If $I$ is an interval, then

Suppose $I=[a,b]$. Then $I\subset (a-ϵ,b+ϵ)$ for all $ϵ>0$. But then we have

Thus we have $m^{\ast }(I)\le b-a$.

Now we show $b-a\le m^{\ast }(I)$. Let $\{I_n{\}}_n$ be a collection of open intervals such that $[a,b]\subset \bigcup _n{I}_n$.

Now, since $[a,b]$ is compact (by the Heine-Borel theorem), there exists a finite subcover $\{J_k{\}}_{k=0}^m\subset \{I_n{\}}_n$ such that $[a,b]\subset \bigcup _{k=0}^m{J}_k$

Since $a\in \bigcup _{k=0}^m{J}_k$, there exists $k_1$ such that $a\in J_{k_1}$. Rearranging intervals, I let $k_1=1$ and $a\in J_1=(a_1,b_1)$. If $b_1<b$, then $b_1\in [a,b]$. So ther eis some $k_2$ such that $b_1\in J_{k_2}$. By rearranging again, I assume $k_2=2$. So $b_1\in J_2=(a_2,b_2)$.

Continuing in the same manner, we conclude that there exists a $K,1\le K\le m$ such that $b<b_K$. And for all $k\in \{1,2,\dots ,K-1\}$ we have $b_k\le b$ and $a_{k+1}\le b_k<b_{k+1}$.

But then we have

Thus we have $m^{\ast }(I)\ge \ell (I)=b-a$

Thus we conclude $m^{\ast }(I)=\ell (I)$

If $I$ is any finite interval $[a,b],[a,b),(a,b],$ or $(a,b)$, we note that for all $ϵ>0$ we have

Taking $ϵ\to 0$, we get the desired result.

An infinite interval cannot be covered by a collection of intervals of finite length

for every $A\subset \mathbb{R}$ and $ϵ>0$, there exist an open set $O$ such that $A\subset O$ and

The result is obvious if $m^{\ast }(A)=\mathrm{\infty }$ (just take $O=\mathbb{R}$), so we assume $m^{\ast }(A)<\mathrm{\infty }$.

Let $\{I_n\}$ be a collection of open intervals that cover $A$ and have total length at most $m^{\ast }(A)+ϵ$. Then $O=\bigcup _n{I}_n$ is open and obviously $A\subset O$. Thus by subadditivity ( theorem from lecture 6 )

A set $E\subset \mathbb{R}$ is Lebesgue measurable if for all $A\subset \mathbb{R}$ we have

Since for all $A,E$ we have $A=(A\cap E)\cup (A\cap E^c)$, we have by subadditivity that $m^{\ast }(A)\le m^{\ast }(A\cap E)+m^{\ast }(A\cap E^c)$.

Thus $E$ is measurable if $m^{\ast }(A\cap E)+m^{\ast }(A\cap E^c)\le m^{\ast }(A)$

( we only need to show one side of the inequality )

The empty set $\varnothing$ and the set of real numbers $\mathbb{R}$ are measurable. And a set $E$ is measurable if and only if $E^c$ is measurable.

These are readily verifiable from the definition of measurability - which is symmetric in $E$ and $E^c$.

If $m^{\ast }(E)=0$ then $E$ is measurable.

Let $A\subset \mathbb{R}$. Then $A\cap E\subset E$ means

But then we see

because $A\cap E^c\subset A$.

Lots of "uninteresting" sets are measurable - it is hard to find interesting sets that are measurable.

• every open set is measurable
• so every closed set is also measurable (by taking complements)
• Most things we can write down are measurable (since taking unions and intersections of basic sets also give us measurable sets)

If $E_1,E_2$ are measurable, then $E_1\cup E_2$ are measurable

Let $A\subset \mathbb{R}$. Since $E_2$ is measurable, we know

Then we also know

Taking outer measure we get

And this gives us the desired result.

If $E_1,\dots ,E_n$ are measurable, then $\bigcup _{k=1}^n{E}_k$ is measurable

by induction.
$n=1$ is trivial. Suppose the statement holds for $n=k$. Then

And then since unions of measurable sets are measurable we get the desired result.

A nonempty collection of sets $\mathcal{A}\subset \mathcal{P}(\mathbb{R})$ is an algebra if

1. (closure under complements) $E\in \mathcal{A}⟹E^c\in \mathcal{A}$
2. (closure under finite unions)$E_1,\dots ,E_n\in \mathcal{A}⟹\bigcup _{k=1}^n{E}_k\in \mathcal{A}$

We say an algebra is a $\sigma -$algebra (sigma-algebra) if also
3. (closure under countable unions) $\{E_n{\}}_{n=1}^{\mathrm{\infty }}\in \mathcal{A}⟹\bigcup _n{E}_n\in \mathcal{A}$

By DeMorgan's Laws, if $E_1,\dots ,E_n\in \mathcal{A}$, then we also have

Thus, if $E\in \mathcal{A}$ then

Similarly, if $\mathcal{A}$ is a sigma-algebra, then $\{E_n{\}}_n\in \mathcal{A}$

We'll show soon that

is a sigma-algebra.

Stupid examples:

• $\mathcal{A}=\{\varnothing ,\mathbb{R}\}$
• $\mathcal{A}=\mathcal{P}(\mathbb{R})$
  Less stupid example:

• obviously, this is closed under complements by definition
• If $\{E_n{\}}_n\subset \mathcal{A}$ with all $E_n$ countable, then the union $\bigcup _n{E}_n$ is a countable union of countable sets, which is countable. Thus it is in $\mathcal{A}$
• If there exists some $N_0$ such that $E_{N_0}^c$ is countable (but $E_{N_0}$ is not), then ${\left(\bigcup _n{E}_n\right)}^c=\bigcap _n{E}_n^c\subset E_{N_0}^c$ is an intersection of sets, one of which is countable $⟹(\bigcup _n{E}_n{)}^c\in \mathcal{A}$

Let

Define the intersection of all such sigma-algebras as

This is the smallest $\sigma -$algebra containing all open subsets of $\mathbb{R}$ and it is called the Borel $\sigma -$algebra

($\mathcal{B}\in \mathrm{\Sigma }$ and for all $\mathcal{A}\in \mathrm{\Sigma }$, we have $\mathcal{B}\subset \mathcal{A}$)

Just need to show that $\mathcal{B}$ is indeed an $\sigma -$algebra.

• Since every open subset is an element of each $\mathcal{A}\in \mathrm{\Sigma }$, we conclude that each is then an element of $\mathcal{B}$.
• And because it is the intersection of all other $\sigma -$algebras in $\mathrm{\Sigma }$, it must indeed be the smallest one.
  Suppose $E\in \mathcal{B}$.
• Then for all $\mathcal{A}\in \mathrm{\Sigma },E\in \mathcal{A}$.
• So for all $\mathcal{A}$, we have $E^c\in \mathcal{A}⟹E^c\in \mathcal{B}$
• Similarly, for any $\{E_n{\}}_n$ such that $E_n\in \mathcal{B}$ for all $n$, we must have $E_n\in \mathcal{A}$
• Then (because they are sigma-algebras) we also have $\bigcup _n{E}_n\in \mathcal{A}$ for all $\mathcal{A}⟹\bigcup _n{E}_n\in \mathcal{B}$