Measure

Measure and integration theory develops the necessary tool for solving PDEs and stochastic processes.

$\sigma$ -Algebra¶

Motivation¶

In physics, the task of finiding “volume” is ubquitous. For instance in computing mass, charge, the integral would resemble:

\int (\cdot ) dV

(1)

In some sense, this $dV$ is a function $dV:E\to \mathbb{R}$ which computes the volume of $E\subset \mathbb{R}^n$ , a tiny control volume we are interested in. It would be very natural to ask for this function to satisfy:

The question would be, when can we find such a function. Trivially we can obtain such property by sending every $A$ to 0. However, for more practical purposes of modeling the world, we would like nicer properties. In particular, let’s consider this compeletly reasonable demand on $\mathbb{R}^n$ .

Exercise 1 (Open Set as a Topology ^[1])

Let $\mathcal{B}$ be the family of all open intervals in $\mathbb{R}$ , i.e. subsets of the form
$(a, b) := \{c \in \mathbb{R} \mid a < c < b\}$
(4)
for all $a, b \in \mathbb{R}$ with $a < b$ . Let $\mathcal{T}(\mathcal{B})$ be the family of all arbitrary unions of elements in $\mathcal{B}$ , i.e.,
$\mathcal{T}(\mathcal{B}) := \left\{ \bigcup_{B \in \mathcal{B}'} B \mathrel{\Big\vert{}} \mathcal{B}' \subseteq \mathcal{B} \right\}$
(5)
Prove that $\mathcal{T}(\mathcal{B})$ is a topology on $\mathbb{R}$ . (This “ $\mathcal{B}$ ” stands for a “base” = “basis” of a topology.)
For $\mathcal{B}$ and $\mathcal{T}(\mathcal{B})$ as above and any $U \in \mathcal{T}(\mathcal{B})$ (i.e., any open set), prove that there exists a countable family $\mathcal{B}'' \subseteq \mathcal{B}$ (of open intervals) such that $U = \bigcup_{B \in \mathcal{B}''} B$ . (Hint: it might help to prove first that the union of any family of open intervals containing a given $x \in \mathbb{R}$ is a generalized open interval in the sense that it is of the form $(a, b)$ , $(-\infty, a)$ or $(a, \infty)$ for some $a, b \in \mathbb{R}$ . Or do this any other way.)
For $\mathcal{T}(\mathcal{B})$ as above, prove that $\mathcal{T}(\mathcal{B}) \subseteq \mathcal{B}_{\mathbb{R}}$ , the Borel $\sigma$ -algebra on $\mathbb{R}$ . (This “ $\mathcal{B}$ ”, confusingly, stands for “Borel”.)
Use the above to give a full proof of Proposition 1.2 on p. 22 in [F].
Also deduce that $\mathcal{M}(\mathcal{B}) = \mathcal{M}(\mathcal{T}(\mathcal{B})) = \mathcal{B}_{\mathbb{R}}$ .

Solution to Exercise 1

We prove that $\mathcal{T}(\mathcal{B})$ is a topology on $\mathbb{R}$ .

Property 1: $\emptyset, \mathbb{R} \in \mathcal{T}(\mathcal{B})$
- $\emptyset = \cup_{B \in \emptyset} B$ (empty union), so $\emptyset \in \mathcal{T}(\mathcal{B})$
- $\mathbb{R} = (-\infty, \infty)$ is a generalized open interval, hence $\mathbb{R} \in \mathcal{T}(\mathcal{B})$ . More explicitly, we can cover it by $\cup_{a=0}^{\infty} \{(a,a+1),(-a-1,-a)\}$
Property 2: Closed under arbitrary unions
- Let $\mathcal{U} \subseteq \mathcal{T}(\mathcal{B})$ be an arbitrary collection of sets in $\mathcal{T}(\mathcal{B})$ . For each $U \in \mathcal{U}$ , we have $U = \cup_{B \in \mathcal{B}_U} B$ for some $\mathcal{B}_U \subseteq \mathcal{B}$ . Then $\cup_{U \in \mathcal{U}} U = \cup_{U \in \mathcal{U}} \left(\cup_{B \in \mathcal{B}_U} B\right) = \cup_{B \in \cup_{U \in \mathcal{U}} \mathcal{B}_U} B$ . Since $\cup_{U \in \mathcal{U}} \mathcal{B}_U \subseteq \mathcal{B}$ , we have $\cup_{U \in \mathcal{U}} U \in \mathcal{T}(\mathcal{B})$
Property 3: Closed under finite intersections
- By induction on the number $n$ of sets.
- Base case ( $n=1$ ): Trivial, $U \in \mathcal{T}(\mathcal{B})$ .
- Inductive step: Assume $U_1 \cap \cdots \cap U_n \in \mathcal{T}(\mathcal{B})$ for any $n$ sets in $\mathcal{T}(\mathcal{B})$ . For $n+1$ sets, write $(U_1 \cap \cdots \cap U_n) \cap U_{n+1}$ . By induction hypothesis, $U_1 \cap \cdots \cap U_n = \cup_{i \in I}(a_i, b_i)$ for some countable collection of open intervals. Then
  $(U_1 \cap \cdots \cap U_n) \cap U_{n+1} = \left(\cup_{i \in I}(a_i, b_i)\right) \cap U_{n+1} = \cup_{i \in I}((a_i, b_i) \cap U_{n+1})$
  (6)
  Each $(a_i, b_i) \cap U_{n+1}$ is an intersection of an open interval with a union of open intervals, which is a finite union of open intervals (or empty). Hence $(U_1 \cap \cdots \cap U_{n+1}) \in \mathcal{T}(\mathcal{B})$ .

