3. Set function

(待修改! ) Given a set and , is a function

Definition 3.0.1. 1. For , we say that is continuous from below at if [ and an increasing sequence and (i.e. and ) ] .

2. Similarly, for , we say that is continuous from above at if [ and an decreasing sequence and (i.e. and ) and such that for some ] . We say that is continuous at if it continuous both from below and above at .

Remark 3.0.2. The symbol means that convergence up to the category of sets or convergence up to the category of topological spaces.

Remark 3.0.3. The condition such that can avoid that the following situation

MT-3-1.png

Assume that and and , then there is a decreasing sequence and . We know that . If there is not exists such that , then .

Lemma 3.0.4. Assume that an algebra and is additive. Then

1. is -additive is continuous at for .

2. is continuous from below is -additive.

3. is continuous from above at and <a id="muisfinite"> is finite</a> is -additive.

Proof. In one hand, let and consider an increasing converges sequence . Set

MT-3-2.png

Then for any and . It is clear that . So and since . Then Consequently, is continuous from below at .

In the other hand, let and an decreasing sequence and for some . Set

MT-3-3.png

Then there is an increasing converges sequence , since . Then . But It conclude that .

Assume that is continuous from below, let where , .

Observation since .

Therefore, , moreover, since is additive.

We claim that Let , is an increasing converges sequence. Then . But which implies that . It is easy to see that and therefore .

Assume that is continuous from above at and (i.e. is finite). Let , and . We construct Then is a decreasing converges sequence. Since is finite, . By the the condition of continuity of at , it conclude that immediately. Then

Example 3.0.5. Consider and is an algebra over , we define Then is additive but not -additive. is not finite and we claim that is continuous from above at .

Let be a decreasing converges sequence and we can write We assume that , hence for all . Since , So the condition is finite is necessary.

Theorem 3.0.6. is a semi-algebra over and a function which is additive, then

1. a function

​ which is additive.

2. is an extension of .

3. The extension of is unique.

Proof. Let , then we have where . We define We claim that is well-defined first.

Assume that where ,, what we need to show is . Since , , the additivity of guarantee that Then By the symmetry of and , we can write Consequently, is well-defined. Next we claim that is additive.

Given and where , and . We write Then It is clear that is an extension of by the definition of .

At the last step, we need to proof that the extension of is unique. In other words, if there are two functions Then Let , where ,

Remark 3.0.7. If the condition is additive replaced to be is -additive, then the extension of is also -additive in the theorem. Assume that where , and and is -additive. It sufficient to show that .

where since , where . Then