Lebesgue Measure¶

This chapter we focus on the measure of bounded set on $\mathbb{R}$.

Our goal: extend the concept of length on intervals to a more general set of real numbers.

Lebesgue Outer Measure¶

Definition of Lebesgue outer measure

Assume $E \subset \mathbb{R}$, we define the Lebesgue outer measure to be

\[ m^*(E)=\inf \left\{\sum_n l(I_n): \{I_n\}_{n\geq 1} \text{ are open intervals and } E\subset \bigcup_n I_n \right\} \]

where $l(I_n)$ denotes the length of intevals.

Example. For denumerable set $E=\{x_n\}_{n\geq 1}$, its Lebesgue outer measure

\[ m^*(E)=0 \]

Proof

Choose open intervals

\[ I_n=\left(x_n-\frac{\varepsilon}{2^{n+1} }, x_n+\frac{\varepsilon}{2^{n+1} } \right) \]

and show the summation of length of $\{I_n\}$ are $\varepsilon$, which could be sufficiently small.

Note that of Lebesgue outer measure of $\varnothing$ is also $0$.

Properties¶

Properties of Lebesgue Outer Measure

(i) Monotonicity.

If $E_1\subset E_2$, then

\[ m^*(E_1)\leq m^*(E_2) \]

(ii) If $I$ is an interval, then

\[ m^*(I)=l(I) \]

(iii) subadditivity

If $\{E_n\}_{n\geq 1}^{\infty}$ is a sequence of subset of $\mathbb{R}$, then

\[ m^*\left(\bigcup_{n=1}^\infty E_n\right) \leq \sum_{n=1}^\infty m^*(E_n) \]

Proof for (i)Proof for (ii)Proof for (iii)

(i) $\forall \varepsilon>0, \exists \{I_n\}$, s.t. $E_2\subset \bigcup\limits_{n=1}^\infty I_n$, and

\[ \begin{equation} \sum_{n=1}^\infty l(I_n)\leq m^*(E_2)+\varepsilon \quad\text{by property of infimum for $E_2$}\label{ieq1} \end{equation} \]

then because $E_1\subset E_2$, we have $E_1\subset \bigcup\limits_{n=1}^\infty I_n$, so

\[ \begin{equation} m^*(E_1)\leq \sum_{n=1}^\infty l(I_n)\quad \text{by property of infimum $E_1$}\label{ieq2} \end{equation} \]

combine inequation $\ref{ieq1}$ and $\ref{ieq2}$, we have

\[ m^*(E_1)\leq m^*(E_2)+\varepsilon \]

because $\varepsilon$ could be sufficiently small, we have

\[ m^*(E_1)\leq m^*(E_2) \]

(ii)

We firstly show that the conclusion for closed intervals $I=[a,b]$.

Closed set has a great property, that is, using Heine-Borel Theorem, if $I \subset \bigcup_{n=1}^\infty I_n$, then $\exists N$, such that $I\subset \bigcup_{n=1}^N I_n$. Thus

\[ l(I)<\sum_{n=1}^N l(I_n) < \sum_{n=1}^\infty l(I_n)\leq m^*(I)+\varepsilon \]

which means $l(I)\leq m^*(I)$ (let $\varepsilon\rightarrow 0$). This is one direction.

The opposite direction could be explained as below. We try to choose a special open interval. $\forall \varepsilon>0$, $I\subset (a-\varepsilon, b+\varepsilon)$, which means

\[ m^*(I)\leq l((a-\varepsilon, b+\varepsilon))=b-a+2\varepsilon \]

So let $\varepsilon\rightarrow 0$, we have

\[ m^*(I)\leq b-a=l(I) \]

Now we consider $I=(a,b]$ (similar logic for $[a,b)$).

Note that we can use a closed interval to contain $I$

\[ I=(a,b]\subset [a,b] \]

So by Monotonicity property of Lebesgue outer measure, we have

\[ m^*(I)\leq m^*([a,b])=l([a,b])=l((a,b])=l(I) \]

For another direction, we can choose a closed set which is contained by $I$, i.e. $\forall \varepsilon>0$, $[a+\varepsilon,b]\subset (a,b]$, by Monotonicity property of Lebesgue outer measure, we have

\[ b-a-\varepsilon=m^*([a+\varepsilon,b])\leq m^*(I) \]

let $\varepsilon\rightarrow 0$, we have

\[ l(I)=b-a\leq m^*(I) \]

Then we consider $(a,b)$, by using

\[ [a+\varepsilon,b-\varepsilon] \subset I \subset [a-\varepsilon,b+\varepsilon] \]

which means

\[ b-a-2\varepsilon=m^*([a+\varepsilon,b-\varepsilon]) \leq m^*(I)\leq m^*([a-\varepsilon,b+\varepsilon])=b-a+2\varepsilon \]

let $\varepsilon\rightarrow 0$, and we are done.

