There are many textbooks, posts, videos and papers about continuity. However, it is hard to find a cohesive introduction to the concept of continuity aimed at what is needed for the basics of measure and probability theory. The textbook by R. M. Dudley , however, is a fundamental introduction, which we recommend for experienced users to this end.
This blog post strives to provide an overview and an easy-to-read introduction to continuity for probability and measure theory by focusing on the motivation behind the definitions and statements. Many examples will illustrate the discussed objects and their logical relationships.
Understanding continuity at a point requires a profound knowledge about Inner Products, Norms and Metrics (i.e. distance functions) since continuity is about ‘small’ changes of function arguments relative to small changes of its (function) values. Distance functions combined with limits (and its underlying topology) of functions provide a measure to determine how ‘small’ might be interpreted.
We restrict our considerations to the Euclidean metric space with the standard metric . Please refer to Inner Products, Norms and Metrics for further details.
Before we actually start, let us think about what we want to achieve with continuous functions. Why is this type of functions so important, not only from a pure theoretical but also from a very practical perspective?
The following heuristic tries to explain that in simple terms. Afterwards, we will start to introduce it formally.
A function behaves as continuous at if a small change in can be reached by a sufficiently small change in the corresponding argument .
Driving a car by operating the steering wheel might serve as a heuristic example of a continuous function . Consider the rotation of the steering wheel as the domain of the function that translates these input variables into the change in direction of the vehicle. You would like that the function is continuous since a small change in direction should be reached by a small change in the rotation angle of the steering wheel.
Ultimately, continuity is all about controlled behavior of the function values relative to its elements of the domain. Even though this might not be a mathematical precise definition of continuity, it might help to understand and remind the actual definition of continuous functions better.
Be aware that we would like to control the behavior of the range since this is by design the area that should behave in a continuous manner.
Keep the following points in your mind:
- Continuity is all about controlled behavior of the function values;
- Different types of continuity basically just require different types of controlled behavior. For instance, one can ask for a controlled behavior at a specific point in the domain or for the entire function;
- Convergent (and Cauchy) sequences are closely interlinked with continuity. Refer to Limits & Topological Spaces for further details;
In the following, we consider the concept of continuity in 1-dimensional metric spaces. This will serve as the basis for the second part of this series, where we will also consider multi-dimensional metric spaces. Nonetheless, we are going to introduce the different concepts in a general way, such that it can be used in one and multi-dimensional metric spaces.
Let us start with the simplest form of continuity.
Continuity at a Point
Keep the heuristic –outlined above– in mind when reading the following definition.
Definition 2. 1(Continuous at a Point):
Let and be metric spaces and let be a function from to . The function is said to be continuous at a point in if for every there is a , such that
If is continuous at every point of a subset of , we say is continuous on . If is continuous on its domain, we say that is continuous.
If is a point in the domain of the function , where is not continuous, we say that is discontinuous at , or that has a discontinuity at .
Definition 2.1 reflects the idea that points close to are mapped by to points sufficiently close to . That is, behaves in a controlled manner. Keeping this heuristic in mind it is also clear why the formulation “for all ” makes sense — if the function should behave in a controlled manner the corresponding range of the function needs to behave controlled in relation to its arguments. Hence, for every a corresponding needs to exist as outlined in the definition.
We can also use balls to provide an equivalent formulation.
A function is continuous at if and only if, for every , there is a such that .
We can even go further and formulate continuity in terms of neighborhoods:
A function with domain is continuous at if and only if, for every neighborhood of , there is a neighborhood of such that .
We can use one of the equivalent formulation of continuity to double-check whether a function is continuous at a point.
Check for Continuity at a point:
One has to conduct several steps to double-check whether a function is continuous at a given point :
- Set ;
- Consider an arbitrary ball around the image point if you want to prove that is continuous at . If you think that is not continuous try to find a suitable ball to contradict the definition in the next step;
- Check whether there is a corresponding ball such that . If for some fixed no such ball can exist, then the function at is not continuous.
Another way of better understanding the definition is to consider what it means if a function is NOT continuous at a point .
According to the Definition 2.1 this would mean that there must be an (sometimes called ‘loser’ ) with the property that, for each , there are such that
Let us consider some examples to illustrate the definition of continuity at a point further.
Example 2.1 (Constant Function):
The constant functions with is continuous at every point in the domain . Let be a bounded subset.
