# Measure Theory/Section 1 Proofs, Measure
## Section 1 Proofs, Measure

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

Proof:
Assume that there is a function $\mu :{\mathcal {P}}({\mathbb {R}})\to {\mathbb {R}}^{*}$ which has the properties of nonnegativity, interval length, translation-invariant, and countably additive.
This implies a contradiction.
* Let $x\in (0,1)$. Then $x-x=0\in {\mathbb {Q}}$ and therefore $x\sim x$ by definition.
* Let $x\sim y$ so $x,y\in (0,1)$ and $x-y\in {\mathbb {Q}}$ by definition.
* Then $y-x=-(x-y)\in {\mathbb {Q}}$ by the closure of ${\mathbb {Q}}$ under multiplication.
* Then $y\sim x$ by definition.
* Let $x\sim y$ and $y\sim z$ so that by definition $x-y,y-z\in {\mathbb {Q}}$ and all three are in (0,1).
* Then $x-z=(x-y)+(y-z)\in {\mathbb {Q}}$ by the closure of ${\mathbb {Q}}$ under addition.
* Then by definition $x\sim z$.
* Let $x=f_{1}+r_{i}$ and $x=f_{2}+r_{j}$.
* Then $f_{2}-f_{1}=r_{i}-r_{j}\in {\mathbb {Q}}$ by the closure of rational numbers under subtraction.
* Then $f_{1}\sim f_{2}$ and therefore $[f_{1}]=[f_{2}]$.
* Because F contains unique representatives of each cell of the partition P, then $f_{1}=f_{2}$.
* Returning to an earlier equation, this implies $f_{2}-f_{1}=0=r_{i}-r_{j}$
* Then $r_{i}=r_{j}$ and therefore since $r_{1},r_{2},\dots$ is an enumeration, $i=j$.
* $(0,1)\subseteq V$
* Let $x\in (0,1)$.
* Let $a\in F$ such that $x\in [a]$ and therefore $x-a\in {\mathbb {Q}}$.
* Also $x-a\in (-1,2)$ since both $x,a\in (0,1)$.
* Therefore there is some $r_{i}$ in the enumeration $r_{1},r_{2},\dots$, such that $r_{i}=x-a$.
* Then $x\in F+r_{i}\subseteq V$.
* $V\subseteq (-1,2)$
* Let $x\in V$, and therefore there is some $r_{i}$ in the enumeration $x\in F+r_{i}$, and then there is some $a\in F$ such that $x=a+r_{i}$.
* Since both $0<a<1$ and $-1<r_{i}<2$ then $x=a+r_{i}\in (-1,2)$.
* Let $A\subseteq B\subseteq {\mathbb {R}}$.
* Then $B=A\sqcup (B\smallsetminus A)$.
* By countable additivity $\mu (B)=\mu (A)+\mu (B\smallsetminus A)$.
* By the nonnegativity of $\mu$, $\mu (B)\geq \mu (A)$
$\Box$