Finally we consider unbounded intervals $I=[a,+\infty]$.

In a similar logic, we choose $[a,b]$ ($\forall b>a$), we have $[a,b] \subset [a,\infty)=I$, so

\[ b-a=m^*([a,b])\leq m^*(I) \]

let $b\rightarrow +\infty$, we have

\[ +\infty\leq m^*(I)\leq +\infty \]

and we are done.

We have a sequence to add up, so we have to use a similar strategy in Example of measure of Denumerable set, i.e. use $\varepsilon/{2^{n}}$.

if $\sum_{n=1}^\infty E_n=\infty$, then (iii) holds naturally.
Consider $\sum_{n=1}^\infty E_n<\infty$. $\forall \varepsilon>0$, $\forall E_n (n\geq 1)$, we have sequence of open intervals $\{I^{(n)}_k\}_{k=1}^\infty$, such that $E_n\subset \bigcup_{n=1}^\infty I_k^{(n)}$ and

\[ \sum_{k=1}^\infty I_k^{(n)}\leq m^*(E_n)+\frac{\varepsilon}{2^n} \]

Show that these sequences of open intervals could compose a covering for $\bigcup_{n=1}^\infty E_n$, i.e.

\[ \bigcup_{n=1}^\infty E_n \subset \bigcup_{n=1}^\infty\bigcup_{k=1}^\infty I_k^{(n)} \]

by Monotonicity property of Lebesgue outer measure, we have

\[ \begin{align*} m^*\left(\bigcup_{n=1}^\infty E_n\right) &\leq m^*\left(\bigcup_{n=1}^\infty\bigcup_{k=1}^\infty I_k^{(n)}\right)\\ &=\sum_{n=1}^\infty \sum_{n=1}^\infty l(I_k^{(n)}) \quad \text{by property of measure for intervals}\\ &\leq \sum_{n=1}^\infty \left[m^*(E_n)+\frac{\varepsilon}{2^n}\right]\\ &=\sum_{n=1}^\infty m^*(E_n) + \varepsilon \end{align*} \]

let $\varepsilon \rightarrow 0$, and we are done.

Example. Prove another definition of Lebesgue outer measure

\[ m^*(E)'=\inf\{m(Q): E\subset Q, Q \text{ is an open set}\} \]

where $m(\cdot)$ means the Lebesgue measure, which we will discuss in more details in the following part. Here we use it because for open intervals, its outer measure and measure are of the same.

Proof

$m^*(E) \leq m^*(E)'$.

\[ \forall \varepsilon>0, \exists \{I_n\} \text{ s.t } E\subset \bigcup_n I_n \text{ & } \sum_n l(I_n)\leq m^*(E)+\varepsilon \]

let $Q=\bigcup_n I_n$, then $E\subset Q$ and $m(Q)=\sum_n l(I_n)\leq m^*(E)+\varepsilon$. Let $\varepsilon\rightarrow 0$, and get

\[ m^*(E)'\leq m^*(E) \]

$m^*(E)\geq m^*(E)'$.

This part is easy. Because we have an arbitrary open set $Q\supset E$, $m^*(E)\leq m^*(Q)=m(Q)$. Apply "inf" to both sides we have

\[ m^*(E)\leq \inf_{Q} {m(Q)}=m^*(E)' \]

Example. Assume $E\subset \mathbb{R}$, $0 < m^*(E)<\infty$, prove:

\[ f(x):=m^*((-\infty,x)\cap E) \]

is a consistently continuous function with respect to $x$.

Proof

$\forall x<y$, we have

\[ (-\infty,y)\cap E=((-\infty,x)\cap E )\cup([x,y)\cap E), \]

thus

\[ \begin{align*} m^*((-\infty,y)\cap E)&=m^*[((-\infty,x)\cap E )\cup([x,y)\cap E) ]\\ f(y)&\leq f(x)+m^*[([x,y)\cap E)]\\ f(y)&\leq f(x)+y-x\\ \Rightarrow 0\leq f(y)-f(x)&\leq y-x. \end{align*} \]

which means $f(x)$ is a consistently continuous function.