Therefore $\mathcal{T}(\mathcal{B})$ is a topology on $\mathbb{R}$ .

Suppose $\mathcal B' \subseteq \mathcal B$ and
$U=\bigcup_{B\in \mathcal B'} B$
(7)
where $\mathcal B'$ can be uncountable. We would like to show $\exists \mathcal B''$ that is countable such that
$U=\bigcup_{B\in \mathcal B^{''}} B$
(8)
We now show a specfic way to construct $\mathcal B''$ from $\mathcal B'$ . Define:
$\mathcal B_x = \{(a,b)\in \mathcal B' : x\in (a,b)\}$
(9)
Clearly $\mathcal B_x \subseteq \mathcal B'$ . Now proceed with the following procedure(an “infinite algorithm”)on the construction of $\mathcal B''$

Algorithm 1

Input: A potentially uncountable set of open intervals $\mathcal B'$ such that $\cup \mathcal B'=U$

Output: A countable set of open intervals $\mathcal B''$ such that $\cup \mathcal B''=U$

$\mathcal B'' = \varnothing$
While $\exists x\in U$ , $x\notin \cup \mathcal B''$ :
1. Compute $(c,d)=\bigcup_{B\in \mathcal B_x} B$ where $c,d$ is in extended real number system.
2. If $c,d\in \mathbb{R}$ :
  1. $\mathcal{B}_x'=\{(c,d)\}$
3. Else if $c\in \mathbb{R}, d=\infty$ :
  1. $\mathcal{B}_x'\gets \{(c,[c])\}\cup \{[c],[c]+j\}_{j=1}^{\infty}$
4. Else Other cases follows as (3)
5. $\mathcal B'' = \mathcal B'' \cup \mathcal B_x '$
Return $\mathcal B''$

where $[c]$ picks the closes integer towards the right of the real axis. Note we implicitly invoked the claim below in computing $(c,d)$ :

$\boxed{\textbf{Cliam 1}}$ The union of arbitrary family of open intervals containing $x\in \mathbb{R}$ is a generalized open interval
$\bigcup_{B\in \mathcal B_x} B\in \{(c,d): c,d\in \mathbb{R}\cup \{\infty, -\infty\}\}$
(10)
- I claim the union is just $(c,d)$ where $c=\inf \{a:(a,b)\in \mathcal B_x\}$ and $d=\sup \{b:(a,b)\in \mathcal B_x\}$ where we allow $\infty$ and $-\infty$ . Clearly $(c,d)$ covers $\cup \mathcal B_x$ . It suffice to show $\cup \mathcal B_x$ covers $(c,d)$ . Suppose $\exists y \in (c,d)$ and $y\notin \cup \mathcal B_x$ . This means $\cup \mathcal B_x$ has a hole and there must exists a pair of open interval in $\mathcal B_x$ that is disjoint, contradicting that all open intervals in $\mathcal B_x$ contains $x$ (in that case no pair should be disjoint).

Now what’s left to show is (1) Only countable number of $\mathcal B_x'$ is needed. (2) each $\mathcal B_x'$ is countable. The later is obvious by construction. The former can be argued by countability of $\mathbb{Q}$ . Pick a $q\in \mathbb{Q}$ in one of the open interval of $\mathcal B_x'$ , this is always possible by density of rationals. Suppose uncountable $\mathcal B_x'$ is needed to cover $\cup \mathcal B''$ then $U=\cup \mathcal B''$ would contain uncountable number of rationals. This finishes the proof.

By definition, $\mathcal B_{\mathbb{R}}$ is the smallest $\sigma$ -algebra containing $\mathcal T(\mathcal B)$ . Therefore the inclusion holds by construction.

We only need to show (b), (c), and (e) since (a) and (d) are generalized open intervals and are actually shown by (5).

Case (b): Closed intervals $\mathcal{E}_2 = \{[a,b]: a < b\}$

$\mathcal{M}(\mathcal{E}_2) \subseteq \mathcal{B}_\mathbb{R}$ : Note that $[a,b] = \cap_{j=1}^{\infty}(a-1/j, b+1/j)$ is a countable intersection of open intervals, so $\mathcal{E}_2 \subset \mathcal{B}_\mathbb{R}$ .
$\mathcal{B}_\mathbb{R} \subseteq \mathcal{M}(\mathcal{E}_2)$ : $(a,b) = \cup_{j=1}^{\infty}[a+1/j, b-1/j]$ is a countable union of closed intervals, so $\mathcal{B} \subset \mathcal{M}(\mathcal{E}_2)$ and hence $\mathcal{B}_\mathbb{R} = \mathcal{M}(\mathcal{B}) \subset \mathcal{M}(\mathcal{E}_2)$ .

