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

Theorem: There Is No Total Measure of Real Numbers

Let $\mu :𝒫(ℝ)\to {ℝ}^{*}$ be any real-valued set function.  Then $\mu$ must not satisfy at least one of the properties: Length measure, nonnegativity, translation-invariance, countable additivity.

Proof:

Assume that there is a function $\mu :𝒫(ℝ)\to {ℝ}^{*}$ which has the properties of nonnegativity, interval length, translation-invariant, and countably additive.

This implies a contradiction. Template:Robelbox Define the relation $\sim$ on $(0,1)$ by $x\sim y$ if $x-y\in \mathbb{Q}$.

$\sim$ is reflexive. Template:Robelbox Let $x\in (0,1)$. Then $x-x=0\in \mathbb{Q}$ and therefore $x\sim x$ by definition. Template:Robelbox/close $\sim$ is symmetric. Template:Robelbox 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. Template:Robelbox/close $\sim$ is transitive. Template:Robelbox 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 $x\sim z$. Template:Robelbox/close

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

Because $\sim$ is an equivalence relation there is a partition P induced by $\sim$. If $x\in (0,1)$ then write the cell of P containing x by $[x]$.

Define F to be any arbitrary set with the following property. $x\in (0,1)$ if and only if there is a unique $a\in F$ such that $x\in [a]$.

Let $r_1,r_2,\dots$ be any enumeration of $(-1,2)\cap \mathbb{Q}$, which must exist because the set is countable.

For any two distinct $1\le i,j$, the sets $F+r_i$ and $F+r_k$ are disjoint. Template:Robelbox Suppose there is some $x\in (F+r_i)\cap (F+r_j)$.

Then $i=j$. Template:Robelbox 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, $i=j$. Template:Robelbox/close

Template:Robelbox/close Define $V=\underset{i=1}{\overset{\mathrm{\infty }}{⨆}}(F+r_i)$.

Then $(0,1)\subseteq V\subseteq (-1,2)$. Template:Robelbox 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 $x\in F+r_i\subseteq V$. Template:Robelbox/close Template:Robelbox 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 $x=a+r_i\in (-1,2)$. Template:Robelbox/close $\mu$ is monotonic because it is countably additive and monotonic. Template:Robelbox Let $A\subseteq B\subseteq ℝ$.

Then $B=A\bigsqcup (B\setminus A)$.

By countable additivity $\mu (B)=\mu (A)+\mu (B\setminus A)$.

By the nonnegativity of $\mu$, $$\mu (B)\ge \mu (A)$$ Template:Robelbox/close By monotonicity and the interval length property, $$\mu ((0,1))=1\le \mu (V)\le \mu ((-1,2))=3$$

By countable additivity and the disjointness of $F+r_1,F+r_2,\dots$, $$\mu (V)=\sum _{i=1}^{\mathrm{\infty }}\mu (F+r_i)$$

By translation-invariance, there is some constant $0\le c$ such that every $F+r_i=c$.

By all of the above, $$\mu (V)=\sum _{i=1}^{\mathrm{\infty }}c$$

If $c=0$ then $\mu (V)=0$, and if $0<c$ then $\mu (V)=\mathrm{\infty }$ from the above.

From all of the above, we have established that

• $1\le \mu (V)\le 3$
• $\mu (V)=0$ or $\mu (V)=\mathrm{\infty }$.

This is a contradiction. Template:Robelbox/close $◻$

Lesson 2: Outer Measure Properties

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

For every $A\subseteq ℝ$ the outer-measure takes a unique extended real number value.  That is to say ${\lambda }^{*}(A)\in {ℝ}^{*}$.  Also this value is nonnegative, $0\le {\lambda }^{*}(A)$.  

Proof:

${\lambda }^{*}(A)$ always exists as an extended real number, and $0\le {\lambda }^{*}(A)$. Template:Robelbox $ℝ$ is an open set and $A\subseteq ℝ$.

Therefore $𝔒=\{ℝ\}$ is an open interval over-approximation of A.

Therefore $\sum _{I\in 𝔒}\ell (I)=\mathrm{\infty }$ 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. Template:Robelbox If $𝔓$ is any open interval over-approximation of A, and $\sum _{I\in 𝔓}\ell (I)=e$ the corresponding over-estimate, then e is a sum of nonnegative numbers.

Therefore $0\le e$. Template:Robelbox/close Therefore the infimum over the set of over-estimates of A exists and is at least zero. Template:Robelbox/close $◻$

Theorem: Outer Measure Is Determined by Countable Sums

For any $A\subseteq ℝ$, 

  ${\lambda }^{*}(A)=\inf \{\sum _{I\in 𝔒}\ell (I):𝔒\text{ is a disjoint countable open interval over-approximation of }A\}$

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 $𝔒$ be an uncountable over-approximation of A.

Either $\sum _{I\in 𝔒}\ell (I)=\mathrm{\infty }$ or there are only countably many nonzero terms, $\ell (I)$.