Lebesgue Measure¶

Definition of Lebesgue Measure

Set $E\subset \mathbb{R}$ is Lebesgue measurable, if $\forall A\subset \mathbb{R}$,

\[ \begin{align} m^*(A)\geq m^*(A\cap E)+m^*(A\cap E^c). \label{caratheo} \end{align} \]

which is also called Carathéodory condition (or criterion). and we have Lebesgue measure denoted by $m(A)=m^*(A)$. Denote $\Omega$ as a set of all measurable set.

Note that $A=(A\cap E)\cup (A\cap E^c)$, so by subadditivity of Lebesgue outer measure, we have equality holds in inequation $\ref{caratheo}$. Here $A$ is also called test set.

Another condition for measurable set

Set $E\subset \mathbb{R}$ and $E\in \Omega$, iff $\forall A\subset E$, $b\subset E^c$, such that

\[ \begin{align} m^*(A\cup B)=m^*(A)+m^*(B) \label{Cara-eq} \end{align} \]

Proof

This is quite simple if we could see the relationship between condition $\ref{Cara-eq}$ and Carathéeodory condition.

$\Rightarrow$. If $E$ satisfies Caratheodory condition, then $\forall T\subset \mathbb{R}$,

\[ m^*(T)\geq m^*(T\cap E)+m^*(T\cap E^c). \]

Let $A=T\cap E\subset E$, $B=T\cap E^c\subset E^c$, then $T=(T\cap E)\cup (T\cap E^c)=A\cup B$, so the above condition $\ref{Cara-eq}$ holds.

$\Leftarrow$. If $E$ satisfies condition $\ref{Cara-eq}$, then $\forall A\subset E$, $B\subset E^c$,

\[ m^*(A\cup B)=m^*(A)+m^*(B) \]

let $T = A\cup B$, then $A = (A\cup B)\cap E=T\cap E$, $B=(A\cup B)\cap E^c=T\cap E^c$, then Carathéodory condition holds.

This condition $\ref{Cara-eq}$ is also quite useful in the following proof.

Measurable Sets¶

Lebesgue Measurable Sets

(i) If $m^*(E)=0$, then $E\in \Omega$, and $m(E)=0$.

(ii) If $E$ is an interval, then $E\in \Omega$, and $m(E)=m^*(E)$.

Proof for (i)Proof for (ii)

Because $E \supset (A\cap E)$, so $0=m^*(E)\geq m^*(A\cap E)\geq 0$, which means $m^*(A\cap E)=0$.

then by $A\supset (A\cap E^c)$, we have

\[ \begin{align*} m^*(A)&\geq m^*(A\cap E^c)\\ &=m^*(A\cap E^c)+0\\ &=m^*(A\cap E^c)+m^*(A\cap E) \end{align*} \]

which satisfies Carathéodory condition.

We need to prove that intervals satisfies Carathéodory condition, i.e. $\forall A\subset \mathbb{R}$,

\[ m^*(A)\geq m^*(A\cap E)+m^*(A\cap E^c) \]

We focus on using open intervals of $A$. That is, $\forall \varepsilon>0$, $\exists {I_n}$ s.t. $A\subset \bigcup_n I_n$ and $\sum_n l(I_n)\leq m^*(A)+\varepsilon$. We want to show that $m^*(A\cap E)+m^*(A\cap E^c)\leq \sum_n l(I_n)$.

Notice that $A\cap E\subset \left[(\bigcup_n I_n)\cap E\right]$, so

\[ \begin{align*} m^*(A\cap E)&\leq m^*\left(\left(\bigcup_n I_n\right)\cap E\right)\\ &=m^*\left(\bigcup_n \left(I_n\cap E\right)\right)\quad\text{by distributive law}\\ &\leq \sum_n m^*(I_n\cap E) \quad \text{by subadditivity}\\ &=\sum_n l(I_n\cap E) \quad\text{$I_n\cap E$ is also intervals} \end{align*} \]

We have $E^c=E_1\cup E_2$ where $E_1$ and $E_2$ are also intervals, because $E$ is an interval. Apply same logic to $E_1$ and $E_2$ gives

\[ m^*(A\cap E_1)\leq \sum_n l(I_n\cap E_1) \]

\[ m^*(A\cap E_2)\leq \sum_n l(I_n\cap E_2) \]

adding the above three inequation up and get

\[ \begin{align*} m^*(A\cap E)+ m^*(A\cap E_1)+m^*(A\cap E_1)&\leq \sum_n l(I_n\cap E)+l(I_n\cap E_1)+l(I_n\cap E_2)\\ &=\sum_n l(I_n) \quad\text{additivity of intervals} \end{align*} \]

