Tutorial Note 8 (Lebesgue inner, outer measures and Lebesgue measure)

Remark. We have gone through the first 3 examples in this tutorial note, after that I put a few remarks in this blog to those examples, interested students may read them.

Remark. In class we also (eeeextremely quickly) mentioned that the measurability so defined in the tutorial notes/lecture notes is equivalent to the following:

Theorem (Caratheodory). A set $E\subseteq \R$ is measurable if and only if the Caratheodory condition holds: for any set $X\subseteq \R$, \begin{equation}
\label{which we gen}
m^*(X)=m^*(X\cap E)+m^*(X\setminus E).
\end{equation}

That said, $E$ is measurable if and only if $E$ and $E^c:=\R\setminus E$ can be used to split the outer measure of any set. Note that due to the symmetry $E$ is measurable if and only if $E^c$ is measurable.

Therefore we may also define measurability as in the statement of above theorem. This formulation is elegant in two senses.

• First, we don't need to worry about the boundedness of $E$, $E$ is measurable whenever the Caratheodory holds.
• Second, this formulation is amenable to a generalization to so-called abstract outer measures and next another generalization to abstract measure spaces, they are found to be appropriate setting for doing measure theory on arbitrary set. 

Many useful measures come from ``abstract'' outer measure, say the Lebesgue measure, or more generally, the Lebesgue-Stieljes measures and the Hausdorff measures, and many more.

In Math3043 (204), if one is interested to audit in the next year, Dr Yan Min will spend one week on Lebesgue measure and delve into the theory of general measure with (\ref{which we gen}) as a starting point. This is good because the abstraction provides a unified way to define Lebesgue measure on $\R^n$, which is perhaps the most interesting and most comfortable space for most people.

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Example 1.  We note that \[
m(C_n)=2^n\times \frac{1}{3^n}.
\] As  $\{C_n\}$ is descending, we have \[
C=\capp_{n=1}^\infty C_n=:\limn C_n.
\] As $\bm{m(C_1)\leq 1<\infty}$, by Monotone Set Theorem we have \[
m(C)=m\brac{\limn C_n}=\limn m(C_n)=\limn \brac{\frac{2}{3}}^n=0.
\]

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Example 2. Recall that for any set $A,B\subseteq \R$ we have \[
A=(A\cap B)\sqcup (A\setminus B).
\] That is, the set $B$ and its complement $B^c$ can be used to split $A$, and vice versa. Therefore when $A,B$ are measurable, Countable Additivity of Lebesgue measure tells us \[
m(A)=m(A\cap B)+m(A\setminus B).
\] Replacing $A$ by $E_1\cup E_2$ and $B$ by $E_1$, we have
\begin{align*}
m(E_1\cup E_2)&= m((E_1\cup E_2)\cap E_1)+m((E_1\cup E_2)\setminus E_1)\\
&=m(E_1)+m(E_2\setminus E_1).
\end{align*}We repeat the process to get \[
m(E_2)=m(E_2\cap E_1)+m(E_2\setminus E_1),
\] combining them to eliminate $m(E_2\setminus E_1)$, we are done.

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Example 3. We use the condition on $E$ to obtain set containment. Let $e\in E$, then $\exists q\in \Q$, $e+q\in C$, i.e., $\exists q\in \Q$, $e\in C-q$, so $e\in \cupp_{q\in \Q}(C-q)$. This is true for each $e\in E$, thus \[
E\subseteq \cupp_{q\in \Q}(C-q).
\] By Subadditivity of outer measure we have \begin{equation}\label{subadd translate}
m^*(E)\leq\sum_{\displaystyle\underbrace{\color{blue}q\color{blue}\in\color{blue} \Q}_{\textbf{countable}}}m^*(C-q)=\sum_{q\in \Q}m(C-q) =\sum_{q\in \Q}m(C)=0.
\end{equation} Therefore $m^*(E)=0$ and hence $E$ is measurable with $m(E)=0$.

Remark. In (\ref{subadd translate}) we cannot drop the $*$ throughout the computation because at that moment we still do not know whether the set $E$ is measurable, in fact we don't have an explicit formula for $E$. More precisely, any subset of $\cupp_{q\in \Q}(C-q)$ can be chosen to be the ``$E$'' in this example.

Remark. In (\ref{subadd translate}) we have used that $m^*(C-q)=m(C-q)=m(C)$, in fact we can also say that \[
m^*(C-q)=m^*(C)=m(C)
\] since $m^*$ is also translation invariant, no matter the set $C$ itself is measurable or not. In fact, $m$ is translation invariant due to the fact that $m^*$ does.

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Example 4. Easy to see that $H_1=\cupp_{n=1}^\infty E_n$. Since $H_1$ is a countable union of measurable sets, $H_1$ is measurable. Subadditivity of Lebesgue measure yields \[
m(H_1)\leq \sum_{n=1}^\infty m(E_n)= \sum_{n=1}^\infty 0=0,
\] so $m(H_1)=0$.

