rangle }
is an increasing sequence of measurable sets, and hence the upward continuity of measure applies to it.
By excision,
lim
n
→
∞
λ
(
F
n
)
=
lim
n
→
∞
(
λ
(
E
1
)
−
λ
(
E
n
)
)
=
λ
(
E
1
)
−
lim
n
→
∞
λ
(
E
n
)
{\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}}}
and
λ
(
⋃
k
=
1
∞
F
n
)
=
λ
(
E
1
∖
⋂
k
=
1
∞
E
k
)
=
λ
(
E
1
)
−
λ
(
⋂
k
=
1
∞
E
k
)
{\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}}}
By upward continuity of measure, therefore the two objects above are equal and therefore
λ
(
E
1
)
−
lim
n
→
∞
λ
(
E
n
)
=
λ
(
E
1
)
−
λ
(
⋂
k
=
1
∞
E
k
)
{\displaystyle \lambda (E_{1})-\lim _{n\to \infty }\lambda (E_{n})=\lambda (E_{1})-\lambda \left(\bigcap _{k=1}^{\infty }E_{k}\right)}
.
Whence
lim
n
→
∞
λ
(
E
n
)
=
λ
(
⋂
k
=
1
∞
E
k
)
{\displaystyle \lim _{n\to \infty }\lambda (E_{n})=\lambda \left(\bigcap _{k=1}^{\infty }E_{k}\right)}
◻
{\displaystyle \Box }