Notice that $(A\cap E^c)=(A\cap E_1)\cup (A\cap E_2)$, then also by subadditivity of Lebesgue outer measure

\[ m^*(A\cap E^c)\leq m^*(A\cap E_1)+m^*(A\cap E_2) \]

So we have

\[ \begin{align*} m^*(A\cap E)+m^*(A\cap E^c)&\leq m^*(A\cap E)+m^*(A\cap E_1)+m^*(A\cap E_2)\\ &\leq \sum_n l(I_n)\\ &\leq m^*(A)+\varepsilon \end{align*} \]

Let $\varepsilon\rightarrow 0$, so $E$ satisfies Carathéodory condition.

From (i) we could know that denumerable sets have zero measure. From (ii) we could know that any open, closed intervals are measurable.

Example. Cantor set has zero measure.

Answer

Because in configuration of Cantor set, the measure of $G$

\[ \begin{align*} m(G)&=\frac{1}{3}+2\cdot \frac{1}{3^2}+2^2\frac{1}{3^3}+\cdots\\ &=\frac{1}{3} \frac{[1-(\frac{2}{3})^n]}{1-\frac{2}{3}}\rightarrow 1 (n\rightarrow \infty) \end{align*} \]

So $m(C)=m([0,1])-m(G)=0$.

Measure Operations¶

This part we aim to show denumerable additivity of measure. So we have to give some foundations.

Measure operations

Assume $E_1\in \Omega$, $E_2\in \Omega$, then

(i) $E_1^c\in \Omega$.

(ii) $E_1\cup E_2\in \Omega$.

(iii) $E_1\cap E_2\in \Omega$.

(iv) $E_1 \backslash E_2 \in \Omega$.

Proof for (i)Proof for (ii)Proof for (iii)Proof for (iv)

This is quite easy, cause we can substitute $E=(E^c)^c$ into Carathéodory condition and we could get the result.

Our goal is $\forall A\subset \mathbb{R}$,

\[ m^*(A)\geq m^*(A\cap (E_1\cup E_2))+m^*(A\cap (E_1\cup E_2)^c) \]

From what we already know, we have

\[ \begin{align*} m^*(A)\geq m^*(A\cap E_1)+m^*(A\cap E_1^c)\\ m^*(A)\geq m^*(A\cap E_2)+m^*(A\cap E_2^c) \end{align*} \]

The following is a little tricky. Let $A=A\cap E_1^c$, then the second of above two becomes

\[ \begin{align*} m^*[(A\cap E_1^c)\cap E_2]+m^*[(A\cap E_1^c)\cap E_2^c]&\leq m^*(A\cap E_1^c)\\ m^*[(A\cap E_1^c)\cap E_2]+m^*[A\cap (E_1\cup E_2)^c]&\leq m^*(A)-m^*(A\cap E_1)\\ m^*[(A\cap E_1^c)\cap E_2]+m^*(A\cap E_1)+m^*[A\cap (E_1\cup E_2)^c]&\leq m^*(A) \end{align*} \]

Now we only need to show that $m^*[(A\cap E_1^c)\cap E_2]+m^*(A\cap E_1)=m^*[A\cap (E_1\cup E_2)]$. Here we could use another definition of measurable set. Notice that $(A\cap E_1)\subset E_1$, $[(A\cap E_1^c)\cap E_2]\subset E_1^c$, $E_1\in \Omega$, so

\[ \begin{align*} m^*[(A\cap E_1^c)\cap E_2]+m^*(A\cap E_1)&=m^*([(A\cap E_1^c)\cap E_2]\cup (A\cap E_1))\\ &=m^*([(A\cap E_1^c)\cup(A\cap E_1)]\cap [E_2\cup (A\cap E_1)])\\ &=m^*(A\cap[(E_2\cup A)\cap(E_2\cup E_1)])\\ &=m^*(A\cap(E_1\cup E_2)) \end{align*} \]

Or we have

\[ \begin{align*} [(A\cap E_1^c)\cap E_2]\cup (A\cap E_1)&=[A\cap (E_1^c\cap E_2)]\cup (A\cap E_1)\\ &=[(E_1^c\cap E_2)\cup E_1]\cap A\\ &=[(E_2\cup E_1)\cap (E_1^c\cup E_1)]\cap A\\ &=[(E_1\cup E_2)\cap \mathbb{R}]\cap A\\ &=(E_1\cup E_2)\cap A \end{align*} \]

which gives the same answer.