Because the length of an open interval is always strictly positive, then $\sum _{I\in 𝔒}\ell (I)=\mathrm{\infty }$.

But also, $\sum _{I\in \{ℝ\}}\ell (I)=\mathrm{\infty }$ is a countable series and is an over-estimate of A.

Therefore if $ℰ$ is the collection of all over-estimates of A, and if ${ℰ}^{\prime }\subseteq ℰ$ is the set of over-estimates which are countable series, then ${ℰ}^{\prime }=ℰ$.

Therefore ${\lambda }^{*}(A)=\inf ℰ=\inf {ℰ}^{\prime }$.

$◻$

Theorem: Outer-measure Is Translation Invariant.

For every subset $A\subseteq ℝ$ and real number $c\in ℝ$ the outer-measure of A is invariant under translation by c, 
  ${\lambda }^{*}(A)={\lambda }^{*}(A+c)$

Proof:

Let $A\subseteq ℝ$ be a set of real numbers, and $c\in ℝ$ any real number.

Let $𝔒$ be any open interval over-approximation of A.

Define

$𝔒+c=\{I+c:I\in 𝔒\}$

Then $𝔒+c$ is an open interval over-approximation of $A+c$. Template:Robelbox If $I=(a,b)\in 𝔒$ then $I+c=(a+c,b+c)$ is an open interval. Hence all the elements of $𝔒+c$ are open intervals.

If $x\in A+c$ then there is some $a\in A$ such that $x=a+c$.

Then there is some $I\in 𝔒$ such that $a\in I$.

Then $x=a+c\in I+c$ and $I+c\in 𝔒+c$.

So $𝔒+c$ covers A. Template:Robelbox/close Let $ℰ=\{\sum _{I\in 𝔒}\ell (I):𝔒\text{ is an open interval over-approximation of }A\}$ be the set of all over-estimates of A.

Likewise define $ℱ$ to be the set of all over-estimates of $A+c$.

For every over-estimate $e=\sum _{I\in 𝔒}\ell (I)$ of A, and for every $\ell (I)$, we have that $\ell (I)=\ell (I+c)$.

Therefore $\sum _{I\in 𝔒}\ell (I)=\sum _{I\in 𝔒}\ell (I+c)=\sum _{I+c\in 𝔒+c}\ell (I+c)$.

Therefore $e\in 𝔒+c$ and so $ℰ\subseteq ℱ$.

Mutatis mutandis the same proof shows $ℱ\subseteq ℰ$.

Therefore $ℰ=ℱ$ and therefore

${\lambda }^{*}(A)=\inf ℰ=\inf ℱ={\lambda }^{*}(A+c)$

$◻$

Theorem: Outer-measure Is Monotonic.

If $A\subseteq B\subseteq ℝ$ then ${\lambda }^{*}(A)\le {\lambda }^{*}(B)$.

Proof:

Let $A\subseteq B\subseteq ℝ$.

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

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

Then $ℱ\subseteq ℰ$.

From elementary analysis, therefore

${\lambda }^{*}(A)=\inf ℰ\le \inf ℱ={\lambda }^{*}(B)$

$◻$

Theorem: Countable Sets Are Null.

If $A\subseteq ℝ$ is a countable set, then ${\lambda }^{*}(A)=0$.

If $N=\mathrm{\varnothing }$ then we can set the open interval over-approximation $𝔒=\{\mathrm{\varnothing }\}$.

Then

${\lambda }^{*}(N)=\sum _{I\in \{\mathrm{\varnothing }\}}\ell (I)=\ell (\mathrm{\varnothing })=0$.

Therefore, for the rest of the proof, assume $N\ne \mathrm{\varnothing }$.

Let $N\subseteq ℝ$ be a countable set, with enumeration $n_1,n_2,\dots$.

Let $\epsilon \in {ℝ}^{+}$.

Then ${\lambda }^{*}(N)\le \epsilon$. Template:Robelbox Define the sequence of open intervals by

$I_1=(n_1-\epsilon /2^2,n_1+\epsilon /2^2),$

$I_2=(n_2-\epsilon /2^3,n_2+\epsilon /2^3),$

...

$I_j=(n_j-\epsilon /2^{j+1},n_j+\epsilon /2^{j+1})$

...

If N is non-finite with final indexed element $n_J$ then for indices $J<k$ set $I_k=\mathrm{\varnothing }$.

Then $\ell (I_j)\le \epsilon /2^j$.

Then as a geometric sum

$\sum _{j=1}^{\mathrm{\infty }}\ell (I_j)\le \sum _{j=1}^{\mathrm{\infty }}\epsilon /2^j=\frac{1}{2}\cdot \epsilon \left(\frac{1}{1-1/2}\right)=\epsilon$

From an elementary argument it is clear that $\{I_j{\}}_{j=1}^{\mathrm{\infty }}$ is an open interval over-approximation of N.

Therefore by definition as an infimum, ${\lambda }^{*}(N)\le \sum _{j=1}^{\mathrm{\infty }}\ell (I_j)=\epsilon$. Template:Robelbox/close Therefore ${\lambda }^{*}(A)=0$.

