Category Archives: An introduction to measure theory

Proof: characterisation of Jordan measurability

The following proof is a solution to exercise 1.1.5 of the book “An introduction to measure theory” by Terence Tao.

Some notation will be established before proceeding with the exercise. Let \mathcal{E}(\mathbb{R}^d) be the set of elementary sets on \mathbb{R}^d. Denote by \mathcal{L}(E):=\left\{m(A):A\in\mathcal{E}(\mathbb{R}^d),A\subseteq E\right\} and by \mathcal{U}(E):=\left\{m(B):B\in\mathcal{E}(\mathbb{R}^d),E\subseteq B\right\} the sets of elementary measures of all elementary subsets and supersets of a bounded set E\subseteq\mathbb{R}^d, respectively. Let m_{*}(E)=\sup{\mathcal{L}(E)} and m^{*}(E)=\inf{\mathcal{U}(E)} be the Jordan inner and Jordan outer measures of E, respectively.

Exercise 1.1.5 requires to prove the equivalence of the following three statements, as a way of characterising Jordan measurability:

  1. E is Jordan measurable, which means that m_{*}(E)=m^{*}(E).
  2. For every \epsilon>0 there exist elementary sets A\subseteq E\subseteq B such that m(B\setminus A)\le\epsilon.
  3. For every \epsilon>0 there exists an elementary set A such that m^{*}(A\triangle E)\le\epsilon.

It suffices to prove that [1]\Rightarrow [2]\Rightarrow [3]\Rightarrow [1]. To provide further practice and familiarity with Jordan inner and outer measures, it will be additionally shown how to prove [1]\Rightarrow [3]\Rightarrow [2]\Rightarrow [1].

\boxed{[1]\Rightarrow [2]}

A reductio ad absurdum argument will be used. Assume that there exists \epsilon_0>0 such that for all elementary sets A,B\in\mathcal{E}(\mathbb{R}^d) with A\subseteq E\subseteq B the inequality m(B\setminus A)>\epsilon_0 holds.

Considering the set equality B=A\cup (B\setminus A), it follows from A\subseteq B that m(B\setminus A)=m(B)-m(A). Hence, m(B\setminus A)=m(B)-m(A)>\epsilon_0.

So, m(A)+\epsilon_0\le m^{*}(E), since m(A)+\epsilon is a lower bound of \mathcal{U}(E) and m^{*}(E)=\inf{\mathcal{U}(E)}. In turn, m^{*}(E)-\epsilon_0 is an upper bound of \mathcal{L}(E) and m_{*}(E)=\sup{\mathcal{L}(E)}, therefore m_{*}(E)\le m^{*}(E)-\epsilon_0.

Thus, m_{*}(E)<m^{*}(E), which contradicts the assumption m_{*}(E)=m^{*}(E).

\boxed{[2]\Rightarrow [3]}

Assume that there exists \epsilon_0>0 such that \forall C\in\mathcal{E}(\mathbb{R}^d) holds m^{*}(C\triangle E)>\epsilon_0.

According to the assumed statement [2], for \epsilon=\epsilon_0, \exists A, B\in\mathcal{E}(\mathbb{R}^d) with A\subseteq E\subseteq B such that m(B\setminus A)\le\epsilon_0.

Pick C=B, so m^{*}(B\triangle E)=m^{*}(C\triangle E)>\epsilon_0. It follows from E\subseteq B and B\triangle E=(B\setminus E)\cup(E\setminus B) that B\triangle E=B\setminus E, so m^{*}(B\setminus E)=m^{*}(B\triangle E)>\epsilon_0.

B\setminus E\subseteq B\setminus A, since A\subseteq E. By also taking into account that B\setminus A is elementary and the inequality m(B\setminus A)\le\epsilon_0, the conclusion is m^{*}(B\setminus E)\le m(B\setminus A)\le\epsilon_0.

A contradiction has been reached, as it has been deduced m^{*}(B\setminus E)>\epsilon_0 and m^{*}(B\setminus E)\le\epsilon_0 on the basis of the negation of statement [3].