Using (i) and (ii)

Notice $E_1\cap E_2=(E_1^c\cup E_2^c)^c$. Because operations of complementation and union both holds for measurable, operation of intersection also holds for measurable.

Same logic with (iii). Notice that $E_1\backslash E_2=E_1\cap E_2^c$, since operation of intersection and complementation holds for measurable, operation of difference also holds for measurable.

Corollary

Assume $E_n\in \Omega$, $k=1,2\cdots$, then $\forall N>0$ and $N<\infty$,

\[ \bigcup_{n=1}^N E_n\in \Omega,\quad \bigcap_{n=1}^N E_n\in \Omega \]

Using the above corollary we could have a great result for operations of sequence of measurable sets.

Theorem for union of sequence of measurable sets

If $\{E_n\}_{n\geq 1}$ are mutually disjoint and $E_n\in \Omega$, then

\[ \bigcup_n E_n\in \Omega \]

Proof

Notice that $\forall N>0$, using corollary

\[ \begin{align*} m^*(A)&\geq m^*\left[A\cap \left(\bigcup_{n=1}^N E_n\right)\right]+m^*\left[A\cap \left(\bigcup_{n=1}^N E_n\right)^c\right]\\ \end{align*} \]

Here we could not let $N\rightarrow \infty$, due to $N$ is embedded in the operation of sets, so we need to get it out. By additivity for mutually disjoint sets, we have

\[ \begin{align*} m^*(A)&=\left[\sum_{n=1}^N m^*(A\cap E_n)\right] + m^*\left[A\cap \left(\bigcup_{n=1}^N E_n\right)^c\right]\quad\\ &\geq \left[\sum_{n=1}^N m^*(A\cap E_n)\right] + m^*\left[A\cap \left(\bigcup_{n=1}^\infty E_n\right)^c\right] \end{align*} \]

let $N\rightarrow \infty$, and get $\infty$ back into operations of sets

\[ \begin{align*} m^*(A)&\geq\left[\sum_{n=1}^\infty m^*(A\cap E_n)\right] + m^*\left[A\cap \left(\bigcup_{n=1}^\infty E_n\right)^c\right]\\ &\geq m^*\left[\bigcup_{n=1}^\infty (A\cap E_n)\right]+ m^*\left[A\cap \left(\bigcup_{n=1}^\infty E_n\right)^c\right]\quad \text{by subadditivity}\\ &=m^*\left[A\cap \left(\bigcup_{n=1}^\infty E_n\right)\right] +m^*\left[A\cap \left(\bigcup_{n=1}^\infty E_n\right)^c\right] \end{align*} \]

and we are done.

In a similar way, we have

Corollary

If $\{E_n\}_{n\geq 1}$ are mutually disjoint and $E_n\in \Omega$, then

\[ \bigcap_n E_n\in \Omega \]

the above two theorems both hold if we remove the condition "mutually disjoint".

Theorem for operations of denumerable sets

Assume $E_n\in \Omega$, $n=1,2,\cdots$, then

\[ \bigcup_n E_n\in \Omega,\quad \bigcap_n E_n\in \Omega \]

Proof

We could use the conclusion from the above two theorem. That is, construct mutually disjoint sets to help prove. We would use some conclusion from Sets.

Define $A_1=E_1$, $A_n=E_n-\bigcup_{k=1}^{n-1}E_n (n\geq 2)$ so we have a sequence of mutually disjoint sets $\{A_n\}$ which satisfies $\bigcup_n A_n=\bigcup_n E_n$.

Because $\bigcup_n A_n$ are measurable due to Theorem for union of sequence of measurable sets, so $\bigcup_n E_n\in \Omega$.

To be more specific, we have a good property for measurable sets, which is denumerable additivity, or $\sigma$-additiviity.

$\sigma$-additivity

If $\{E_n\}_{n\geq 1}$ are mutually disjoint and $E_n\in \Omega$, then $\forall A\subset \mathbb{R}$,

\[ \begin{align} m^*\left[A\cap \left(\bigcup_{n=1}^\infty E_n\right)\right]=\sum_{n=1}^\infty m^*(A\cap E_n)\label{sigma-additivity} \end{align}\]

If we let $A=\mathbb{R}$, then equation $\ref{sigma-additivity}$ becomes

\[ m^*\left(\bigcup_{n=1}^\infty E_n\right)=\sum_{n=1}^\infty m^*(E_n) \]

Proof

The proof has been given in proof of Theorem for union of sequence of measurable sets.