Case (c): Half-open intervals $\mathcal{E}_3 = \{(a,b]: a < b\}$

$\mathcal{M}(\mathcal{E}_3) \subseteq \mathcal{B}_\mathbb{R}$ : $(a,b] = (a,b) \cup \{b\}$ where $(a,b) \in \mathcal{B}_\mathbb{R}$ and $\{b\} = \cap_{j=1}^{\infty}(b-1/j, b+1/j) \in \mathcal{B}_\mathbb{R}$ .
$\mathcal{B}_\mathbb{R} \subseteq \mathcal{M}(\mathcal{E}_3)$ : $(a,b) = \cup_{j=1}^{\infty}(a, b-1/j]$ is a countable union of half-open intervals, so $\mathcal{B} \subset \mathcal{M}(\mathcal{E}_3)$ and hence $\mathcal{B}_\mathbb{R} \subset \mathcal{M}(\mathcal{E}_3)$ .

Case (e): Closed rays $\mathcal{E}_7 = \{[a,\infty): a \in \mathbb{R}\}$

$\mathcal{M}(\mathcal{E}_7) \subseteq \mathcal{B}_\mathbb{R}$ : $[a,\infty) = \cap_{j=1}^{\infty}(a-1/j, \infty)$ is a countable intersection of open rays (generalized open intervals), so $\mathcal{E}_7 \subset \mathcal{B}_\mathbb{R}$ .
$\mathcal{B}_\mathbb{R} \subseteq \mathcal{M}(\mathcal{E}_7)$ : $(a,b) = (a,\infty) \cap (-\infty,b)$ where $(a,\infty) = \cup_{j=1}^{\infty}[a+1/j,\infty)$ and $(-\infty,b) = \mathbb{R} \setminus [b,\infty)$ are in $\mathcal{M}(\mathcal{E}_7)$ , so $\mathcal{B} \subset \mathcal{M}(\mathcal{E}_7)$ and hence $\mathcal{B}_\mathbb{R} \subset \mathcal{M}(\mathcal{E}_7)$ .

Cases (c’), (e’) follow by symmetry.

We only need to show
$\begin{align*} \mathcal B &\subseteq \mathcal M(\mathcal T(\mathcal B)) \\ \mathcal T(\mathcal B) &\subseteq \mathcal M( \mathcal B) \end{align*}$
(11)

The former is trivial since $\mathcal B \subseteq \mathcal T(\mathcal B)\subseteq \mathcal M(\mathcal T(\mathcal B)$
The later can be shown using (2). $\mathcal M(\mathcal B)$ contains all countable unions of open intervals $\mathcal B$ while $\mathcal T(\mathcal B)$ contains arbitrary union. (2) bridges the gap by claiming the arbitrary union can actually be replaced by countable union.

Construction of Measure¶

The big picture¶

The general idea of constructing measure is, in crude terms:

Find a collection of set that has a clear notion of size
Approximate arbitrary subsets using element in the elementary set, hence get a notion of size on arbtirary subsets
Make the notion of size truely a measure by restricting to a special collection of subset.

Observe how the notion of size ripens:

\begin{align*} \text{Stage 1: $\boxed{|\cdot|:\mathcal E \to \mathbb{R}^+}$} &\longrightarrow \text{Stage 2: $\boxed{\mu^* : P(X)\to \mathbb{R}^+}$} \\ &\longrightarrow \text{Stage 3: $\boxed{\mu: \mathcal M|_{\mu^*} \to \mathbb{R}^{+}}$} \end{align*}

(12)

Stage 1 to 2: Create a notion of size on arbitrary sets $A\subset P(X)$ by approximating them using “building blocks” $(A\in \mathcal E)$ which has a notion of size $|\cdot |$ . This scarifices additivity and we instead of countable sub-additivity
Stage 2 to 3: Shrink the domain by restrict to a particular family $\mathcal M |_{\mu^*}\subset P(X)$ . This gives us countable additivity and $\mu$ is now a complete measure

Example 1

Concretely, think $\mathcal E$ as the set of all open rectangles in $\mathbb{R}^2$ :

\mathcal E= \{(a,b)\times (c,d) | a<b,c<d; a,b,c,d\in \mathbb{R}\}

(13)

These clearly “canonical” notion of size, namingly:

|A|=(d-c)(b-a)

(14)

Now using these as “building blocks”, we can measure arbitrary sets $E\in P(\mathbb{R}^2)$ , for instance, a unit circle. Here is a quick numerical demo for approximating a unit circle denoted by $E$ with squares on a grid. The resolution $n$ is defined as the number of intervals in $(0,1)$ .

As we take finer and finer squares, we approach a limit(or the liminf). This number can be used as a notion of size for the set $E$ . This motivates the definition of outer measure.