### Lesson 2: Outer Measure Properties

Proof:
$\lambda ^{*}(A)$ always exists as an extended real number, and $0\leq \lambda ^{*}(A)$.
* If ${\mathfrak {P}}$ is any open interval over-approximation of A, and $\sum _{I\in {\mathfrak {P}}}\ell (I)=e$ the corresponding over-estimate, then e is a sum of nonnegative numbers.
* Therefore $0\leq e$.
$\Box$
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 ${\mathfrak {O}}$ be an uncountable over-approximation of A.
Either $\sum _{I\in {\mathfrak {O}}}\ell (I)=\infty$ or there are only countably many nonzero terms, $\ell (I)$.
Because the length of an open interval is always strictly positive, then $\sum _{I\in {\mathfrak {O}}}\ell (I)=\infty$.
But also, $\sum _{I\in \{{\mathbb {R}}\}}\ell (I)=\infty$ is a countable series and is an over-estimate of A.
Therefore if ${\mathcal {E}}$ is the collection of all over-estimates of A, and if ${\mathcal {E}}'\subseteq {\mathcal {E}}$ is the set of over-estimates which are countable series, then ${\mathcal {E}}'={\mathcal {E}}$.
Therefore $\lambda ^{*}(A)=\inf {\mathcal {E}}=\inf {\mathcal {E}}'$.
$\Box$
Proof:
Let $A\subseteq {\mathbb {R}}$ be a set of real numbers, and $c\in {\mathbb {R}}$ any real number.
Let ${\mathfrak {O}}$ be any open interval over-approximation of A.
Define
    ${\mathfrak {O}}+c=\{I+c:I\in {\mathfrak {O}}\}$
Then ${\mathfrak {O}}+c$ is an open interval over-approximation of $A+c$.
* If $I=(a,b)\in {\mathfrak {O}}$ then $I+c=(a+c,b+c)$ is an open interval. Hence all the elements of ${\mathfrak {O}}+c$ are open intervals.
* If $x\in A+c$ then there is some $a\in A$ such that $x=a+c$.
* Then there is some $I\in {\mathfrak {O}}$ such that $a\in I$.
* Then $x=a+c\in I+c$ and $I+c\in {\mathfrak {O}}+c$.
* So ${\mathfrak {O}}+c$ covers A.
Let ${\mathcal {E}}=\left\{\sum _{I\in {\mathfrak {O}}}\ell (I):{\mathfrak {O}}{\text{ is an open interval over-approximation of }}A\right\}$ be the set of all over-estimates of A.
Likewise define ${\mathcal {F}}$ to be the set of all over-estimates of $A+c$.
For every over-estimate $e=\sum _{I\in {\mathfrak {O}}}\ell (I)$ of A, and for every $\ell (I)$, we have that $\ell (I)=\ell (I+c)$.
Therefore $\sum _{I\in {\mathfrak {O}}}\ell (I)=\sum _{I\in {\mathfrak {O}}}\ell (I+c)=\sum _{I+c\in {\mathfrak {O}}+c}\ell (I+c)$.
Therefore $e\in {\mathfrak {O}}+c$ and so ${\mathcal {E}}\subseteq {\mathcal {F}}$.
Mutatis mutandis the same proof shows ${\mathcal {F}}\subseteq {\mathcal {E}}$.
Therefore ${\mathcal {E}}={\mathcal {F}}$ and therefore
    $\lambda ^{*}(A)=\inf {\mathcal {E}}=\inf {\mathcal {F}}=\lambda ^{*}(A+c)$
$\Box$
Proof:
Let $A\subseteq B\subseteq {\mathbb {R}}$.
Any open interval over-approximation of B is immediately also an open interval over-approximation of A, which follows trivially by definitions.
Set ${\mathcal {E}}$ as the set of over-estimates of A and ${\mathcal {F}}$ the over-estimates of B.
Then ${\mathcal {F}}\subseteq {\mathcal {E}}$.
From elementary analysis, therefore
    $\lambda ^{*}(A)=\inf {\mathcal {E}}\leq \inf {\mathcal {F}}=\lambda ^{*}(B)$
$\Box$
If $N=\emptyset$ then we can set the open interval over-approximation ${\mathfrak {O}}=\{\emptyset \}$.
Then
    $\lambda ^{*}(N)=\sum _{I\in \{\emptyset \}}\ell (I)=\ell (\emptyset )=0$.
Therefore, for the rest of the proof, assume $N\neq \emptyset$.
Let $N\subseteq {\mathbb {R}}$ be a countable set, with enumeration $n_{1},n_{2},\dots$.
Let $\varepsilon \in {\mathbb {R}}^{+}$.
Then $\lambda ^{*}(N)\leq \varepsilon$.
* Define the sequence of open intervals by
* $I_{1}=(n_{1}-\varepsilon /2^{2},n_{1}+\varepsilon /2^{2}),$
* $I_{2}=(n_{2}-\varepsilon /2^{3},n_{2}+\varepsilon /2^{3}),$
* $I_{j}=(n_{j}-\varepsilon /2^{j+1},n_{j}+\varepsilon /2^{j+1})$
* If N is non-finite with final indexed element $n_{J}$ then for indices $J<k$ set $I_{k}=\emptyset$.
* Then $\ell (I_{j})\leq \varepsilon /2^{j}$.
* Then as a geometric sum
* $\sum _{j=1}^{\infty }\ell (I_{j})\leq \sum _{j=1}^{\infty }\varepsilon /2^{j}={\frac {1}{2}}\cdot \varepsilon \left({\frac {1}{1-1/2}}\right)=\varepsilon$
* From an elementary argument it is clear that $\{I_{j}\}_{j=1}^{\infty }$ is an open interval over-approximation of N.
* Therefore by definition as an infimum, $\lambda ^{*}(N)\leq \sum _{j=1}^{\infty }\ell (I_{j})=\varepsilon$.
Therefore $\lambda ^{*}(A)=0$.
$\Box$