Limit Operation¶

Now we introduce limit operation which might be useful in future proof.

Limit Operations

If a sequence of measurable sets $\{E_n\}_{n\geq 1}$ satisfies either of the following condition

(i) $\{E_n\}$ is monotonically increasing.

(ii) $\{E_n\}$ is monotonically decreasing, and $m(E_1)<\infty$.

then

\[ \lim_{n\rightarrow \infty} m(E_n) = m\left(\lim_{n\rightarrow \infty} E_n\right) \]

Proof for (i)Proof for (ii)

(i) means $E_1\subset E_2\subset\cdots$, so

\[ m\left( \lim_{n\rightarrow \infty} E_n\right)=m\left( \bigcup_{n=1}^\infty E_n \right) \]

and

\[ \lim_{n\rightarrow \infty} m(E_n)=\lim_{N\rightarrow \infty}m\left( \bigcup_{n=1}^N E_n \right) \]

and the proposition to be proved equals to

\[ m\left( \bigcup_{n=1}^\infty E_n \right)=\lim_{N\rightarrow \infty}m\left( \bigcup_{n=1}^N E_n \right) \]

Let $A_1=E_1$, $A_n=E_n-\bigcup_{k=1}^{n-1}E_n (n\geq 2)$, so $\{A_n\}$ are mutually disjoint and satisfies $\bigcup_n A_n=\bigcup_n E_n$. So

\[ \begin{align*} m\left( \bigcup_{n=1}^\infty E_n \right)&=m\left( \bigcup_{n=1}^\infty A_n \right)\\ &=\sum_{n=1}^\infty m(A_n)\quad\text{by $\sigma$-additivity}\\ &=\lim_{N\rightarrow \infty}\sum_{n=1}^N m(A_n)\quad\text{holds for series}\\ &=\lim_{N\rightarrow \infty}m\left( \bigcup_{n=1}^N A_n \right)\\ &=\lim_{N\rightarrow \infty}m\left( \bigcup_{n=1}^N E_n \right)\\ \end{align*} \]

and we are done.

(ii) means $E_1\supset E_2\supset \cdots$. We could use the conclusion from (i), that is, let $A_n=E_1\backslash E_n (n\geq 1)$ so it is easy to see that

\[ A_1\subset A_2\subset \cdots \text{ and } \bigcup_n A_n=E_1\backslash \left(\bigcup_n E_n\right) \]

So by operation of difference,

\[ \begin{align*} m(E_1)-m\left(\bigcup_n E_n\right)&=m\left(\bigcup_n A_n\right)\\ m(E_1)&=\lim_n m(A_n)+m\left(\bigcup_n A_n\right)\\ \Rightarrow \lim_n [m(E_1)-m(A_n)]&=m\left(\bigcup_n A_n\right) \\ \lim_n [m(E_1\backslash A_n)]&=m\left(\bigcup_n A_n\right)\quad\text{by operation of difference}\\ \lim_n m(E_n)&=m\left(\bigcup_n A_n\right) \end{align*} \]

and we are done.

Translation Invariance¶

Definition of Translation

Assume $E\subset \mathbb{R}$ and $y\in \mathbb{R}$, then

\[ E_y=\{x+y: x\in E\} \]

is called a translation of $E$ about $y$.

About translation, we have the following property.

Property of Translation

Assume $E, F\subset \mathbb{R}$, then $\forall y\in \mathbb{R}$,

(i) $E\cap E_y=(E_{-y}\cap F)_y$.

(ii) $(E^c)_y=(E_y)^c$.

(iii) $m^*(E)=m^*(E_y)$.

Translation Invariance of Measure

Assume $E \in \Omega$, then $\forall y\in \mathbb{R}$, $E_y\in \Omega$ and $m(E_y)=m(E)$.

Proof

Use the above property to transfer Carathéodory condition of $E$ into $E_y$.

Approximate Measurable set with open & closed sets¶

Theorem of Approximation

The following proposition are equivalent.

(i) $E\in \Omega$.

(ii) $\forall \varepsilon>0$, $\exists \text{ open set }G\supset E$, such that $m^*(G-E)<\varepsilon$.

(iii) $\forall \varepsilon>0$, $\exists \text{ closed set } F \subset E$, such that $m^*(E-F)<\varepsilon$.

(i) $\Rightarrow$ (ii)(ii) $\Rightarrow$ (i)(i) $\Rightarrow$ (iii)

We could prove this on $\mathbb{R}^n$.