$◻$

Lesson 3: Outer Measure Interval Length

Theorem: Outer Measure Interval Length.

For every interval $I\subseteq ℝ$, its outer-measure is its length, ${\lambda }^{*}(I)=\ell (I)$.

Proof:

Let $I\subseteq ℝ$ be any interval with end-points a and b. Note that the bounds are extended real numbers, so a may be $-\mathrm{\infty }$ and b may be $\mathrm{\infty }$.

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 $I=\subseteq ℝ$ is any closed, bounded interval of real numbers, then ${\lambda }^{*}(I)=\ell (I)$. Template:Robelbox Let $\epsilon \in {ℝ}^{+}$ be a positive real number less than $\frac{b-a}{2}$.

Then ${\lambda }^{*}(I)\le b-a-\epsilon$. Template:Robelbox Define the open interval over-approximation $𝔒=\{(a-\epsilon /2,b+\epsilon /2)\}$.

Then

$\sum _{J\in 𝔒}\ell (J)=b-a-\epsilon$

Template:Robelbox/close Letting $\epsilon \to 0$ we have ${\lambda }^{*}(I)\le b-a$.

Also $b-a$ is a lower bound on the set of all over-estimates of I. Template:Robelbox Let $e=\sum _{J\in 𝔒}\ell (J)$ be any over-estimate of I.

Let ${𝔒}^{\prime }\subseteq 𝔒$ be a finite subcover of I, which must exist because [a,b] is a compact set and $𝔒$ an open cover.

Let ${𝔒}^{″}\subseteq {𝔒}^{\prime }$ be the subset which results from successively removing from ${𝔒}^{\prime }$ any interval which is a subset of some other interval.

${𝔒}^{″}$ is still a cover of I.

List the elements ${𝔒}^{″}=\{(a_1,b_1),(a_2,b_2),\dots ,(a_n,b_n)\}$ in increasing order of the left end-point. $a_i\le a_{i+1}$ for $1\le i\le n$.

For any $1\le i\le n$, 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 $1\le i\le n$, if $b_i\le a_{i+1}$ then $b_i$ is not covered by any interval in ${𝔒}^{″}$. Therefore $a_{i+1}<b_i$.

Then by merely shifting parentheses for regrouping terms,

$\begin{array}{cc}\sum _{j=1}^n\ell (I_j)& =(b_1-a_1)+\cdots +(b_n-a_n)\\ & =-a_1+\stackrel{>0}{\overbrace{(b_1-a_2)}}+\stackrel{>0}{\overbrace{(b_2-a_3)}}+\cdots +\stackrel{>0}{\overbrace{(b_{n-1}-a_n)}}+b_n\\ & >-a+0+\cdots +0+b\\ & =b-a\end{array}$

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

$\begin{array}{cc}e& =\sum _{J\in 𝔒}\ell (J)\\ & \ge \sum _{J\in {𝔒}^{\prime }}\ell (J)\\ & \ge \sum _{J\in {𝔒}^{″}}\ell (J)\end{array}$

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

Combining all of the above,

$b-a\le e$

Template:Robelbox/close By definition of the outer measure as an infimum, then $b-a\le {\lambda }^{*}(I)$

Therefore

${\lambda }^{*}(I)=b-a$

If $I=(a,b)$ is a bounded open interval then ${\lambda }^{*}(I)=b-a$. Template:Robelbox By monotonicity ${\lambda }^{*}(I)\le \lambda ([a,b])=b-a$.

Let $\epsilon \in {ℝ}^{+}$ be any positive real number less than $\frac{b-a}{2}$.

Then by monotonicity,

${\lambda }^{*}([a+\epsilon /2,b-\epsilon /2])=b-a-\epsilon \le {\lambda }^{*}(I)$

Letting $\epsilon \to 0$ we have

$b-a\le {\lambda }^{*}(I)$

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

$b-a\le {\lambda }^{*}(I)\le b-a$

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

${\lambda }^{*}((a,n))=n-a\le {\lambda }^{*}(I)\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$

Mutatis mutandis the same argument shows that if I is finite on the right and unbounded on the left, or if $I=ℝ$, then ${\lambda }^{*}(I)=\mathrm{\infty }$.

In every case, therefore, ${\lambda }^{*}(I)=\ell (I)$.

$◻$

Lesson 4: Outer Measure Subadditivity

Theorem: Outer-measure not Additive

${\lambda }^{*}$ is not countably additive.

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 $𝒫(ℝ)$, has all four properties, Nonnegativity, Interval length, Translation invariance, and Countable additivity.

Since ${\lambda }^{*}$ is defined on $𝒫(ℝ)$ then it cannot be countably additive.

$◻$

Theorem: Outer-measure Subadditivity.