\boxed{[3]\Rightarrow [1]}

Before proceeding with the proof of [3]\Rightarrow [1], four lemmas will be proved.

Lemma 1

The Jordan inner measure m_{*}(E) of any bounded set E\subseteq\mathbb{R}^d is less than or equal to its Jordan outer measure m^{*}(E), i.e. m_{*}(E)\le m^{*}(E).

Proof of lemma 1

For any elementary sets A,B with A\subseteq E\subseteq B, the set relation A\subseteq B yields m(A)\le m(B). This means that any m(A)\in\mathcal{L}(E) is a lower bound of \mathcal{U}(E), therefore m(A)\le m^{*}(E). Since the last inequality holds for any m(A)\in\mathcal{L}(E), it follows that m^{*}(E) is an upper bound of \mathcal{L}(E), thus m_{*}(E)\le m^{*}(E).

Lemma 2

The elementary and Jordan measures of any elementary set X coincide, that is m_{*}(X)=m^{*}(X)=m(X).

Proof of lemma 2

It suffices to notice that X\subseteq X, whence m^{*}(X)\le m(X) and m(X)\le m_{*}(X), thereby m^{*}(X)\le m_{*}(X). It is also known from lemma 1 that m_{*}(X)\le m^{*}(X), so m_{*}(X)=m^{*}(X). Finally, m^{*}(X)\le m(X), m(X)\le m_{*}(X) and m_{*}(X)=m^{*}(X) yield m(X)=m_{*}(X)=m^{*}(X).

Lemma 3

Let E\subseteq\mathbb{R}^d be a bounded set and B\in\mathcal{E}(\mathbb{R}^d) with E\subseteq B. It then holds that m(B)-m_{*}(E)\le m^{*}(B\setminus E).

Proof of lemma 3

Recall that for a set X\subseteq\mathbb{R} the following equivalences hold:

  • \ell=\inf{X}\Leftrightarrow(\forall\epsilon>0)(\exists x\in X)x<\ell+\epsilon.
  • u=\sup{X}\Leftrightarrow(\forall\epsilon>0)(\exists x\in X)u-\epsilon < x.

According to the former equivalence, for any \epsilon>0 there exists an elementary set B\setminus E\subseteq C such that

m(C)<m^{*}(B\setminus E)+\epsilon /2\ \ \ \ (1).

By using the latter equivalence, there exists an elementary set A\subseteq B such that

m_{*}(B)-\epsilon /2<m(A)\ \ \ \ (2).

It follows from lemma 2 and inequality (2) that m(B)-m_{*}(E)<m(A)-m_{*}(E)+\epsilon /2. B\setminus E\subseteq C and A\subseteq B lead to A\setminus C\subseteq E, so m(A\setminus C)\le m_{*}(E), which in turn gives

m(B)-m_{*}(E)<m(A)-m(A\setminus C)+\epsilon /2\ \ \ \ (3).

Inequality (3) connects the statement of lemma 3 to an analogous simpler statement for elementary sets; in particular, it will be shown that

m(A)-m(C)\le m(A\setminus C)\ \ \ \ (4).

Indeed, (A\setminus C)\cup (A\cap C)=A, so m(A)-m(A\cap C)=m(A\setminus C). Moreover, A\cap C\subseteq C, which means m(A\cap C)\le m(C), therefore m(A)-m(C)\le m(A)-m(A\cap C). Inequality (4) has thus been reached.

(3) and (4) yield m(B)-m_{*}(E)<m(C)+\epsilon/2, which is combined with (1) to give

m(B)-m_{*}(E)<m^{*}(B\setminus E)+\epsilon\ \ \ \ (5).

For two reals x, y, if (\forall\epsilon>0) x\le y+\epsilon, then x\le y. This can be shown by assuming x>y, whence 0<x-y\le\epsilon\Rightarrow x=y, contradiction. As inequality (5) holds for any \epsilon>0, the conclusion m(B)-m_{*}(E)\le m^{*}(B\setminus E) follows.