### Lesson 3: Outer Measure Interval Length

Proof:
Let $I\subseteq {\mathbb {R}}$ be any interval with end-points a and b. Note that the bounds are extended real numbers, so a may be $-\infty$ and b may be $\infty$.
Note that if $a=b$ 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 $a<b$.
If $I=\subseteq {\mathbb {R}}$ is any closed, bounded interval of real numbers, then $\lambda ^{*}(I)=\ell (I)$.
* Define the open interval over-approximation ${\mathfrak {O}}=\{(a-\varepsilon /2,b+\varepsilon /2)\}$.
* Then
* $\sum _{J\in {\mathfrak {O}}}\ell (J)=b-a-\varepsilon$
* Let $e=\sum _{J\in {\mathfrak {O}}}\ell (J)$ be any over-estimate of I.
* Let ${\mathfrak {O}}'\subseteq {\mathfrak {O}}$ be a finite subcover of I, which must exist because [a,b] is a compact set and ${\mathfrak {O}}$ an open cover.
* Let ${\mathfrak {O}}''\subseteq {\mathfrak {O}}'$ be the subset which results from successively removing from ${\mathfrak {O}}'$ any interval which is a subset of some other interval.
* ${\mathfrak {O}}''$ is still a cover of I.
* List the elements ${\mathfrak {O}}''=\{(a_{1},b_{1}),(a_{2},b_{2}),\dots ,(a_{n},b_{n})\}$ in increasing order of the left end-point. $a_{i}\leq a_{i+1}$ for $1\leq i\leq n$.
* For any $1\leq i\leq n$, if $a_{i}=a_{i+1}$ then either $(a_{i},b_{i})\subseteq (a_{i+1},b_{i+1})$ or $(a_{i+1},b_{i+1})\subseteq (a_{i},b_{i})$.
* Either case is impossible because no interval can be a subset of any other, hence we must have the strict inequality $a_{i}<a_{i+1}$.
* Because a is covered by the set, then it is in one of the intervals. Whichever one it is, we must then have $a_{1}<a$.
* Mutatis mutandis, the same argument shows $b<b_{n}$.
* For each $1\leq i\leq n$, if $b_{i}\leq a_{i+1}$ then $b_{i}$ is not covered by any interval in ${\mathfrak {O}}''$. Therefore $a_{i+1}<b_{i}$.
* Then by merely shifting parentheses for regrouping terms,
* ${\begin{aligned}\sum _{j=1}^{n}\ell (I_{j})&=(b_{1}-a_{1})+\cdots +(b_{n}-a_{n})\\&=-a_{1}+\overbrace {(b_{1}-a_{2})} ^{>0}+\overbrace {(b_{2}-a_{3})} ^{>0}+\cdots +\overbrace {(b_{n-1}-a_{n})} ^{>0}+b_{n}\\&>-a+0+\cdots +0+b\\&=b-a\end{aligned}}$
* where the inequalities indicated by curly braces are due to $a_{i+1}<b_{i}$. The inequalities $-a_{1}>-a$ and $b_{n}>b$ are due to earlier inequalities.
* Also
* ${\begin{aligned}e&=\sum _{J\in {\mathfrak {O}}}\ell (J)\\&\geq \sum _{J\in {\mathfrak {O}}'}\ell (J)\\&\geq \sum _{J\in {\mathfrak {O}}''}\ell (J)\end{aligned}}$
* because each series is of nonnegative terms, and contains a subsequence of the terms in the series before it.
* Combining all of the above,
* $b-a\leq e$
If $I=(a,b)$ is a bounded open interval then $\lambda ^{*}(I)=b-a$.
* By monotonicity $\lambda ^{*}(I)\leq \lambda ([a,b])=b-a$.
* Let $\varepsilon \in {\mathbb {R}}^{+}$ be any positive real number less than ${\frac {b-a}{2}}$.
* Then by monotonicity,
* $\lambda ^{*}([a+\varepsilon /2,b-\varepsilon /2])=b-a-\varepsilon \leq \lambda ^{*}(I)$
* Letting $\varepsilon \to 0$ we have
* $b-a\leq \lambda ^{*}(I)$
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
    $(a,b)\subseteq I\subseteq [a,b]$
so that by monotonicity
    $b-a\leq \lambda ^{*}(I)\leq b-a$
and so $\lambda ^{*}(I)=b-a$.
If I is any interval unbounded on the right but finite on the left, then for every $n\in {\mathbb {N}}^{+}$ we have $(a,n)\subseteq I$.
By monotonicity
    $\lambda ^{*}((a,n))=n-a\leq \lambda ^{*}(I)\to \infty$ as $n\to \infty$
Mutatis mutandis the same argument shows that if I is finite on the right and unbounded on the left, or if $I={\mathbb {R}}$, then $\lambda ^{*}(I)=\infty$.
In every case, therefore, $\lambda ^{*}(I)=\ell (I)$.
$\Box$

