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 be the set of elementary sets on . Denote by and by the sets of elementary measures of all elementary subsets and supersets of a bounded set , respectively. Let and be the Jordan inner and Jordan outer measures of , respectively.

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

- is Jordan measurable, which means that .
- For every there exist elementary sets such that .
- For every there exists an elementary set such that .

It suffices to prove that . To provide further practice and familiarity with Jordan inner and outer measures, it will be additionally shown how to prove .

A reductio ad absurdum argument will be used. Assume that there exists such that for all elementary sets with the inequality holds.

Considering the set equality , it follows from that . Hence, .

So, , since is a lower bound of and . In turn, is an upper bound of and , therefore .

Thus, , which contradicts the assumption .

Assume that there exists such that holds .

According to the assumed statement [2], for , with such that .

Pick , so . It follows from and that , so .

, since . By also taking into account that is elementary and the inequality , the conclusion is .

A contradiction has been reached, as it has been deduced and on the basis of the negation of statement [3].

Before proceeding with the proof of , four lemmas will be proved.

**Lemma 1**

The Jordan inner measure of any bounded set is less than or equal to its Jordan outer measure , i.e. .

**Proof of lemma 1**

For any elementary sets with , the set relation yields . This means that any is a lower bound of , therefore . Since the last inequality holds for any , it follows that is an upper bound of , thus .

**Lemma 2**

The elementary and Jordan measures of any elementary set coincide, that is .

**Proof of lemma 2**

It suffices to notice that , whence and , thereby . It is also known from lemma 1 that , so . Finally, , and yield .

**Lemma 3**

Let be a bounded set and with . It then holds that .

**Proof of lemma 3**

Recall that for a set the following equivalences hold:

- .
- .

According to the former equivalence, for any there exists an elementary set such that

.

By using the latter equivalence, there exists an elementary set such that

.

It follows from lemma 2 and inequality (2) that . and lead to , so , which in turn gives

.

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

.

Indeed, , so . Moreover, , which means , therefore . Inequality (4) has thus been reached.

(3) and (4) yield , which is combined with (1) to give

.

For two reals , if , then . This can be shown by assuming , whence , contradiction. As inequality (5) holds for any , the conclusion follows.

**Lemma 4**

The Jordan outer measure of the union of two bounded sets and is less than or equal to the sum of the Jordan outer measures of the two sets, i.e. .

**Proof of lemma 4**

Let be elementary sets with .

Start by noticing the set equality . So, . Moreover, implies that . Thus,

.

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

, since . By taking into account inequality (6), . So, is a lower bound of and . Furthermore, is a lower bound of , so , quod erat demonstrandum.

**Proof of** **using the lemmas**

The main idea of the proof of is to show that for any the inequality holds.

It is known from statement [3] that for any such that

.

Using the relevant property of infimum, for any there exists an elementary superset of such that

.

Introduce the set . Obviously is an elementary set and , so . It thus follows from lemma 3 that

.

, so by lemma 4

.

, so . Consequently, inequality (10) gives

.

In a similar way, implies

.

Finally, means that

.

All the components of the proof have been established. More concretely, inequalities (9), (11), (7), (13), (8) and (12) produce for any , so .

Assume that .

As , it is deduced that there exists an elementary set such that

.

Since , there exists an elementary set such that

.

, so . Moreover, means , hence

.

Combining (14), (15) and (16) gives . By the assumed statement [1], , thus . This is a contradiction, as the assumed negation of statement [3] gives for .

The proof of is similar in spirit to the proof of . It will be shown that for any there exist with such that .

It is known from statement [3] that for any such that

.

According to the relevant property of infimum, for any there exists an elementary set with such that

.

Consider the set , which is an elementary set and , so . By application of lemma 3,

.

It is noted that , so by lemma 4

.

, so . Thus, inequality (20) leads to

.

Similarly, implies

.

Moreover, gives

.

Finally, there exists an elementary set with such that

.

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

.

Combining inequalities (24) and (25) confirms that for any there exist elementary sets with such that , which completes the proof of .

It is known from lemma 1 that . Assume the negation of statement [1], that is assume . So, .

Set . From the assumed statement [2], it is known that there exist with such that

.

Note that and , which leads to

.

It follows from equation (26) that , which is combined with equation (27) to give , and finally . Moreover, equation (27) yields , i.e. . Thus, a contradiction has been reached.