Lemma 4

The Jordan outer measure of the union of two bounded sets X\subseteq\mathbb{R}^d and Y\subseteq\mathbb{R}^d is less than or equal to the sum of the Jordan outer measures of the two sets, i.e. m^{*}(X\cup Y)\le m^{*}(X)+m^{*}(Y).

Proof of lemma 4

Let V\subseteq\mathbb{R}^d,W\subseteq\mathbb{R}^d be elementary sets with X\subseteq V,Y\subseteq W.

Start by noticing the set equality V\cup W=V\cup(W\setminus V). So, m(V\cup W)=m(V)+m(W\setminus V). Moreover, W\setminus V\subseteq W implies that m(W\setminus V)\le m(W). Thus,

m(V\cup W)\le m(V)+m(W)\ \ \ \ (6).

Inequality (6) holds for elementary sets, thereby it is a special case of lemma 4.

m^{*}(X\cup Y)\le m(V\cup W), since X\cup Y \subseteq V\cup W. By taking into account inequality (6), m^{*}(X\cup Y)\le m(V)+m(W). So, m^{*}(X\cup Y)-m(V) is a lower bound of \mathcal{U}(Y) and m^{*}(X\cup Y)-m(V)\le m^{*}(Y). Furthermore, m^{*}(X\cup Y)-m^{*}(Y) is a lower bound of \mathcal{U}(X), so m^{*}(X\cup Y)-m^{*}(Y)\le m^{*}(X), quod erat demonstrandum.

Proof of \boldsymbol{[3]\Rightarrow [1]} using the lemmas

The main idea of the proof of \boldsymbol{[3]\Rightarrow [1]} is to show that for any \epsilon>0 the inequality m^{*}(E)-m_{*}(E)\le \epsilon holds.

It is known from statement [3] that for any (\epsilon>0)(\exists X\in\mathcal{E}(\mathbb{R}^d)) such that

m^{*}(X\triangle E)\le\epsilon/3\ \ \ \ (7).

Using the relevant property of infimum, for any \epsilon>0 there exists an elementary superset E\setminus X\subseteq Y of E\setminus X such that

m(Y)<m^{*}(E\setminus X)+\epsilon/3\ \ \ \ (8).

Introduce the set B:=X\cup Y. Obviously B is an elementary set and E\subseteq B, so m^{*}(E)\le m(B). It thus follows from lemma 3 that

m^{*}(E)-m_{*}(E)\le m^{*}(B\setminus E)\ \ \ \ (9).

B\setminus E=(X\cup Y)\setminus E=(X\setminus E)\cup (Y\setminus E), so by lemma 4

m^{*}(B\setminus E)\le m^{*}(X\setminus E)+m^{*}(Y\setminus E)\ \ \ \ (10).

X\setminus E\subseteq X\triangle E\Rightarrow \mathcal{U}(X\triangle E)\subseteq\mathcal{U}(X\setminus E), so m^{*}(X\setminus E)=\inf{\mathcal{U}(X\setminus E)}\le \inf{\mathcal{U}(X\triangle E)}=m^{*}(X\triangle E). Consequently, inequality (10) gives

m^{*}(B\setminus E)\le m^{*}(X\triangle E)+m^{*}(Y\setminus E)\ \ \ \ (11).

In a similar way, E\setminus X\subseteq X\triangle E implies

m^{*}(E\setminus X)\le m^{*}(X\triangle E)\ \ \ \ (12).

Finally, Y\setminus E\subseteq Y means that

m^{*}(Y\setminus E)\le m(Y)\ \ \ \ (13).

All the components of the proof have been established. More concretely, inequalities (9), (11), (7), (13), (8) and (12) produce m^{*}(E)-m_{*}(E)\le \epsilon for any \epsilon>0, so m_{*}(E)=m^{*}(E).

\boxed{[1]\Rightarrow [3]}

Assume that (\exists\epsilon_0>0)(\forall X\in\mathcal{E}(\mathbb{R}^d))m^{*}(X\triangle E)>\epsilon_0.