### Lesson 4: Outer Measure Subadditivity

Proof:
In earlier theorems we have proved that $\lambda ^{*}$ satisfies the properties Nonnegativity, Interval length, and Translation invariance.
We also know from an earlier theorem that no function, defined on ${\mathcal {P}}({\mathbb {R}})$, has all four properties, Nonnegativity, Interval length, Translation invariance, and Countable additivity.
Since $\lambda ^{*}$ is defined on ${\mathcal {P}}({\mathbb {R}})$ then it cannot be countably additive.
$\Box$
Proof:
Suppose that for some $i\in {\mathbb {N}}^{+}$ we have $\lambda ^{*}(A_{i})=\infty$.
The sum of $\infty$ with any nonnegative number is again $\infty$.
Therefore, regardless of the value of the left-hand side, we have
    $\lambda ^{*}\left(\bigcup _{k=1}^{\infty }A_{k}\right)\leq \infty =\sum _{k=1}^{\infty }\lambda ^{*}(A_{k})$
Therefore throughout the rest of the proof, assume that each $A_{i}$ has finite outer measure.
Let $\varepsilon \in {\mathbb {R}}^{+}$ be any positive real number.
For each $1\leq k$ there is an open interval over-approximation of $A_{k}$, call it ${\mathfrak {O}}_{k}$, such that
    $\sum _{I\in {\mathfrak {O}}_{k}}\ell (I)<\lambda ^{*}(A_{k})+\varepsilon /2^{k}$
Then ${\mathfrak {O}}=\bigcup _{k=1}^{\infty }{\mathfrak {O}}_{k}$ is an open interval over-approximation of $\bigcup _{k=1}^{\infty }A_{k}$.
Also,
    ${\begin{aligned}\sum _{I\in {\mathfrak {O}}}\ell (I)&=\sum _{k\in {\mathbb {N}}^{+}}\sum _{I\in {\mathfrak {O}}_{k}}\ell (I)\\&<\sum _{k\in {\mathbb {N}}^{+}}(\lambda ^{*}(A_{k})+\varepsilon /2^{k})\\&=\sum _{k=1}^{\infty }\lambda ^{*}(A_{k})+\varepsilon \end{aligned}}$
Therefore by definition of the infimum,
    $\lambda ^{*}\left(\bigcup _{k=1}^{\infty }I_{k}\right)<\sum _{k=1}^{\infty }\lambda ^{*}(A_{k})+\varepsilon$
Letting $\varepsilon \to 0$
    $\lambda ^{*}\left(\bigcup _{k=1}^{\infty }I_{k}\right)\leq \sum _{k=1}^{\infty }\lambda ^{*}(A_{k})$
$\Box$
Proof:
By monotonicity, and the fact that $\lambda ^{*}(E)=0$,
    $\lambda ^{*}(A)\leq \lambda ^{*}(A\cup E)\leq \lambda ^{*}(A)+\lambda ^{*}(E)=\lambda ^{*}(A)$
Therefore $\lambda ^{*}(A)=\lambda ^{*}(A\cup E)$.
From this result,
    $\lambda ^{*}(A\smallsetminus E)=\lambda ^{*}([A\smallsetminus E]\sqcup E)=\lambda ^{*}(A\cup E)$
$\Box$