How did we compute that?

A simple way is to count grid points that is inside the unit circle. A quick julia implementation gives the above plot:

function compute_area_of_circle(n)
    cover::Int = 0
    for i in -n:n
        for j in -n:n
            if ((i/n)^2 + (j/n)^2) < 1
                if i == 0 && j == 0
                    cover += 4
                elseif i == 0 || j == 0 
                    cover += 2
                else 
                    cover += 1
                end 
            end
        end
    end
    return cover * (1/n)^2
end

Outer Measure¶

So how do we actually construct measures no matter what set we are working on?(Though in this book we are really interested in just $\mathbb{R}^n$ ). A general guidline is:

Find some elementary set $\mathcal{E}$ where premeasure is defined on the disjoint uniont of $\mathcal{E}$ .
Define the measure of arbitrary subset $A\in P(X)$ as the the size of “smallest” cover of $A$ using the elementary set:

\inf \left\{\sum_{\nu=1}^{\infty} \rho(E_{\nu}): E_{\nu}\in \mathcal{E}, A\subset \bigcup_{\nu=1}^{\infty} E_{\nu}\right\}

(15)

More generally we can define an outer measure as follows:

We note that property (2),(3) are both derived properties of measure. So we can think the definition of outer measure as a relaxation of measure: take out countable additvity and adds back two derived property. In addition, We note that (15) obeys Definition 2

Carathéodory Family $\mathcal M_{\mu^*}$ ¶

How do we then obtain measure from outer measure? We note that what outer measure lacks is finite additivity, so we can further ask, is there a collection of subset $\mathcal {M}\subset P(X)$ such that $\mu^*$ is actually finitey additive? The answer is actually YES! However this seems rather arbitrary:

\mathcal M = \{A\subset X: \mu^*(F)=\mu^*(F-A)+\mu^*(A\cap F), \forall F \subset X\}

(16)

We call elements of $\mathcal M$ $\mu^*$ -measureable. In plain English, this saying $A$ can “split measure” arbitrary $F\subset X$ by measuring the intersection and the complement separetly and add them together.

Premeasure and The Extension Theorem¶

THe workflow (12) is already very promising except one key issue:

The next theorem prevents this by starting with an algebra $\mathcal A$ instead of just a plain family of sets $\mathcal E$ . The plain notion of size on $\mathcal E$ is replaced by the so-called premeasure.

Solution to Exercise 2

Proof 1

Note that $\mu(N)=0\Longrightarrow \mu(N')=0$ . $\forall E\subseteq X$ , by subadditivity:

\mu^*(E) \leq \mu^*(E\cap N') + \mu^*(E\cap N'^c)

(20)

By monotoncity and positivity of $\mu^*$ we have $\mu^*(E\cap N')=0$ (since $E\cap N' \subseteq N'$ so $\mu^*(E\cap N')\leq 0$ ). But again by monotonicity $\mu^*(E\cap N'^c) \leq \mu^*(E)$ hence we have:

\mu^*(E) \leq \mu^*(E\cap N') + \mu^*(E\cap N'^c) \leq \mu^*(E)

(21)

Hence:

\mu^*(E) = \mu^*(E\cap N') + \mu^*(E\cap N'^c)

(22)

As desired.

Exercise 3

Let $\mathcal{A} \subset \mathcal{P}(X)$ be an algebra, $\mathcal{A}_\sigma$ the collection of countable unions of sets in $\mathcal{A}$ , and $\mathcal{A}_{\sigma\delta}$ the collection of countable intersections of sets in $\mathcal{A}_\sigma$ . Let $\mu_0$ be a premeasure on $\mathcal{A}$ and $\mu^*$ the induced outer measure.

For any $E \subset X$ and $\epsilon > 0$ there exists $A \in \mathcal{A}_\sigma$ with $E \subset A$ and $\mu^*(A) \le \mu^*(E) + \epsilon$ .
If $\mu^*(E) < \infty$ , then $E$ is $\mu^*$ -measurable iff there exists $B \in \mathcal{A}_{\sigma\delta}$ with $E \subset B$ and $\mu^*(B \setminus E) = 0$ .
If $\mu_0$ is $\sigma$ -finite, the restriction $\mu^*(E) < \infty$ in (2) is superfluous.

Solution to Exercise 3

1.This is just the definition of outer measure(namingly, infimum). Note since

\mu^*(E)= \inf C

(23)

where

C= \left\{\sum \mu_0(A_j) | A=\cup A_j\in \mathcal A_{\sigma}, E\subset A \right\}

(24)

Hence $\exists c\in C$ such that $c < \mu^*(E)+\epsilon$ by the definition of infimum. Now the map between open covers and their area is clearly surjective by construction:

\begin{align*} \{A\in \mathcal A_{\sigma} : E\subset A\} &\to \mathbb{R} \\ A &\mapsto \underbrace{\sum_j \mu_0(A_j)}_{\mu^*(A)} \end{align*}

(25)

hence we can find $A\in A_{\sigma}$ that correponds to $c$ .

