Continuity
We are restricting ourselves into endowed with normal topology. Recall that a function is continuous if and only if for any open set , we have
to be open. One can rewrite this statement using language. To say a function continuous at , we mean for any , there exists some such that for , we have
is continuous on if and only if is continuous at every point of .
If is replaced with or , we get left continuous and right continuous, one of which plays an important role in probability theory.
But the problem is, sometimes continuity is too strong for being a restriction, but the ‘direction’ associated with left/right continuous functions are unnecessary as well. For example the function
is neither left nor right continuous (globally), but it is a thing. Left/right continuity is not a perfectly weakened version of continuity. We need something different.
Definition of semicontinuous
Let be a real (or extended-real) function on . The semicontinuity of is defined as follows.
If
is open for all real , we say is lower semicontinuous.
If
is open for all real , we say is upper semicontinuous.
Is it possible to rewrite these definitions à la ? The answer is yes if we restrict ourselves in metric space.
is upper semicontinuous at if, for every , there exists some such that for , we have
is lower semicontinuous at if, for every , there exists some such that for , we have
Of course, is upper/lower semicontinuous on if and only if it is so on every point of . One shall find no difference between the definitions in different styles.
Relation with continuous functions
Here is another way to see it. For the continuity of , we are looking for arbitrary open subsets of , and is expected to be open. For the lower/upper semicontinuity of , however, the open sets are restricted to be like and . Since all open sets of can be generated by the union or intersection of sets like and , we immediately get
is continuous if and only if is both upper semicontinuous and lower semicontinuous.
Proof. If is continuous, then for any , we see is open, and therefore
has to be open. The upper semicontinuity is proved. The lower semicontinuity of is proved in the same manner.
If is both upper and lower semicontinuous, we see
is open. Since every open subset of can be written as a countable union of segments of the above types, we see for any open subset of , is open. (If you have trouble with this part, it is recommended to review the definition of topology.)
Examples
There are two important examples.
- If is open, then is lower semicontinuous.
- If is closed, then is upper semicontinuous.
We will prove the first one. The second one follows in the same manner of course. For , the set is equal to , which is open. For , since , we see . For however, the set of where has to be , which is still open.
When checking the semicontinuity of a function, we check from bottom to top or top to bottom. The function is defined by
Addition of semicontinuous functions
If and are upper/lower semicontinuous, then so is .
Proof. We are going to prove this using different tools. Suppose now both and are upper semicontinuous. For , there exists some and such that
Proof. If we pick , then we see for all , we have
The upper semicontinuity of is proved by considering all .
Now suppose both and are lower semicontinuous. We have an identity by
The set on the right side is always open. Hence is lower semicontinuous.
However, when there are infinite many semicontinuous functions, things are different.
Let be a sequence of nonnegative functions on , then
- If each is lower semicontinuous, then so is .
- If each is upper semicontinuous, then is not necessarily upper semicontinuous.
Proof. To prove this we are still using the properties of open sets. Put . Now suppose all are lower. Since is a finite sum of lower functions, we see each is lower. Let . As are non-negative, we see if and only if there exists some such that . Therefore
The set on the right hand is open already.
For the upper semicontinuity, it suffices to give a counterexample, but before that, we shall give the motivation.
As said, the characteristic function of a closed set is upper semicontinuous. Suppose is a sequence of almost disjoint closed set, then is not necessarily closed, therefore (a.e.) is not necessarily upper semicontinuous. Now we give a concrete example. Put and for where
For , we have . Meanwhile, , which is not open.
Notice that can be defined on any topological space here.
Maximum and minimum
There is one fact we already know about continuous functions.
If is compact, is continuous, then there exists some such that , .
In fact, is compact still. But for semicontinuous functions, things will be different but reasonable. For upper semicontinuous functions, we have the following fact.
If is compact and is upper semicontinuous, then there exists some such that .
Notice that is not assumed to hold any other topological property. It can be Hausdorff or Lindelöf, but we are not asking for restrictions like this. The only property we will be using is that every open cover of has a finite subcover. Of course, one can replace with any compact subset of , for example, .
Proof. Put , and define
If attains no maximum, then for any , there exists some such that . That is, for some . Therefore covers . But this cover has no finite subcover of . A contradiction since is compact.
Approximating integrable functions
This is a comprehensive application of several properties of semicontinuity.
(Vitali–Carathéodory theorem) Suppose , where is a real-valued function. For , there exist some functions and on such that , is an upper semicontinuous function bounded above, and is lower semicontinuous bounded below, and
It suffices to prove this theorem for (of course is not identically equal to since this case is trivial). Since is the pointwise limit of an increasing sequence of simple functions , can to write as
By putting , for , we get . We can write as
where is measurable for all . Also, we have
and the series on the right hand converges (since . By the properties of Lebesgue measure, there exists a compact set and an open set such that and . Put
(now you can see is lower semicontinuous and is upper semicontinuous). The is chosen in such a way that
Since , we have . Therefore . Similarly, . Now we need to check the desired integral inequality. A simple recombination shows that
If we integrate the function above, we get
This proved the case when . In the general case, we write . Attach the semicontinuous functions to and respectively by and . Put , . As we can see, is upper semicontinuous and is lower semicontinuous. Also, with the desired property since
and the theorem follows.
Generalisation
Indeed, the only property about measure used is the existence of and . The domain here can be replaced with for , and be replaced with the respective . Much more generally, the domain can be replaced by any locally compact Hausdorff space and the measure by any measure associated with the Riesz-Markov-Kakutani representation theorem on .
Is the reverse approximation always possible?
The answer is no. Consider the fat Cantor set , which has Lebesgue measure . We shall show that can not be approximated below by a lower semicontinuous function.
If is a lower semicontinuous function such that , then .
Proof. Consider the set . Since , we have . We will show that has to be empty.
Pick . Since is open, there exists some neighbourhood containing such that . But since and has an empty interior. Therefore . That is, for all .
Suppose is an upper semicontinuous function such that . For , we have
This example shows that there exist some integrable functions that are not able to reversely approximated in the sense of the Vitali–Carathéodory theorem.