Next consider $H_2$. We note that \begin{align*}
x\in H_2&\iff x\in H_i,\exists !\, i\\
&\iff \exists i, \forall j\neq i, x\in H_i,x\not\in H_j,
\end{align*} so we have \[
H_2=\cupp_{i=1}^\infty \capp_{j=1\atop j\neq i}^\infty (H_i\setminus H_j)=\underbrace{\cupp_{i=1}^\infty \bigg(\overbrace{H_i\setminus \overbrace{\cupp_{j=1\atop j\neq i}^\infty H_j}^{\text{measurable}}}^{\text{measurable}}\bigg)}_{\text{measurable}}.
\] Finally, since $H_2\subseteq H_1$, we have $m(H_2)=0$ by either Subadditivity or Monotonicity of Lebesgue measure.

Remark. You may find the proof of measurability of $H_2$ complicated since we already know that $H_2\subseteq H_1$ and $m(H_1)=0$ (this implies $H_2$ is measurable with measure 0), that's right. The emphasis in the solution of this example is the way to rewrite the expression of $H_2$.

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Example 5. (Uploaded on 26/11/2013) We understand union (due to the Subadditivity) more than intersection, so let's translate the quantity in the following way: \[
1-m\bigg(\capp_{i=1}^k E_i\bigg)=m\bigg([0,1]\setminus \capp_{i=1}^k E_i\bigg)=m\bigg(\cupp_{i=1}^k([0,1]\setminus  E_i)\bigg).
\] By Subadditivity we have \[
1-m\bigg(\capp_{i=1}^k E_i\bigg)\leq \sum_{i=1}^k m([0,1]\setminus  E_i)=\sum_{i=1}^k(1-m(E_i))=k-\sum_{i=1}^k m(E_i).
\] Since $\sum_{i=1}^k m(E_i)>k-1$, we obtain \[
1-m\bigg(\capp_{i=1}^k E_i\bigg)<k-(k-1)=1,
\]  therefore \[
m\bigg(\capp_{i=1}^k E_i\bigg)>0.
\]

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Example 6. Let \[P_n=\{0.a_1\dots a_n \dots_{}\in [0,1]:a_1,\dots, a_n\text{ are prime}\}.\]
Then we have \[P=\capp_{n=1}^\infty P_n,\] to show $P$ is measurable, it is enough to show each $P_n$ is measurable.

Let's denote $\mathcal P=\{2,3,5,7\}$ the set of primes in $[0,9]$, for simplicity.  Note that \begin{align}
\nonumber P_n&=\cupp_{(a_1,\dots,a_n)\in \mathcal P^n} \{0.a_1\dots a_n\dots_{} :a_{n+1},a_{n+2},\dots  \in \{0,\dots,9\}\}\\
\label{this step disjoint}&= \cupp_{(a_1,\dots,a_n)\in \mathcal P^n}[0.a_1\dots a_n,0.a_1\dots a_n +10^{-n}].
\end{align}
So each $P_n$ is a union of finitely many intervals, $P_n$ is measurable, so is $P$.

Let's compute $m(P)$. Since $P_n$ is descending, we have \[
P=\capp_{n=1}^\infty P_n=:\limn P_n.
\] Since $\bm{m(P_1)<\infty}$, by Montone Set Theorem we obtain \[m(P)=m\brac{\limn P_n}=\limn m(P_n).\] On the other hand, by Subadditivity we have \[
m(P_n)\leq \sum_{(a_1,\dots,a_n)\in \mathcal P^n} m([0.a_1\dots a_n,0.a_1\dots a_{n}+10^{-n}])=\sum_{(a_1,\dots,a_n)\in \mathcal P^n} \frac{1}{10^n}=\frac{4^n}{10^n}=\frac{2^n}{5^n}.
\] So $\limn m(P_n)=0$, i.e., $m(P)=0$.

Remark. Actually the union in (\ref{this step disjoint}) is almost disjoint such that $m(P_n)=\frac{2^n}{5^n}$. Indeed, when \[(a_1,\dots,a_{n-1}),(b_1,\dots,b_{n-1})\in \mathcal P^{n-1}\] are distinct, no matter what $a_n,b_n\in \P$ are, the following \[
[0.a_1\dots a_n, 0.a_1\dots a_n+10^{-n}]\quad \text{and} \quad [0.b_1\dots b_n, 0.b_1\dots b_n +10^{-n}]
\] are disjoint (imagine that if $k=1,\dots ,n-1$ is the first index that $a_k\neq b_k$, let's say $a_k<b_k$, then every element of the form $0.b_1\dots b_k\dots$ are strictly bigger than $0.a_1\dots a_k\dots$ since $0.a_1\dots a_k999\dots$ never appears). 

The ``trouble'' arises when $a_1=b_1,\dots,a_{n-1}=b_{n-1}$, in this case \[ [0.a_1\dots a_n, 0.a_1\dots a_n+10^{-n}]\cap [0.b_1\dots b_n, 0.b_1 \dots b_n +10^{-n})]\neq \emptyset\] happens when $a_n=2$ and $b_n=3$, this intersection is a singleton of measurable zero, therefore contribute no measure to the whole union, and we have $m(P_n)=\frac{2^n}{5^n}$.