$(\Longrightarrow)$ Suppose $E$ is $\mu^*$ -measurable. Now by (1), for each $n$ we can find $A_n\in \mathcal A_{\sigma}$ with $E\subset A_n$ and $\mu^*(A_n)\leq \mu^*(E)+\frac1n$ . The $A_n$ need not be decreasing, so replace them by $\tilde A_n:=\bigcap_{k=1}^n A_k$ ; since $\mathcal A_\sigma$ is closed under finite intersections (intersection distributes over the countable unions defining each $A_k$ ), $\tilde A_n\in\mathcal A_\sigma$ , and now $\tilde A_n$ is a decreasing sequence with $E\subset \tilde A_n\subset A_n$ , so still $\mu^*(E)\leq \mu^*(\tilde A_n)\leq \mu^*(A_n)\leq \mu^*(E)+\frac1n$ . Define:
$B:=\bigcap_{n}\tilde A_{n} = \bigcap_n A_n \in \mathcal A_{\sigma \delta}$
(26)
By construction $E\subset B$ . Since $\tilde A_n \in \mathcal A_\sigma \subset \mathcal M_{\mu^*}$ is a decreasing sequence with $\mu^*(\tilde A_1)\leq \mu^*(E)+1<\infty$ , continuity-from-above on $\mathcal M_{\mu^*}$ gives:
$\mu^*(B)=\lim_{n \to \infty}\mu^*(\tilde A_n)=\mu^*(E)$
(27)
Now use that $E$ is $\mu^*$ -measurable, testing it against $B$ : since $E\subset B$ ,
$\mu^*(B) = \mu^*(B\cap E)+\mu^*(B\cap E^c) = \mu^*(E) + \mu^*(B\setminus E)$
(28)
Combining with $\mu^*(B)=\mu^*(E)<\infty$ and subtracting the finite common value $\mu^*(E)$ from both sides gives $\mu^*(B\setminus E)=0$ , as desired.

$(\Longleftarrow)$ Suppose there exists $B\in \mathcal A_{\sigma \delta}$ such that $E\subset B$ and $\mu^*(B\setminus E)=0$ . Note any $B\in \mathcal A_{\sigma \delta}$ is $\mu^*$ -measurable since $\mathcal A \subset \mathcal M_{\mu^*}$ (by extension theorem). We would like for all $F\subset X$

\mu^*(F) \geq \mu^*(F\cap E) +\mu^*(F \cap E^c)

(29)

But now $E\subset B$ hence:

\begin{align*} \mu^*(F) &= \mu^*(F\cap B) + \mu^*(F\cap B^c) \\ &\geq \mu^*(F\cap E) + \mu^*(F\cap B^c) \end{align*}

(30)

Note all we need now is $\mu^*(F \cap B^c) \geq \mu^*(F\cap E^c)$ .

But now

\mu^*(F\cap E^c) = \mu^*(F\cap (B^c \cup (B\cap E^c))) \leq \mu^*(F\cap B^c) + \mu^*(F\cap (B\cap E^c))

(31)

Again, by monotonicity and the assumption

\mu^*(F\cap (B\cap E^c))\leq \mu^*(B\cap E^c) = 0

(32)

Therefore $\mu^*(F \cap B^c) \geq \mu^*(F\cap E^c)$ . As desired.

Suppose $\mu_0$ is $\sigma$ -finite then $\exists \{X_n\}$ , $\mu_0(X_n)<\infty$ such that $X = \cup_n X_n$ . Now define $E_n= X_n\cap E$ , we can apply (2) to each $E_n$ (There existss $B_n\in \mathcal A_{\sigma\delta}$ satisfying the property iff $E_n\in \mathcal M$ ).

( $\Longrightarrow$ ) Say we already have desired set $B_n$ for each $E_n$ (each $B_n\in \mathcal A_{\sigma\delta}$ , $E_n\subset B_n$ , $\mu^*(B_n\setminus E_n)=0$ )^[2]. Take the union $B=\cup_n B_n$ ; it covers $E$ since $E_n \subset B_n$ and $E=\cup_n E_n$ . It also satisfies the null-excess property: since $E_n \subset E$ we have $B_n\setminus E \subset B_n \setminus E_n$ for each $n$ , so $B\setminus E = \bigcup_n(B_n\setminus E) \subset \bigcup_n(B_n\setminus E_n)$ , and by countable subadditivity:

\mu^*(B\setminus E) \leq \sum_n \mu^*(B_n\setminus E_n) = \sum_n 0 = 0

(33)

( $\Longleftarrow$ ) Suppose there is such $B$ for $E$ (possibly of infinite outer-measure). Define $E_n = X_n\cap E$ as before, and now also $B_n := X_n \cap B \in \mathcal A_{\sigma\delta}$ (intersecting with $X_n \in \mathcal A$ preserves membership in $\mathcal A_{\sigma\delta}$ ). Then $E_n \subset B_n$ , and