As m^{*}(E)=\inf{\mathcal{U}(E)}, it is deduced that there exists an elementary set E\subseteq B such that

m(B)<m^{*}(E)+\epsilon_0/2\ \ \ \ (14).

Since m_{*}(E)=\sup{\mathcal{L}(E)}, there exists an elementary set A\subseteq E such that

m_{*}(E)-\epsilon_0/2<m(A)\ \ \ \ (15).

A\subseteq E\subseteq B\Rightarrow E\setminus A\subseteq B\setminus A, so m^{*}(E\setminus A)\le m(B\setminus A). Moreover, A\subseteq B means m(B\setminus A)=m(B)-m(A), hence

m_{*}(E\setminus A)\le m(B)-m(A)\ \ \ \ (16).

Combining (14), (15) and (16) gives m_{*}(E\setminus A)<m^{*}(E)-m_{*}(E)+\epsilon_0. By the assumed statement [1], m_{*}(E)=m^{*}(E), thus m_{*}(E\setminus A)<\epsilon_0. This is a contradiction, as the assumed negation of statement [3] gives m_{*}(A\triangle E)=m_{*}(E\setminus A)>\epsilon_0 for A\subseteq E.

\boxed{[3]\Rightarrow [2]}

The proof of [3]\Rightarrow [2] is similar in spirit to the proof of [3]\Rightarrow [1]. It will be shown that for any \epsilon >0 there exist A,B\in\mathcal{E}(\mathbb{R}^d) with A\subseteq E\subseteq B such that m(B\setminus A)\le \epsilon.

It is known from statement [3] that for any (\epsilon>0)(\exists X\in\mathcal{E}(\mathbb{R}^d)) such that

m^{*}(X\triangle E)\le\epsilon/4\ \ \ \ (17).

According to the relevant property of infimum, for any \epsilon>0 there exists an elementary set Y with E\setminus X\subseteq Y such that

m(Y)<m^{*}(E\setminus X)+\epsilon/4\ \ \ \ (18).

Consider the set B:=X\cup Y, which is an elementary set and E\subseteq B, so m^{*}(E)\le m(B). By application of lemma 3,

m(B)-m_{*}(E)\le m^{*}(B\setminus E)\ \ \ \ (19).

It is noted that B\setminus E=(X\cup Y)\setminus E=(X\setminus E)\cup (Y\setminus E), so by lemma 4

m^{*}(B\setminus E)\le m^{*}(X\setminus E)+m^{*}(Y\setminus E)\ \ \ \ (20).

X\setminus E\subseteq X\triangle E, so m^{*}(X\setminus E)\le m^{*}(X\triangle E). Thus, inequality (20) leads to

m^{*}(B\setminus E)\le m^{*}(X\triangle E)+m^{*}(Y\setminus E)\ \ \ \ (21).

Similarly, E\setminus X\subseteq X\triangle E implies

m^{*}(E\setminus X)\le m^{*}(X\triangle E)\ \ \ \ (22).

Moreover, Y\setminus E\subseteq Y gives

m^{*}(Y\setminus E)\le m(Y)\ \ \ \ (23).

Finally, there exists an elementary set A with A\subseteq E such that

m_{*}(E)-\epsilon /4<m(A) \ \ \ \ (24).

Inequalities (19), (21), (17), (23), (18) and (22) produce

m(B)-m_{*}(E)\le 3\epsilon /4 \ \ \ \ (25).

Combining inequalities (24) and (25) confirms that for any \epsilon > 0 there exist elementary sets A,B with A\subseteq E\subseteq B such that m(B)-m(A)\le \epsilon, which completes the proof of [3]\Rightarrow [2].

\boxed{[2]\Rightarrow [1]}

It is known from lemma 1 that m_{*}(E)\le m^{*}(E). Assume the negation of statement [1], that is assume m_{*}(E)\neq m^{*}(E). So, m_{*}(E)< m^{*}(E).