Let $A_1,A_2,\dots \subseteq ℝ$ be any countable sequence of subsets of real numbers.  Then 
  ${\lambda }^{*}\left(\bigcup _{k=1}^{\mathrm{\infty }}{A}_k\right)\le \sum _{k=1}^{\mathrm{\infty }}{\lambda }^{*}(A_k)$

Proof:

Suppose that for some $i\in {\mathbb{N}}^{+}$ we have ${\lambda }^{*}(A_i)=\mathrm{\infty }$.

The sum of $\mathrm{\infty }$ with any nonnegative number is again $\mathrm{\infty }$.

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

${\lambda }^{*}\left(\bigcup _{k=1}^{\mathrm{\infty }}{A}_k\right)\le \mathrm{\infty }=\sum _{k=1}^{\mathrm{\infty }}{\lambda }^{*}(A_k)$

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

Let $\epsilon \in {ℝ}^{+}$ be any positive real number.

For each $1\le k$ there is an open interval over-approximation of $A_k$, call it ${𝔒}_k$, such that

$\sum _{I\in {𝔒}_k}\ell (I)<{\lambda }^{*}(A_k)+\epsilon /2^k$

Then $𝔒=\bigcup _{k=1}^{\mathrm{\infty }}{𝔒}_k$ is an open interval over-approximation of $\bigcup _{k=1}^{\mathrm{\infty }}{A}_k$.

Also,

$\begin{array}{cc}\sum _{I\in 𝔒}\ell (I)& =\sum _{k\in {\mathbb{N}}^{+}}\sum _{I\in {𝔒}_k}\ell (I)\\ & <\sum _{k\in {\mathbb{N}}^{+}}({\lambda }^{*}(A_k)+\epsilon /2^k)\\ & =\sum _{k=1}^{\mathrm{\infty }}{\lambda }^{*}(A_k)+\epsilon \end{array}$

Therefore by definition of the infimum,

${\lambda }^{*}\left(\bigcup _{k=1}^{\mathrm{\infty }}{I}_k\right)<\sum _{k=1}^{\mathrm{\infty }}{\lambda }^{*}(A_k)+\epsilon$

Letting $\epsilon \to 0$

${\lambda }^{*}\left(\bigcup _{k=1}^{\mathrm{\infty }}{I}_k\right)\le \sum _{k=1}^{\mathrm{\infty }}{\lambda }^{*}(A_k)$

$◻$

Theorem: Null Adding and Subtracting.

Let $A,E\subseteq ℝ$ be two subsets and E a null set.  Then 
  ${\lambda }^{*}(A)={\lambda }^{*}(A\cup E)={\lambda }^{*}(A\setminus E)$

Proof:

By monotonicity, and the fact that ${\lambda }^{*}(E)=0$,

${\lambda }^{*}(A)\le {\lambda }^{*}(A\cup E)\le {\lambda }^{*}(A)+{\lambda }^{*}(E)={\lambda }^{*}(A)$

Therefore ${\lambda }^{*}(A)={\lambda }^{*}(A\cup E)$.

From this result,

${\lambda }^{*}(A\setminus E)={\lambda }^{*}([A\setminus E]\bigsqcup E)={\lambda }^{*}(A\cup E)$

$◻$

Lesson 5: Measurable Sets

Theorem: Null Sets Are Measurable.

Every null set is measurable.

Proof:

Let $E\subseteq ℝ$ be a null set.

Let $A\subseteq ℝ$ be any set.

By monotonicity ${\lambda }^{*}(A\cap E)\le {\lambda }^{*}(E)=0$ therefore

${\lambda }^{*}(A\cap E)=0$

By a previous result, because E is null, ${\lambda }^{*}(A\setminus E)={\lambda }^{*}(A)$.

Therefore

${\lambda }^{*}(A\cap E)+{\lambda }^{*}(A\setminus E)={\lambda }^{*}(A)$

so E splits A cleanly.

Since A was arbitrary, therefore E is measurable.

$◻$

Theorem: Measurable Sets Closed Under Complement.

If $E\subseteq ℝ$ is a measurable subset then $E^c$ is measurable.

Proof:

Let $E\subseteq ℝ$ be a measurable set.

Let $A\subseteq ℝ$ be any set of real numbers.

Then by definition of the set difference,

$\begin{array}{cc}{\lambda }^{*}(A)& ={\lambda }^{*}(A\cap E)+{\lambda }^{*}(A\setminus E)\\ & ={\lambda }^{*}(A\setminus E^c)+{\lambda }^{*}(A\cap E^c)\end{array}$

So $E^c$ splits A cleanly, and since A was arbitrary, then $E^c$ is measurable.

$◻$

Theorem: Open Rays Are Measurable.

For every $a\in ℝ$ the open ray $(a,\mathrm{\infty })$ is measurable.

Proof:

Let $a\in ℝ$, and let $A\subseteq ℝ$ be any set of real numbers. Let $I=(a,\mathrm{\infty })$ be the open ray to the right of a.

If $a\notin A$ then I splits A cleanly. Template:Robelbox By monotonicity, ${\lambda }^{*}(A)\le {\lambda }^{*}(A\cap I)+{\lambda }^{*}(A\setminus I)$.