\mu^*(B_n\setminus E_n) = \mu^*\big(X_n\cap(B\setminus E)\big) \leq \mu^*(B\setminus E) = 0

(34)

Since $\mu^*(E_n)\leq \mu_0(X_n)<\infty$ , part (2) applies to $E_n$ and gives $E_n \in \mathcal M$ . Now we want to show $E \in \mathcal M$ but this is trivial as $\mathcal M$ is a $\sigma$ -algebra closed under countable union and $E= \cup_n E_n$

Solution to Exercise 4

$(\Longrightarrow)$ Suppose $E$ is $\mu^*$ -measurable. Then so is $E^c$ , and since $X=E\sqcup E^c \in \mathcal A$ , additivity gives $\mu_0(X)=\mu^*(X)=\mu^*(E)+\mu^*(E^c)$ , i.e. $\mu^*(E)=\mu_*(E)$ .

$(\Longleftarrow)$ Suppose $\mu^*(E)+\mu^*(E^c)=\mu_0(X)=\mu^*(X)<\infty$ . Fix $\epsilon>0$ ; by Exercise 3 part (1), pick $A,A'\in \mathcal A_\sigma\subset \mathcal M_{\mu^*}$ with $E\subset A$ , $E^c\subset A'$ , $\mu^*(A)\leq \mu^*(E)+\epsilon$ , $\mu^*(A')\leq \mu^*(E^c)+\epsilon$ . Then $A\cup A'=X$ , so

\mu^*(X)+\mu^*(A\cap A') = \mu^*(A)+\mu^*(A') \leq \mu^*(X)+2\epsilon \implies \mu^*(A\cap A')\leq 2\epsilon

(35)

Since $A\setminus E \subset A\cap A'$ , $\mu^*(A\setminus E)\leq 2\epsilon$ . Taking $\epsilon=\frac1n$ gives $A_n\supset E$ with $\mu^*(A_n\setminus E)\leq \frac2n$ ; let $B:=\bigcap_n A_n\in \mathcal M_{\mu^*}$ . Then $E\subset B$ and $\mu^*(B\setminus E)\leq \mu^*(A_n\setminus E)\to 0$ , so $\mu^*(B\setminus E)=0$ .

This is exactly the hypothesis of the $(\Longleftarrow)$ direction of Exercise 3 part (2) (only measurability of $B$ was used there, not $B\in\mathcal A_{\sigma\delta}$ ), which concludes $E$ is $\mu^*$ -measurable. As desired.

The following exercise is assumed everywhere in mathematics and is not directly related to outer measure. But they are invokded multiple times, for instance, in subadditivity of outer measure.

Exercise 5 (Order invariance for Series of Positive Terms ^[1])

Prove that any generalized series with summands in $[0, \infty]$ can be summed in any order. Specifically, if $a_j \in [0, \infty]$ for $j \in \mathbb{N} = \{1, 2, 3, \dots\}$ , then for any bijection $\mathbb{N} \to \mathbb{N},\ j \mapsto k_j$ , the equality $\sum_{j=1}^\infty a_j = \sum_{j=1}^\infty a_{k_j}$ holds. (That is, a nonnegative generalized series can be summed in any order.)
Prove that any two generalized series with summands in $[0, \infty]$ can be summed as one series. Specifically, if $a_j, b_j \in [0, \infty]$ for $j \in \mathbb{N}$ , then the equality $\sum_{j=1}^\infty a_j + \sum_{j=1}^\infty b_j = \sum_{j=1}^\infty (a_j + b_j)$ holds.

Solution to Exercise 5

We first make a claim
$\sum_{j\in \mathbb{N}}a_j = \sup_J \sum_{j\in J} a_j$
(36)
where $J\subset \mathbb{N}, |J|<\infty$ .

To show $\sum_{j\in \mathbb{N}} a_j \leq \sup \sum_{j\in J} a_j$ :
$\sum_{j\in \mathbb{N}}a_j=\lim_{N\to\infty} \sum_{j=1}^N a_j \leq \sup \sum_{j\in J} a_j$
(37)
To show $\sum_{j\in \mathbb{N}} a_j \geq \sup \sum_{j\in J} a_j$ , we first note $\forall J$ as above:
$\sum_{j\in \mathbb{N}} a_j = \sup_m s_m \geq s_{\max J} \geq \sum_{j\in J} a_j$
(38)
Now $\sup$ perserves non-strict inequality so we can take $\sup$ on both sides to get:
$\sup_{J} \sum_{j\in \mathbb{N}} a_j = \sum_{j\in \mathbb{N}} a_j \geq \sup_J \sum_{j\in J} a_j$
(39)
as desired. Now (1) becomes trivial since $\forall J$ defined as above, suppose we define:
$k(J)=\{k(j): j\in J\}$
(40)
now by the claim
$\sum_{j\in \mathbb{N}} a_{k(j)} = \sup_{J} \sum_{j\in J'} a_j = \sup_{J} \sum_{j\in k(J)} a_j$
(41)
where the last equality follows as $J'$ is also in the set $\{J\subset N: |J|<\infty\}$ .