Define

\[ B_n:=\{x\in \mathbb{R} : n-1 \leq |x|< n\},\quad n\in \mathbb{N}^+ \]

because open set is measurable, whose subtraction is also measurable, i.e. $B_n \in \Omega$.

We here want to use these sequence of half-open cubes. That is, define $E_n:=B_n\cap E \in \Omega$, and it is easy to see that $\{E_n\}$ are mutually disjoint. If we assume $E(n)<\infty$ (if not, the proposition also holds), then by $E=\bigcup_n E_n$, we have sigma-additivity

\[ m(E)=\sum_n m(E_n) \]

Then we want to use open covering of $E_n$ to construct the so-called open set $G$. That is, $\forall \varepsilon>0$, $\exists \{I^{(n)}_k\}_{k\geq 1}$ such that $E_n\subset \bigcup_k I^{(n)}_k$ and $\sum_k l(I_k^{(n)})\leq m(E_n)+\varepsilon/2^n$. Then the following construction is quite natural.

Define $G_n:=\bigcup_k I_k^{(n)}\supset E_n$, so by operation of difference, we have

\[ \begin{align*} m(G_n\backslash E_n)&=m(G_n)-m(E_n)\\ &\leq \sum_km(I_k^{(n)})-m(E_n) \quad \text{by subadditivity}\\ &< \frac{\varepsilon}{2^n} \end{align*} \]

Define $G:=\bigcup_n G_n \supset \bigcup_n E_n=E$, so by operation of difference, we have

\[ \begin{align*} m(G)-m(E)&=m(G\backslash E)\\ &=m\left[ \left(\bigcup_n G_n\right) \backslash \left(\bigcup_n E_n\right)\right]\\ &\leq m\left[\bigcup_n (G_n \backslash E_n)\right]\quad \text{by subtraction relation}\\ &\leq \sum_n m(G_n\backslash E_n)<\varepsilon \end{align*} \]

and we are done.

Here we utilize $\varepsilon$ to formulate a covering of open sets and hope to show $G\backslash E$ has zero measure. That is, $\forall \varepsilon=\frac{1}{n}$, $\exists G_n \supset E$ s.t. $m^*(G_n \backslash E)<\varepsilon=\frac{1}{n}$. Define $G:=\bigcap_n G_n$ which is measurable. By $G\backslash E \subset G_n\backslash E$, we have

\[ m^*(G\backslash E) = m^*(G_n\backslash E)\leq \frac{1}{n} \]

Let $n\rightarrow \infty$, we have $m^*(G\backslash E)=0$. Because $G\backslash E \in \Omega$ and $G\in \Omega$, we have $E\in \Omega$.

WE could utilize conclusions from (i) and (ii). Assume $E\in \Omega$, then $E^c\in \Omega$. By (ii), $\forall \varepsilon>0$, $\exists \text{ open set }G\supset E^c$, such that $m(G\backslash E^c)<\varepsilon$. Since $G$ is an open set, $G^c$ is a closed set. If we let $F=G^c$, then

\[ E\backslash F=E\backslash G^c=E\cap G = G\backslash E^c \]

which means $m(E\backslash F)=m(G\backslash E^c)<\varepsilon$.

Corollary

The following statements are equivalent.

(i) $E\in \Omega$.

(ii) $\forall \varepsilon>0$, $\exists \text{ open set }G$ and $\text{ closed set } F$, such that

\[ F\subset E\subset G \text{ and } m(G\backslash F)<\varepsilon. \]

(iii) $\exists G=G_\delta \supset E$, such that $m^*(G-E)=0$

(iv) $\exists F=F_\sigma \subset E$, such that $m^*(E-F)=0$

Borel Set¶

Check the definition of $\sigma$-algebra and its properties in elementary probability theory, and $\sigma$-algebra generated by set.

Definition of Borel Algebra and Borel Set

A $\sigma$-algebra generated by the whole open intervals on $\mathbb{R}$ is called Borel Algebra, denoted by $\mathscr{B}$, or $\mathscr{B}(\mathbb{R})$. The element of $\mathscr{B}$ is called Borel Set. Apparently, Borel set is measurable.

Properties of Borel Set

(i) Open set $G\in \mathscr{B}$

(ii) Closed set $F\in \mathscr{B}$.

(iii) single point set $\{a\}\in \mathscr{B}$.

(iv) half open intervals $H\in \mathscr{B}$.

Proof

(i) $G$ is denumerable union of open intervals.