If ${\lambda }^{*}(A)=\mathrm{\infty }$ then regardless of the right-hand side we must have ${\lambda }^{*}(A)\ge {\lambda }^{*}(A\cap I)+{\lambda }^{*}(A\setminus I)$.

Therefore for the rest of this proof, assume ${\lambda }^{*}(A)$ is finite.

Let $\epsilon \in {ℝ}^{+}$ be any positive real number.

Then ${\lambda }^{*}(A)+\epsilon >{\lambda }^{*}(A\cap I)+{\lambda }^{*}(A\setminus I)$. Template:Robelbox By definition as a finite infimum, there is an open interval over-approximation $𝔒$ of A such that

$\sum _{J\in 𝔒}\ell (J)<{\lambda }^{*}(A)+\epsilon$

Define ${𝔒}_l=\{J\cap (-\mathrm{\infty },a):J\in 𝔒\}$ and ${𝔒}_r=\{J\cap (a,\mathrm{\infty }):J\in 𝔒\}$.

Then ${𝔒}_l$ and ${𝔒}_r$ are open interval over-approximations of $A\setminus I$ and $A\cap I$ respectively.

Also

$\begin{array}{cc}{\lambda }^{*}(A\cap I)+{\lambda }^{*}(A\setminus I)& \le \sum _{J\in {𝔒}_l}\ell (J)+\sum _{J\in {𝔒}_r}\ell (J)\\ & =\sum _{J\in 𝔒}\ell (J\cap I)+\sum _{J\in 𝔒}\ell (J\cap (-\mathrm{\infty },a))\\ & =\sum _{J\in 𝔒}(\ell (J\cap I)+\ell (J\cap (-\mathrm{\infty },a)))\end{array}$

Also $\ell (J\cap I)+\ell (J\cap (-\mathrm{\infty },a))=\ell (J)$. Template:Robelbox Let $J=(x,y)$ where the end-points $x\le y$ are real numbers.

If $a<x$ then $J\cap I=J$ and $J\cap (-\mathrm{\infty },a)=\mathrm{\varnothing }$ and then $\ell (J\cap I)=y-x$ and $\ell (J\cap (-\mathrm{\infty },a))=0$.

Then

$\ell (J)=\ell (J\cap I)+\ell (J\cap (-\mathrm{\infty },a))$

Mutatis mutandis the same proof handles the case where $y<a$.

If $x<a<y$ then $J\cap I=(a,y)$ and $J\cap (-\mathrm{\infty },a)=(x,a)$. Then $\ell (J\cap I)=y-a$ and $\ell (J\cap (-\mathrm{\infty },a))=a-x$.

$\ell (J)=y-x=(y-a)+(a-x)=\ell (J\cap (-\mathrm{\infty },a))$

So we have shown in all cases that $\ell (J)=\ell (J\cap I)+\ell (J\cap (-\mathrm{\infty },a))$. Template:Robelbox/close Therefore

$\begin{array}{cc}{\lambda }^{*}(A\cap I)+{\lambda }^{*}(A\setminus I)& \le \sum _{J\in 𝔒}\ell (J)\\ & <{\lambda }^{*}(A)+\epsilon \end{array}$

Template:Robelbox/close Letting $\epsilon \to 0$

${\lambda }^{*}(A)\ge {\lambda }^{*}(A\cap I)+{\lambda }^{*}(A\setminus I)$

Therefore ${\lambda }^{*}(A)={\lambda }^{*}(A\cap I)+{\lambda }^{*}(A\setminus I)$.

Therefore I splits A cleanly. Template:Robelbox/close If $a\in A$ then I splits A cleanly. Template:Robelbox 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 $A\setminus \{a\}=B$.

Therefore, by all of the above,

$\begin{array}{cc}{\lambda }^{*}(A)& ={\lambda }^{*}(B)\\ & ={\lambda }^{*}(B\cap I)+{\lambda }^{*}(B\setminus I)\\ & ={\lambda }^{*}(A\cap I)+{\lambda }^{*}(A\setminus I)\end{array}$

Since I splits A cleanly in all cases, and A was arbitrary, then I is measurable. Template:Robelbox/close $◻$

Lesson 6: Measurable Sets Are a Sigma-algebra

Theorem: The Measurable Sets Form a $\sigma$-algebra.

The collection $ℳ$ is a $\sigma$-algebra.

Proof:

From two earlier theorems, $\mathrm{\varnothing }\in ℳ$ because countable sets are null and null sets are measurable.

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

If $E,F\in ℳ$ are two measurable sets then their union is measurable, $E\cup F\in ℳ$. Template:Robelbox Let

$E\cap F=E_{11}$

$E\cap F^c=E_{10}$

$E^c\cap F=E_{01}$

$E^c\cap F^c=E_{00}$

Let $A\subseteq ℝ$ be any set of real numbers.