### Lesson 5: Measurable Sets

Proof:
Let $E\subseteq {\mathbb {R}}$ be a null set.
Let $A\subseteq {\mathbb {R}}$ be any set.
By monotonicity $\lambda ^{*}(A\cap E)\leq \lambda ^{*}(E)=0$ therefore
    $\lambda ^{*}(A\cap E)=0$
By a previous result, because E is null, $\lambda ^{*}(A\smallsetminus E)=\lambda ^{*}(A)$.
Therefore
    $\lambda ^{*}(A\cap E)+\lambda ^{*}(A\smallsetminus E)=\lambda ^{*}(A)$
so E splits A cleanly.
Since A was arbitrary, therefore E is measurable.
$\Box$
Proof:
Let $E\subseteq {\mathbb {R}}$ be a measurable set.
Let $A\subseteq {\mathbb {R}}$ be any set of real numbers.
Then by definition of the set difference,
    ${\begin{aligned}\lambda ^{*}(A)&=\lambda ^{*}(A\cap E)+\lambda ^{*}(A\smallsetminus E)\\&=\lambda ^{*}(A\smallsetminus E^{c})+\lambda ^{*}(A\cap E^{c})\end{aligned}}$
So $E^{c}$ splits A cleanly, and since A was arbitrary, then $E^{c}$ is measurable.
$\Box$
Proof:
Let $a\in {\mathbb {R}}$, and let $A\subseteq {\mathbb {R}}$ be any set of real numbers. Let $I=(a,\infty )$ be the open ray to the right of a.
If $a\notin A$ then I splits A cleanly.
* Let $J=(x,y)$ where the end-points $x\leq y$ are real numbers.
* If $a<x$ then $J\cap I=J$ and $J\cap (-\infty ,a)=\emptyset$ and then $\ell (J\cap I)=y-x$ and $\ell (J\cap (-\infty ,a))=0$.
* Then
* $\ell (J)=\ell (J\cap I)+\ell (J\cap (-\infty ,a))$
* Mutatis mutandis the same proof handles the case where $y<a$.
* If $x<a<y$ then $J\cap I=(a,y)$ and $J\cap (-\infty ,a)=(x,a)$. Then $\ell (J\cap I)=y-a$ and $\ell (J\cap (-\infty ,a))=a-x$.
* $\ell (J)=y-x=(y-a)+(a-x)=\ell (J\cap (-\infty ,a))$
* So we have shown in all cases that $\ell (J)=\ell (J\cap I)+\ell (J\cap (-\infty ,a))$.
If $a\in A$ then I splits A cleanly.
* Because $\{a\}$ is countable, therefore it is a null set, due to a theorem from an earlier lesson.
* Therefore we may union and subtract $\{a\}$ from a set without changing its outer measure.
* Denote $A\smallsetminus \{a\}=B$.
* Therefore, by all of the above,
* ${\begin{aligned}\lambda ^{*}(A)&=\lambda ^{*}(B)\\&=\lambda ^{*}(B\cap I)+\lambda ^{*}(B\smallsetminus I)\\&=\lambda ^{*}(A\cap I)+\lambda ^{*}(A\smallsetminus I)\end{aligned}}$
* Since I splits A cleanly in all cases, and A was arbitrary, then I is measurable.
$\Box$