We first make a claim: For monotone non-decreasing sequences $s_n, t_n$ :
$\sup(s_n + t_n) = \sup s_n + \sup t_n$
(42)

$\sup(s_n + t_n) \leq \sup s_n + \sup t_n$ is true since $\forall m, s_m + t_m \leq \sup s_n + \sup t_n$ and taking $\sup_m$ on both sides get us desired.
$\sup(s_n + t_n) \geq \sup s_n + \sup t_n$ can be shown by arguing $\forall \epsilon > 0$ ,
$\sup(s_n + t_n) \geq \sup s_n + \sup t_n - \epsilon$
(43)
Denote $s=\sup s_n, t = \sup t_n$ . By definition of supremum, we can pick $N_1, N_2$ such that $s_{N_1} > s - \epsilon/2$ and $t_{N_2} > t - \epsilon/2$ . Let $N=\max(N_1,N_2)$ ; since $s_n, t_n$ are non-decreasing, $\forall n \geq N$ we have $s_n \geq s_{N_1} > s-\epsilon/2$ and $t_n \geq t_{N_2} > t-\epsilon/2$ , hence:
$s_n + t_n > s + t - \epsilon$
(44)
Taking $\sup_n$ on the left gives $\sup(s_n+t_n) \geq s+t-\epsilon = \sup s_n + \sup t_n - \epsilon$ . As desired, since $\epsilon>0$ was arbitrary. (2) then easily follows by taking the claim in (1) and then let
$\begin{align*} s_n &=\sum_{j=1}^n a_j \\ t_n &=\sum_{j=1}^m b_j \end{align*}$
(45)
Apply the above claim gets desired.

Casestudy: Lebesgue-Stieltjes Measure¶

The most useful set that we would like to equip a measure with is, unarguably, the real numbers $\mathbb{R}$ .

Starting Point: The half-intervals¶

Recall that we would start with an elementary collection that has a clear notion of size. We define the h-intervals:

H=\{(a,b]:a<b, a,b\in \mathbb{R}\}

(46)

We define a terminology here:

Another terminology:

We define size on $H$ as follows:

The next step is to extend this notion of size to $\mathcal A=\mathcal M(H)$ .

Step 1: Construction of a premeasure¶

Now we have an algebra $\mathcal A$ where elements of the algebra $A\in \mathcal A$ has an elementary decomposition $A=\{I_j\}$ to h-intervals $I=(a_j,b_j]$ , which is equipped with a notion of size $b_j-a_j$ . A natural way to define the notion of size for $A$ is hence:

\mu_0(A) = \sum_{j=1}^N |I_j| =\sum_{j=1}^N F(b_j)-F(a_j)

(49)

We claim that $\mu_0$ is actually a pre-measure on $\mathcal A$ .

Solution to Exercise 7

Recall that:

m^*([a,b])=\inf\left\{\sum_{j=1}^{\infty} b_j - a_j {\Huge |} [a,b]\subseteq \bigcup (a_j, b_j] \right\}

(51)

For $\leq$ it suffices to find one cover $A=\cup (a_j,b_j]$ with $\sum b_j - a_j \leq b-a$ . We note $\forall \epsilon > 0$ , $(a-\epsilon/2,b]$ would cover $[a,b]$ and
$\inf\left\{\sum_{j=1}^{\infty} b_j - a_j {\Huge |} [a,b]\subseteq \bigcup (a_j, b_j] \right\} \leq b - a + \epsilon/2 \leq b-a -\epsilon$
(52)
where the first inequality follows since $(a-\epsilon/2, b)$ is in the set of $A$ we are taking infimum over.
For $\geq$ we need to show for all cover $A$ as above, $\forall \epsilon>0$ ,
$\epsilon + \sum_{j=1}^{\infty} (b_j - a_j) \geq b-a$
(53)
Let $A=\cup A_j, A_j=(a_j,b_j]$ be an arbitary, potentially infinite cover. Consider $A'_j=(a_j,b_j+\frac{\epsilon}{2^j})$ , clearly $[a,b]\subset \cup A'_j$ . By compactness, we can pick a finite subcover:
$[a,b]=\bigcup_{k=1}^n (a_{{j_k}},b_{{j_k}}+\delta_{j_k})$
(54)
where $\delta_{j_k} \leq \epsilon / 2^{j_k}$ . Denote $c_k = a_{j_k}, d_k = b_{j_k}+\delta_{j_k}$ . We have $[a,b]=\bigcup_{k=1}^n (c_k, d_k)$ . Note since implies $d_n>b, c_n < a$ . Without loss of generality, we can assume $d_{k+1}> c_k$ . Hence:
$\begin{align*} \sum_{k=1}^n (d_j - c_j) &= d_n - c_1 + \sum_{k=1}^{n-1}(d_k-c_{k+1}) \\ &\geq d_n - c_1 \\ &\geq b - a \end{align*}$
(55)
It remains to show
$\epsilon + \sum_{j=1}^{\infty} (b_j - a_j) \geq \sum_k d_k - c_k$
(56)
but
$\begin{align*} \sum_k (d_k - c_k) &\leq \sum_{k=1}^n (b_{j_k} + \epsilon / 2^{j_k}-a_{j_k}) \\ &\leq \epsilon + \sum_{k=1}^n (b_{j_k} -a_{j_k}) \\ &\leq \epsilon + \sum_{k=j}^{\infty} (b_{j} -a_{j}) \end{align*}$
(57)