Set \epsilon_0=(m^{*}(E)-m_{*}(E))/2>0. From the assumed statement [2], it is known that there exist A,B\in\mathcal{E}(\mathbb{R}^d) with A\subseteq E\subseteq B such that

m(B\setminus A)=m(B)-m(A)\le  \epsilon_0=(m^{*}(E)-m_{*}(E))/2 \ \ \ \ (26).

Note that A\subseteq E\Rightarrow m(A)\le m_{*}(E) and E\subseteq B\Rightarrow m^{*}(E)\le m(B), which leads to

m^{*}(E)-m_{*}(E)\le m(B)-m(A) \ \ \ \ (27).

It follows from equation (26) that 2(m(B)-m(A))\le  m^{*}(E)-m_{*}(E), which is combined with equation (27) to give 2(m(B)-m(A))\le m(B)-m(A), and finally m(B)-m(A)\le 0. Moreover, equation (27) yields 0 < 2\epsilon_0=m^{*}(E)-m_{*}(E)\le m(B)-m(A), i.e. m(B)-m(A)>0. Thus, a contradiction has been reached.

Proof: uniqueness of elementary measure

The following proof is a solution to exercise 1.1.3 of the book “An introduction to measure theory” by Terence Tao.

A box B\in\mathbb{R}^d, d\in\mathbb{N}, is a Cartesian product B:={\sf X}_{i=1}^d I_i, where each interval I_i is I_i=(a, b) or I_i=(a, b] or I_i=[a, b) or I_i=[a, b] for a,b\in\mathbb{R} with a\le b. An elementary set E=\cup_{i=1}^n B_i\subseteq\mathbb{R}^d is a finite union of disjoint boxes B_i\in\mathbb{R}^d. Let \mathcal{E}(\mathbb{R}^d) denote the collection of elementary sets in \mathbb{R}^d. The measure m:\mathcal{E}(\mathbb{R}^d)\rightarrow R^{+}\cup\left\{0\right\} is defined as m(E)=\displaystyle\lim_{N\rightarrow\infty}\frac{1}{N^d}\#\left(E\cap\frac{1}{N}\mathbb{Z}^d\right), where \#(\cdot) denotes set cardinality and \displaystyle\frac{1}{N}\mathbb{Z}^d:=\left\{\frac{\mathbf{z}}{N}:\mathbf{z}\in\mathbb{Z}^d\right\}.

Let m^{'}:\mathcal{E}(\mathbb{R}^d)\rightarrow R^{+}\cup\left\{0\right\} be a function satisfying the non-negativity (m^{'}(E)\ge 0 for any elementary set E), finite additivity (m^{'}(\displaystyle\cup_{i=1}^n E_i)\le \sum_{i=1}^{n}m^{'}(E_i) for disjoint elementary sets E_i) and translation invariance (m^{'}(E+\mathbf{x})=m^{'}(E) for any elementary set E and any \mathbf{x} \in \mathbb{R}^d) properties.

It will be shown that there exists a positive constant c\in\mathbb{R}^+ such that m^{'}=cm, i.e. the functions m^{'} and m are equal up to a positive normalization constant c.

Observe that \left[0,1\right)=\displaystyle\cup_{i=0}^{n-1}\left[\frac{i}{n},\frac{i+1}{n}\right). Due to translation invariance, m^{'}\left(\displaystyle\left[\frac{i}{n},\frac{i+1}{n}\right)\right)=m^{'}\left(\displaystyle\left[\frac{i}{n},\frac{i+1}{n}\right)-\frac{i}{n}\right)=m^{'}\left(\left[0,\frac{1}{n}\right)\right). Using finite additivity, it follows that m^{'}\left(\left[0,1\right)^d\right)=n^dm^{'}\left(\left[0,\frac{1}{n}\right)^d\right). So, m^{'}\left(\left[0,\frac{1}{n}\right)^d\right)=\displaystyle\frac{c}{n^d} for c:=m^{'}\left(\left[0,1\right)^d\right). Since m\left(\left[0,\frac{1}{n}\right)^d\right)=\displaystyle\frac{1}{n^d}, it follows that m^{'}\left(\left[0,\frac{1}{n}\right)^d\right)=c m\left(\left[0,\frac{1}{n}\right)^d\right).