Then $E\cup F$ splits A cleanly. Template:Robelbox By the measurability of E followed by the measurability of F, followed by elementary set theory,

$\begin{array}{cc}{\lambda }^{*}(A)& ={\lambda }^{*}(A\cap E)+{\lambda }^{*}(A\setminus E)\\ & ={\lambda }^{*}(A\cap E)+{\lambda }^{*}([A\setminus E]\cap F)+{\lambda }^{*}([A\setminus E]\setminus F)\\ & ={\lambda }^{*}(A\cap E)+{\lambda }^{*}(A\cap E_{01})+{\lambda }^{*}(A\cap E_{00})\end{array}$

By monotonicity and elementary set theory,

$\begin{array}{cc}{\lambda }^{*}(A\cap E)+{\lambda }^{*}(A\cap E_{01})& \ge {\lambda }^{*}([A\cap E]\bigsqcup [A\cap E_{01}])\\ & ={\lambda }^{*}(A\cap [E\bigsqcup E_{01}])\end{array}$

With the further observation by elementary set theory that $A\cap E_{00}=A\setminus (E\cup F)$ and $E\cup E_{01}=E\cup F$, then all of the above now shows

$\begin{array}{cc}{\lambda }^{*}(A)& \ge {\lambda }^{*}(A\cap [E\bigsqcup E_{01}])+{\lambda }^{*}(A\cap E_{00})\\ & ={\lambda }^{*}(A\cap [E\cup F])+{\lambda }^{*}(A\setminus [E\cup F])\end{array}$

Template:Robelbox/close Therefore $E\cup F\in ℳ$. Template:Robelbox/close If $E_1,E_2,\dots ,E_n\in ℳ$ is any finite sequence of measurable sets then their union is measurable, $\bigcup _{k=1}^n{E}_k\in ℳ$ Template:Robelbox 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 $\bigcup _{k=1}^n{E}_k\in ℳ$.

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

$\bigcup _{k=1}^{n+1}{E}_k=\bigcup _{k=1}^n{E}_k\cup E_{n+1}\in ℳ$

Thus the proof by induction is complete. Template:Robelbox/close If $E_1,E_2,\dots \in ℳ$ is any countable sequence of measurable sets then $\bigcup _{k=1}^{\mathrm{\infty }}{E}_k\in ℳ$. Template:Robelbox Let $E=\bigcup _{k=1}^{\mathrm{\infty }}{E}_k$.

Let $A\subseteq ℝ$.

Also let $F_1=E_1$ and for $1<k$ define $F_k=E_k\setminus \bigcup _{j=1}^{k-1}{E}_j$ so that $⟨F_k⟩$ forms a sequence of disjoint sets.

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

Also $E=\bigcup _{k=1}^{\mathrm{\infty }}{F}_k$.

By subadditivity,

${\lambda }^{*}(A)\le {\lambda }^{*}(A\cap E)+{\lambda }^{*}(A\setminus E)$

For each $n\in {\mathbb{N}}^{+}$

$\begin{array}{cc}{\lambda }^{*}(A)& ={\lambda }^{*}(A\cap \underset{k=1}{\overset{n}{⨆}}{F}_k)+{\lambda }^{*}(A\setminus \underset{k=1}{\overset{n}{⨆}}{F}_k)\text{ by the earlier result}\\ & \ge {\lambda }^{*}(A\cap \underset{k=1}{\overset{n}{⨆}}{F}_k)+{\lambda }^{*}(A\setminus \underset{k=1}{\overset{\mathrm{\infty }}{⨆}}{F}_k)\text{ by monotonicity}\end{array}$

Moreover, ${\lambda }^{*}(A\cap \underset{k=1}{\overset{n}{⨆}}{F}_k)=\sum _{k=1}^n{\lambda }^{*}(A\cap F_k)$. Template:Robelbox If $G_1,G_2,\dots ,G_n\in ℳ$ are any finite sequence of disjoint measurable sets and $B\subseteq ℝ$ is any set of real numbers, then

${\lambda }^{*}(B\cap \underset{k=1}{\overset{n}{⨆}}{G}_k)=\sum _{k=1}^n{\lambda }^{*}(B\cap G_k)$

Proof:

If $G,H\in ℳ$ are any two disjoint measurable sets and $B\subseteq ℝ$ is any set of real numbers, then ${\lambda }^{*}(B\cap (G\bigsqcup H))={\lambda }^{*}(B\cap G)+{\lambda }^{*}(B\cap H)$. Template:Robelbox Because H is measurable then

$\begin{array}{cc}{\lambda }^{*}(B\cap [G\bigsqcup H])& ={\lambda }^{*}(B\cap [G\bigsqcup H]\cap H)+{\lambda }^{*}(B\cap [G\bigsqcup H]\setminus H)\\ & ={\lambda }^{*}(B\cap H)+{\lambda }^{*}(B\cap G)\end{array}$

Template:Robelbox/close If $G_1,G_2,\dots ,G_n\in ℳ$ is any finite sequence of disjoint measurable sets, and $B\subseteq ℝ$ is any set of real numbers, then

${\lambda }^{*}(B\cap \underset{k=1}{\overset{n}{⨆}}{G}_k)=\sum _{k=1}^n{\lambda }^{*}(B\cap G_k)$

Template:Robelbox The case for $n=1$ is trivial.