Solution to Exercise 8

Suppose $E\subset \mathbb{R}$ , we show

m^*(cE + d) = |c|m^*(E)

(58)

First we isolate the special case when $c=0$ . In that case $cE+d = \{d\}$ and outer measure of singleton is 0 so the equality is trivial. Now we assume $c\neq 0$ . we claim that every cover $A=\cup A_j$ for $cE+d$ can be obtained by transforming a cover for $E$ . Specifically, suppose $A=\bigcup (a_j,b_j]$ is a cover for $cE+d$ then

B:= \begin{cases} \bigcup_j \left( \frac{a_j-d}{c}, \frac{b_j-d}{c}\right] & c > 0\\ \bigcup_j \left( \frac{b_j-d}{c}, \frac{a_j-d}{c}\right] & c < 0 \end{cases}

(59)

is a cover for $E$ , furthermore $cB+d=A$ . We consider the positive case first, the negative case follows similarily. Since for all $e\in E$ , $ce+d\in cE+d$ hence $\exists (a_t, b_t]$ such that $ce+d \in (a_t, b_t]$ . Hence $e \in \left( \frac{a_t-d}{c}, \frac{b_t-d}{c}\right]$ . Now

\begin{align*} m^*(E) &=\inf \left\{ \sum_j \frac{b_j - a_j}{c} {\Huge{|}} cE+d \subset \bigcup_j \left( \frac{a_j-d}{c}, \frac{b_j-d}{c}\right] \right\} \end{align*}

(60)

Hence $m^*(E)= m^*(cE+d)/c$ when $c>0$ . If $c<0$ , then by the same reason, we get:

\begin{align*} m^*(E) &=\inf \left\{ \sum_j \frac{a_j - b_j}{c} {\Huge{|}} cE+d \subset \bigcup_j \left( \frac{b_j-d}{c}, \frac{a_j-d}{c}\right] \right\} \end{align*}

(61)

hence $m^*(E)=-m^*(cE+d)/c$ . As desired.

Solution to Exercise 9

Consider $\mathbb{Q}\subset \mathbb{R}$ which is dense. Since $\mathbb{Q}$ is denumerable, consider an enumeration $Q = q_1,q_2,...$ . Consider the following construction(psuedo-algorithm):

where $B^o_{r}(x)$ denotes the open ball of radius $r$ around $x$ in $\mathbb{R}^d$ . Now clearly $A_{\epsilon}$ is dense since $\mathbb{Q}\subset A_{\epsilon}$ and

m(A_{\epsilon})\leq \sum_{j=1}^{\infty}\frac{\epsilon}{2^j } = \epsilon

(62)

The set is open because it is a countable union of open balls. Threrefore its complement is closed and has measure $\infty -\epsilon=\infty$ (by additivity of measure). It’s left to show that the complement is nowhere dense. We need $(\overline{\mathbb{R}\setminus A_{\epsilon}})^{\circ}=\varnothing$ , since $\mathbb{R}\setminus A_{\epsilon}$ is closed it suffices to show $\mathbb{R}\setminus A_{\epsilon}$ has empty interior, i.e., contains no open interval. Suppose $(a,b)\subset \mathbb{R}\setminus A_{\epsilon}$ , then $\forall j, B_{\epsilon/2^{j+1}}^o(q_j) \cap (a,b) = \varnothing$ . But rationals are dense in $\mathbb{R}$ hence there must exists $q\in (a,b), q\in \mathbb{Q}$ , contradicts.

Footnotes¶

This problem is selected from Professor Igor Mineyev’s homework from MATH 540(Real Analysis) in Fall 2026 at UIUC.
↩↩↩↩↩↩
Strictly, $\mathcal A_{\sigma\delta}$ need not be closed under countable unions (only finite ones) — the analogous fact for $G_\delta$ sets in $\mathbb R$ fails: $\mathbb Q=\bigcup_q \{q\}$ is a countable union of $G_\delta$ singletons yet is not itself $G_\delta$ . So $B=\cup_n B_n$ below should be read as living in $\mathcal M_{\mu^*}$ (a genuine $\sigma$ -algebra), which is all that is actually used afterward.
↩
Note that in definiting a notion of size $\mu_0$ on $A\in \mathcal A$ , we requires $A$ to have a concrete representation $\cup_j I_j$ . If the result is independent on how we choose this representation(i.e. different elementary decomposition $\{I_j\}$ ), then this notion of size is indeed fundamentally correct. Otherwise if $\mu_0$ is dependent on how we choose representation, then it is not a credibe notion of size. Such “representational invariance” is called well-definedness in mathematics.
↩