This result generalizes to intervals \left[0,q\right) for any rational q=\displaystyle\frac{s}{n} with s\in\mathbb{N} andn\in\mathbb{N}. Since \left[0,q\right)=\displaystyle\cup_{i=0}^{s-1}\left[\frac{i}{n},\frac{i+1}{n}\right), finite additivity and translation invariance lead to m^{'}\left(\left[0,q\right)^d\right)=c m\left(\left[0,q\right)^d\right)=cq^d.

It will be shown that the result holds also for intervals \left[0,p\right) for any irrational p\in\mathbb{P}.

The set of rationals is dense, which means that (\forall \epsilon>0)(\forall x\in\mathbb{R})(\exists q\in\mathbb{Q})|x-q|<\epsilon. For some irrational x=p and for each n\in\mathbb{N}, set \epsilon=\displaystyle\frac{1}{n}, so (\forall n\in\mathbb{N})(\exists q_n\in\mathbb{Q})|p-q_n|<\displaystyle\frac{1}{n}. Pick some n_0\in\mathbb{N} with n_0>\displaystyle\frac{2}{p}. For all n\in\mathbb{N} with n>n_0, it holds that \displaystyle \frac{2}{n}<\frac{2}{n_0}<p, whence 0<\displaystyle\frac{1}{n}<p-\frac{1}{n}<q_n. So, (\exists n_0\in\mathbb{N})(\forall n\in\mathbb{N}) with n>n_0, it is true that 0<q_n-\displaystyle\frac{1}{n}<p<q_n+\frac{1}{n} and consequently \left[0,q_n-\displaystyle\frac{1}{n}\right)^d\subseteq\left[0,p\right)^d\subseteq\left[0,q_n+\displaystyle\frac{1}{n}\right)^d.

For any two elementary sets E\subseteq F, it can be shown that m^{'}(E)\le m^{'}(F) via the equality F=E\cup(F\setminus E), non-negativity and finite additivity. Hence, m^{'}\left(\left[0,q_n-\displaystyle\frac{1}{n}\right)^d\right)\le m^{'}\left(\left[0,p\right)^d\right)\le m^{'}\left(\left[0,q_n+\displaystyle\frac{1}{n}\right)^d\right).

Since q_n\pm\displaystyle\frac{1}{n} are rationals, it is deduced that m^{'}\left(\left[0,q_n\pm\displaystyle\frac{1}{n}\right)^d\right)=c\left(q_n\pm\displaystyle\frac{1}{n}\right)^d. Thus, c\left(q_n-\displaystyle\frac{1}{n}\right)^d\le m^{'}\left(\left[0,p\right)^d\right)\le c\left(q_n+\displaystyle\frac{1}{n}\right)^d.

0<q_n-\displaystyle\frac{1}{n}<p<q_n+\frac{1}{n} gives 0<p-\displaystyle\frac{1}{n}<q_n<p+\frac{1}{n}, hence the sandwich theorem yields \displaystyle\lim_{n\rightarrow\infty}q_n=p.

Combining c\left(q_n-\displaystyle\frac{1}{n}\right)^d\le m^{'}\left(\left[0,p\right)^d\right)\le c\left(q_n+\displaystyle\frac{1}{n}\right)^d and \displaystyle\lim_{n\rightarrow\infty}q_n=p gives cp^d\le m^{'}\left(\left[0,p\right)^d\right)\le cp^d, so m^{'}\left(\left[0,p\right)^d\right)=cp^d for any irrational p.

This effectively completes the proof. There remains to show that m^{'}=cm is true for Cartesian products of unequal intervals \left[0,x_i\right) in each coordinate i=1,2,\dots,d, for any x_i\in\mathbb{R}, and for any subinterval of the real line. These are all trivial given the existing foundations.