Suppose the claim holds for some $1\le n$ and now consider the claim for $n+1$ sets.

$\begin{array}{cc}{\lambda }^{*}(B\cap \underset{k=1}{\overset{n+1}{⨆}}{G}_k)& ={\lambda }^{*}([B\cap \underset{k=1}{\overset{n}{⨆}}{G}_k]\bigsqcup [B\cap G_{n+1}])\\ & ={\lambda }^{*}(B\cap \underset{k=1}{\overset{n}{⨆}}{G}_k)+{\lambda }^{*}(B\cap G_{n+1})\text{ by the result for pairs}\\ & =\sum _{k=1}^n{\lambda }^{*}(B\cap G_k)+{\lambda }^{*}(B\cap G_{n+1})\\ & =\sum _{k=1}^{n+1}{\lambda }^{*}(B\cap G_k)\end{array}$

Template:Robelbox/close Template:Robelbox/close Therefore

${\lambda }^{*}(A)\ge \sum _{k=1}^n{\lambda }^{*}(A\cap F_k)+{\lambda }^{*}(A\setminus E)$

Now taking the limit $n\to \mathrm{\infty }$ on each side of this last inequality,

${\lambda }^{*}(A)\ge \sum _{k=1}^{\mathrm{\infty }}{\lambda }^{*}(A\cap F_k)+{\lambda }^{*}(A\setminus E)$

By subadditivity,

$\begin{array}{cc}{\lambda }^{*}(A\cap E)& ={\lambda }^{*}(A\cap \underset{k=1}{\overset{\mathrm{\infty }}{⨆}}{F}_k)\\ & ={\lambda }^{*}(\underset{k=1}{\overset{\mathrm{\infty }}{⨆}}(A\cap F_k))\\ & \le \sum _{k=1}^{\mathrm{\infty }}{\lambda }^{*}(A\cap F_k)\end{array}$

Putting together results from above,

${\lambda }^{*}(A)\ge {\lambda }^{*}(A\cap E)+{\lambda }^{*}(A\setminus E)$

Since A was arbitrary, this shows that E splits every set cleanly, and therefore E is measurable. Template:Robelbox/close The above shows that

• $\mathrm{\varnothing }\in ℳ$.
• $ℳ$ is closed under complement and countable unions.

Therefore $ℳ$ is a $\sigma$-algebra.

$◻$

Theorem: $\sigma$-algebra Space and Intersection

Let X be any set and $𝒮$ a $\sigma$-algebra on X.  Then $X\in 𝒮$ and $𝒮$ is closed under countable intersections.

Proof:

By closure under complements, $X={\mathrm{\varnothing }}^c\in 𝒮$.

By de Morgan's law, if $E_1,E_2,\dots \in 𝒮$ then

$\begin{array}{cc}\bigcap _{k=1}^{\mathrm{\infty }}{E}_k& ={\left(\bigcup _{k=1}^{\mathrm{\infty }}{E}_k^c\right)}^c\end{array}$

By the closure under complements, each $E_k^c$ is measurable.

By closure under unions, $\bigcup _{k=1}^{\mathrm{\infty }}{E}_k^c$ is measurable.

By closure under complements, ${\left(\bigcup _{k=1}^{\mathrm{\infty }}{E}_k^c\right)}^c$ is measurable.

$◻$

Theorem: Intervals Are Measurable.

If $I\subseteq ℝ$ is an interval then $I\in ℳ$.

Proof:

From an earlier result, all open rays to the right, $(a,\mathrm{\infty })\subseteq ℝ$, are measurable.

Let $I=(-\mathrm{\infty },a]\subseteq ℝ$ be any closed ray to the left.

Then $I=(a,\mathrm{\infty }{)}^c$ is the complement of an open ray to the right.

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

Now let $J=(a,b]\subseteq ℝ$ a bounded interval open on the left, for $a<b$ two real numbers.

Then $J=(-\mathrm{\infty },b]\cap (a,\mathrm{\infty })$, 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 $K=\bigcup _{n=1}^{\mathrm{\infty }}(a,b-1/n]$ 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.

$◻$

Lesson 8: Properties of Length Measure

Theorem: Length-measure Is Countably Additive.

Let $E_1,E_2,\dots \in ℳ$ be a disjoint sequence of measurable sets.  Then 
  $\lambda \left(\underset{k=1}{\overset{\mathrm{\infty }}{⨆}}{E}_k\right)=\sum _{k=1}^{\mathrm{\infty }}\lambda (E_k)$

Proof:

By subadditivity,

$\lambda \left(\underset{k=1}{\overset{\mathrm{\infty }}{⨆}}{E}_k\right)\le \sum _{k=1}^{\mathrm{\infty }}\lambda (E_k)$

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 $\mathrm{\varnothing }$.