(ii) $F$ is the complementary set of open set $G$.

(iii) single point set is the denumerable intersection of open intervals, i.e.

\[ \{a\}=\bigcap_{n=1}^\infty\left(a-\frac{1}{n}, a+\frac{1}{n}\right). \]

(iv) half open intervals are union of an interval and a single point set.

Equivalent definition of Borel Algebra

The following are equivalent definition of $\mathscr{B}$.

(i) \sigma({(-\infty,x]: x\in \mathbb{R}}),

(ii) \sigma({(-\infty,x): x\in \mathbb{R}}),

(iii) \sigma({(-\infty,x]: x\in \mathbb{Q}}).

Actually, $\mathscr{B}$ could be generated by the whole closed set, or the whole half open sets, but not single point set. Check the following example.

Example. (Simgle point set) Assume $Q$ is a non-empty set and $S$ is a set composed by all single point in $Q$, i.e.

\[ S=\{\{x\}: x\in Q\}. \]

Show that $\sigma(S)=\mathscr{A}$, where $\mathscr{A}$ is defined by

\[ \mathscr{A}:=\{A\subset Q: \text{either }A \text{ or } A^c \text{ is denumerable.}\} \]

Proof

We use $\mathscr{C}$ to be the whole denumerable sets on $Q$.

"$\subset$". Firstly show that $\mathscr{A}$ is a $\sigma$-algebra. Apparently $\varnothing, Q\in \mathscr{A}$. We show the following two condition.

(i) $\forall A\in \mathscr{A}$, $A^c \in\mathscr{A}$. This is apparently, because $\forall A\in \mathscr{A}$, there are two cases:

if $A\in \mathscr{C}$, then $(A^c)^c=A\in \mathscr{C}$, so $A^c\in\mathscr{A}$;

else if $A^c\in \mathscr{C}$, then $A^c\in \mathscr{A}$.

(ii) $\forall \{A_n\}_{n\geq 1}\in \mathscr{A}$, $\bigcup_nA_n\in\mathscr{A}$. There are also two cases.

If for all $A_n$, we have $A_n\in\mathscr{C}$, then by union of denumerable sets is denumerable, $\bigcup_nA_n\in\mathscr{C}$, which means $\bigcup_nA_n\in\mathscr{A}$;

else if there exists $j$ such that $A_j^c\in \mathscr{C}$, then $(\bigcup_n A_n)^c=\bigcap_n A_n^c\in \mathscr{C}$, so $\bigcup_n A_n\in \mathscr{A}$.

Combining the above two, we have $A$ is a $sigma$-algebra. By definition of $\sigma(S)$, we have $\sigma(C)\subset \mathscr{A}$.

"$\supset$". $\forall A\in \mathscr{A}$, there are two cases.

If $A\in \mathscr{C}$, then $A=\{x_n\}_{n\geq 1}=\bigcup_n x_n$. Since $\{x_n\}\in S\in \sigma(S)$, and $\sigma(S)$ is closed for denumerable union, so $A\in \sigma(S)$.

Else if $A^c\in \mathscr{C}$, then $A^c\in \sigma(S)$. Since $\sigma(S)$ is closed for complementation, we have $A\in \sigma(S)$.

In a nutshell, we have $\mathscr{A}\subset \sigma(S)$. And we are done.

Measure of Direct Product¶

This part is useful in deducing multiple integral formula.

Theorem for measurability of Direct Product

Assume $P$ and $Q$ are measurable sets in $\mathbb{R}^p$ and $\mathbb{R}^q$ respectively, then $P\times Q$ is a measurable set in $\mathbb{R}^{p+q}$, and

\[ m(P\times Q)=m(P)\cdot m(Q). \]

Proof

(i) $P$, $Q$ are cubes in $\mathbb{R}^p$ and $\mathbb{R}^q$ respectively.

(ii) $P$, $Q$ are open sets in $\mathbb{R}^p$ and $\mathbb{R}^q$ respectively.

Use Configuration of open sets in high dimension space.

(iii) $P$, $Q$ are measure-finite sets in $\mathbb{R}^p$ and $\mathbb{R}^q$ respectively.

Use Approximation with open and closed sets.

(iv) $P$, $Q$ are general measurable sets in $\mathbb{R}^p$ and $\mathbb{R}^q$ respectively.

Use series of measure-finite sets like $\{x\in \mathbb{R}^p: k-1\leq \|x\|< k\}$ to intersect to construct cases that could use conclusion in (iii).