### Lesson 6: Measurable Sets Are a Sigma-algebra

Proof:
From two earlier theorems, $\emptyset \in {\mathcal {M}}$ because countable sets are null and null sets are measurable.
Also from an earlier theorem, ${\mathcal {M}}$ is closed under taking complements.
If $E,F\in {\mathcal {M}}$ are two measurable sets then their union is measurable, $E\cup F\in {\mathcal {M}}$.
* By the measurability of E followed by the measurability of F, followed by elementary set theory,
* ${\begin{aligned}\lambda ^{*}(A)&=\lambda ^{*}(A\cap E)+\lambda ^{*}(A\smallsetminus E)\\&=\lambda ^{*}(A\cap E)+\lambda ^{*}([A\smallsetminus E]\cap F)+\lambda ^{*}([A\smallsetminus E]\smallsetminus F)\\&=\lambda ^{*}(A\cap E)+\lambda ^{*}(A\cap E_{01})+\lambda ^{*}(A\cap E_{00})\end{aligned}}$
* By monotonicity and elementary set theory,
* ${\begin{aligned}\lambda ^{*}(A\cap E)+\lambda ^{*}(A\cap E_{01})&\geq \lambda ^{*}([A\cap E]\sqcup [A\cap E_{01}])\\&=\lambda ^{*}(A\cap [E\sqcup E_{01}])\end{aligned}}$
* With the further observation by elementary set theory that $A\cap E_{00}=A\smallsetminus (E\cup F)$ and $E\cup E_{01}=E\cup F$, then all of the above now shows
* ${\begin{aligned}\lambda ^{*}(A)&\geq \lambda ^{*}(A\cap [E\sqcup E_{01}])+\lambda ^{*}(A\cap E_{00})\\&=\lambda ^{*}(A\cap [E\cup F])+\lambda ^{*}(A\smallsetminus [E\cup F])\end{aligned}}$
If $E_{1},E_{2},\dots ,E_{n}\in {\mathcal {M}}$ is any finite sequence of measurable sets then their union is measurable, $\bigcup _{k=1}^{n}E_{k}\in {\mathcal {M}}$
* If $n=1$ the case is trivial.
* Suppose the theorem is true for some $n\in \mathbb {N} ^{+}$, and consider the case for $n+1$ sets.
* Then $\bigcup _{k=1}^{n}E_{k}\in {\mathcal {M}}$.
* Then by the result which we proved above for two sets,
* $\bigcup _{k=1}^{n+1}E_{k}=\bigcup _{k=1}^{n}E_{k}\cup E_{n+1}\in {\mathcal {M}}$
* Thus the proof by induction is complete.
If $E_{1},E_{2},\dots \in {\mathcal {M}}$ is any countable sequence of measurable sets then $\bigcup _{k=1}^{\infty }E_{k}\in {\mathcal {M}}$.
* Because H is measurable then
* ${\begin{aligned}\lambda ^{*}(B\cap [G\sqcup H])&=\lambda ^{*}(B\cap [G\sqcup H]\cap H)+\lambda ^{*}(B\cap [G\sqcup H]\smallsetminus H)\\&=\lambda ^{*}(B\cap H)+\lambda ^{*}(B\cap G)\end{aligned}}$
* The case for $n=1$ is trivial.
* Suppose the claim holds for some $1\leq n$ and now consider the claim for $n+1$ sets.
* ${\begin{aligned}\lambda ^{*}\left(B\cap \bigsqcup _{k=1}^{n+1}G_{k}\right)&=\lambda ^{*}\left(\left[B\cap \bigsqcup _{k=1}^{n}G_{k}\right]\sqcup \left[B\cap G_{n+1}\right]\right)\\&=\lambda ^{*}\left(B\cap \bigsqcup _{k=1}^{n}G_{k}\right)+\lambda ^{*}(B\cap G_{n+1})\quad {\text{ by the result for pairs}}\\&=\sum _{k=1}^{n}\lambda ^{*}(B\cap G_{k})+\lambda ^{*}(B\cap G_{n+1})\\&=\sum _{k=1}^{n+1}\lambda ^{*}(B\cap G_{k})\end{aligned}}$
The above shows that
* $\emptyset \in {\mathcal {M}}$.
* ${\mathcal {M}}$ is closed under complement and countable unions.
Therefore ${\mathcal {M}}$ is a $\sigma$-algebra.
$\square$
Proof:
By closure under complements, $X=\emptyset ^{c}\in {\mathcal {S}}$.
By de Morgan's law, if $E_{1},E_{2},\dots \in {\mathcal {S}}$ then
    ${\begin{aligned}\bigcap _{k=1}^{\infty }E_{k}&=\left(\bigcup _{k=1}^{\infty }E_{k}^{c}\right)^{c}\end{aligned}}$
By the closure under complements, each $E_{k}^{c}$ is measurable.
By closure under unions, $\bigcup _{k=1}^{\infty }E_{k}^{c}$ is measurable.
By closure under complements, $\left(\bigcup _{k=1}^{\infty }E_{k}^{c}\right)^{c}$ is measurable.
$\Box$
Proof:
From an earlier result, all open rays to the right, $(a,\infty )\subseteq {\mathbb {R}}$, are measurable.
Let $I=(-\infty ,a]\subseteq {\mathbb {R}}$ be any closed ray to the left.
Then $I=(a,\infty )^{c}$ is the complement of an open ray to the right.
By the closure of measurable sets under complements, therefore I is measurable.
Now let $J=(a,b]\subseteq {\mathbb {R}}$ a bounded interval open on the left, for $a<b$ two real numbers.
Then $J=(-\infty ,b]\cap (a,\infty )$, and by closure of measurable sets under intersections, then J is measurable.
Now let $K=(a,b)$ be any bounded open interval.
Then define the sequence of intervals open on the left, $(a,b-1/n]$ for $n\in {\mathbb {N}}^{+}$.
Then $K=\bigcup _{n=1}^{\infty }(a,b-1/n]$ and each $(a,b-1/n]$ 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.
$\Box$