$\begin{array}{cc}\lambda \left(\underset{k=1}{\overset{\mathrm{\infty }}{⨆}}{E}_k\right)& \ge \lambda \left(\underset{k=1}{\overset{n}{⨆}}{E}_k\right)\\ & =\sum _{k=1}^n\lambda (E_k)\end{array}$

Taking $n\to \mathrm{\infty }$ in the above shows $\lambda \left(\underset{k=1}{\overset{\mathrm{\infty }}{⨆}}{E}_k\right)\ge \sum _{k=1}^{\mathrm{\infty }}\lambda (E_k)$.

Therefore $\lambda \left(\underset{k=1}{\overset{\mathrm{\infty }}{⨆}}{E}_k\right)=\sum _{k=1}^{\mathrm{\infty }}\lambda (E_k)$.

$◻$

Theorem: Length-measure Excision.

Let $E,F\in ℳ$ be two measurable sets, and $E\subseteq F$, and assume that $\lambda (F)<\mathrm{\infty }$.  Then $\lambda (F\setminus E)=\lambda (F)-\lambda (E)$.

Proof:

By monotonicity $\lambda (E)\le \lambda (F)<\mathrm{\infty }$ and hence $\lambda (E)$ is a finite real number. Therefore arithmetic operations are well-defined for $\lambda (E)$ and $\lambda (F)$.

By additivity,

$\begin{array}{cc}\lambda (F)& =\lambda (F\cap E)+\lambda (F\setminus E)\\ & =\lambda (E)+\lambda (F\setminus E)\end{array}$

Whence

$\lambda (F\setminus E)=\lambda (F)-\lambda (E)$

$◻$

Theorem: Upward Continuity of Measure.

Let $E_1,E_2,\dots \in ℳ$ be an ascending sequence of measurable sets.  Then 

  $\underset{n\to \mathrm{\infty }}{\lim }\lambda (E_n)=\lambda \left(\bigcup _{k=1}^{\mathrm{\infty }}{E}_k\right)$

Proof:

Define $F_1=E_1$ and for $2\le n$, then $F_n=E_n\setminus E_{n-1}$.

Then $⟨F_n⟩$ is a sequence of disjoint measurable sets, and $\bigcup _{k=1}^{\mathrm{\infty }}{E}_k=\underset{k=1}{\overset{\mathrm{\infty }}{⨆}}{F}_k$.

Therefore by additivity,

$\begin{array}{cc}\lambda \left(\bigcup _{k=1}^{\mathrm{\infty }}{E}_k\right)& =\lambda \left(\underset{k=1}{⨆}{F}_k\right)\\ & =\sum _{k=1}^{\mathrm{\infty }}\lambda (F_k)\\ & =\underset{n\to \mathrm{\infty }}{\lim }\sum _{k=1}^n\lambda (F_k)\\ & =\underset{n\to \mathrm{\infty }}{\lim }\sum _{k=1}^n(\lambda (E_k)-\lambda (E_{k-1}))\text{ by excision}\\ & =\underset{n\to \mathrm{\infty }}{\lim }\lambda (E_n)\text{ by telescoping sum}\end{array}$

$◻$

Theorem: Downward Continuity of Measure.

Let $E_1,E_2,\dots \in ℳ$ be a descending sequence of measurable sets.  If $\lambda (E_1)$ is finite then 
  $\underset{n\to \mathrm{\infty }}{\lim }\lambda (E_m)=\lambda \left(\bigcap _{k=1}^{\mathrm{\infty }}{E}_k\right)$

Proof:

By monotonicity, every $\lambda (E_n)<\mathrm{\infty }$ and therefore arithmetic operations on these are well-defined.

Define $F_n=E_1\setminus E_n$ for each $n\in {\mathbb{N}}^{+}$.

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

By excision,

$\begin{array}{cc}\underset{n\to \mathrm{\infty }}{\lim }\lambda (F_n)& =\underset{n\to \mathrm{\infty }}{\lim }(\lambda (E_1)-\lambda (E_n))\\ & =\lambda (E_1)-\underset{n\to \mathrm{\infty }}{\lim }\lambda (E_n)\end{array}$

and

$\begin{array}{cc}\lambda \left(\bigcup _{k=1}^{\mathrm{\infty }}{F}_n\right)& =\lambda (E_1\setminus \bigcap _{k=1}^{\mathrm{\infty }}{E}_k)\\ & =\lambda (E_1)-\lambda \left(\bigcap _{k=1}^{\mathrm{\infty }}{E}_k\right)\end{array}$

By upward continuity of measure, therefore the two objects above are equal and therefore $\lambda (E_1)-\underset{n\to \mathrm{\infty }}{\lim }\lambda (E_n)=\lambda (E_1)-\lambda \left(\bigcap _{k=1}^{\mathrm{\infty }}{E}_k\right)$.

Whence

$\underset{n\to \mathrm{\infty }}{\lim }\lambda (E_n)=\lambda \left(\bigcap _{k=1}^{\mathrm{\infty }}{E}_k\right)$

$◻$

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