### Lesson 8: Properties of Length Measure

Proof:
By subadditivity,
    $\lambda \left(\bigsqcup _{k=1}^{\infty }E_{k}\right)\leq \sum _{k=1}^{\infty }\lambda (E_{k})$
Fix any $n\in {\mathbb {N}}^{+}$.
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 $\emptyset$.
    ${\begin{aligned}\lambda \left(\bigsqcup _{k=1}^{\infty }E_{k}\right)&\geq \lambda \left(\bigsqcup _{k=1}^{n}E_{k}\right)\\&=\sum _{k=1}^{n}\lambda (E_{k})\end{aligned}}$
Taking $n\to \infty$ in the above shows $\lambda \left(\bigsqcup _{k=1}^{\infty }E_{k}\right)\geq \sum _{k=1}^{\infty }\lambda (E_{k})$.
Therefore $\lambda \left(\bigsqcup _{k=1}^{\infty }E_{k}\right)=\sum _{k=1}^{\infty }\lambda (E_{k})$.
$\Box$
Proof:
By monotonicity $\lambda (E)\leq \lambda (F)<\infty$ and hence $\lambda (E)$ is a finite real number. Therefore arithmetic operations are well-defined for $\lambda (E)$ and $\lambda (F)$.
By additivity,
    ${\begin{aligned}\lambda (F)&=\lambda (F\cap E)+\lambda (F\smallsetminus E)\\&=\lambda (E)+\lambda (F\smallsetminus E)\end{aligned}}$
Whence
    $\lambda (F\smallsetminus E)=\lambda (F)-\lambda (E)$
$\Box$
Proof:
Define $F_{1}=E_{1}$ and for $2\leq n$, then $F_{n}=E_{n}\smallsetminus E_{n-1}$.
Then $\langle F_{n}\rangle$ is a sequence of disjoint measurable sets, and $\bigcup _{k=1}^{\infty }E_{k}=\bigsqcup _{k=1}^{\infty }F_{k}$.
Therefore by additivity,
    ${\begin{aligned}\lambda \left(\bigcup _{k=1}^{\infty }E_{k}\right)&=\lambda \left(\bigsqcup _{k=1}F_{k}\right)\\&=\sum _{k=1}^{\infty }\lambda (F_{k})\\&=\lim _{n\to \infty }\sum _{k=1}^{n}\lambda (F_{k})\\&=\lim _{n\to \infty }\sum _{k=1}^{n}(\lambda (E_{k})-\lambda (E_{k-1}))\quad {\text{ by excision}}\\&=\lim _{n\to \infty }\lambda (E_{n})\quad {\text{ by telescoping sum}}\end{aligned}}$
$\Box$
Proof:
By monotonicity, every $\lambda (E_{n})<\infty$ and therefore arithmetic operations on these are well-defined.
Define $F_{n}=E_{1}\smallsetminus E_{n}$ for each $n\in {\mathbb {N}}^{+}$.
Then $\langle F_{n}\rangle$ is an increasing sequence of measurable sets, and hence the upward continuity of measure applies to it.
By excision,
    ${\begin{aligned}\lim _{n\to \infty }\lambda (F_{n})&=\lim _{n\to \infty }(\lambda (E_{1})-\lambda (E_{n}))\\&=\lambda (E_{1})-\lim _{n\to \infty }\lambda (E_{n})\end{aligned}}$
and
    ${\begin{aligned}\lambda \left(\bigcup _{k=1}^{\infty }F_{n}\right)&=\lambda \left(E_{1}\smallsetminus \bigcap _{k=1}^{\infty }E_{k}\right)\\&=\lambda (E_{1})-\lambda \left(\bigcap _{k=1}^{\infty }E_{k}\right)\end{aligned}}$
By upward continuity of measure, therefore the two objects above are equal and therefore $\lambda (E_{1})-\lim _{n\to \infty }\lambda (E_{n})=\lambda (E_{1})-\lambda \left(\bigcap _{k=1}^{\infty }E_{k}\right)$.
Whence
    $\lim _{n\to \infty }\lambda (E_{n})=\lambda \left(\bigcap _{k=1}^{\infty }E_{k}\right)$
$\Box$