Let and let be an arbitrary ball around the only image point . We can choose to be an arbitrarily (large) positive real figure such as , then the following holds true:
for any . Given that any point of the domain will always be mapped on the distance between the corresponding image points will always be zero.
If we pick a small instead, the implication above would still hold true since the implication from a wrong statement is always correct.
Hence, the function is continuous on its domain.
Note that this type of continuity of at a point is by definition a local property:
If is an isolated point of the domain (i.e. a point of which is not an accumulation point of ), then every function defined at will be continuous at . The reason is quite simple: for a sufficiently small there is only one satisfying , namely , and .
The following example shows why jumps are usually not compatible with continuity.
Example 2.2 (Heavyside Function):
Let us consider the so-called Heavyside Function defined via
Constant functions are continuous as shown in Example 2.1. In addition, continuity is a local property which is why we can restrict our focus to the point since there is a jump in the graph from 0 to 1.
The set is a ball of . Note that the image set of only contains and .
There cannot be a ball of the domain, such that since there must be a negative that will be mapped to 0 via . Apparently, , which is why the Heaviside function is not continuous at zero but continuous everywhere else.
Maybe you want to have a look on the second part of the following video by 3Blue1Brown where limits and the – definition is explained.
Theorem 2.1 (Limits & Continuity):
Let be a function from one metric space to the metric space . Then is continuous at if, and only if, for every sequence in convergent to , the corresponding sequence in converges to , i.e.
Hence, continuity can also be interpreted as convergence-preserving.
Let be continuous at and let be a convergent sequence to . Due to the continuity the following holds:
This, however, already implies the convergence of to since gets arbitrarily close to .
Let us now assume that for every convergent sequence the following holds:
Let us further assume that is not continuous at . This means that there exists a ball around such that we cannot find a with . That is, there is an (that corresponds with the ball ), such that for all . Let us now consider the balls for the convergent sequence with . However, we cannot find a (or equivalent an integer ) with . This means that does not converge to since the ball would not contain any element. This contradicts the initial assumption (4) and proves the assertion.
At last, we will also look into global properties of continuous functions.
Theorem 2.2 (Continuity & Open Sets):
Let be a function from one metric space to the metric space . Then is continuous on if, and only if, for every open set in .
Continuity therefore preserves open sets. This should not be a big surprise since limits can also be expressed using open sets (e.g. balls).
Suppose is continuous on and is an open set in . We have to show that every point of is an interior point of . Suppose and . Since is open, there exists an such that if . Applying the continuity of at we know that there exists a such that if . Hence, as soon as .
Conversely, suppose that is open in for every open set in . Fix and , let be the set of all such that . Since is open is also an open set. Hence, there exists a such that .
This completes the proof.
The following video introduces continuity at a point on the real line by using limits (from the left and right). Hence, it might serve as a nice warm-up for this section.
Let us restrict in this section to the metric space
Definition 3.1 (Left- & Right-Continuous)
Let and . Let . If is continuous at as a function on , we say it is right-continuous at .
Let and . Let . If is continuous at as a function on , we say it is left-continuous at .
We can characterize both limit-types based on one-sided limits. Then, the definition of left- and right-continuity is equivalent to
In a 1-dimensional vector space such as , there are two possibilities to approach an element .
In a 2-dimensional space, however, it is possible to approach from infinite many directions since you can approach a point from any possible angle .
Continuity in multi-dimensions will be treated further below in this article. Hence, let us get back to the 1-dimensional metric space with as distance function.
Example 3.1 (Signum Function and One-Sided Continuity):
Let us consider the so-called Signum Function defined via
The domain of the function is the real line and the corresponding graph is shown as follows.
The function does not have a limit at , because if you approach 0 from the right the value is 1 while if you approach from the left the value is -1. We then write and . The actual value at is, however, .
Note that the definition makes the signum function continuous from neither side at , but a different convention would allow us to have one or the other but not both.
Example 3.1 illustrated that there might be different types of discontinuities. There are three kinds of discontinuities at a point :
- Removable Discontinuity:
If one-sided limit exists and is finite and can be removed by re-defining the function. That is, the function is either undefined at or .
- Jump or Step Discontinuity:
If one-sided limit exists and is finite but not equal. It is not possible to re-define the function such that the one-sided limits are all the same.
- Infinite or Essential Discontinuity:
If one-sided limits do not exist or are infinite.
Please also refer to the classification of singularities, where it is about differentiability of (complex) functions. Both classifications are related to each other closely.
Let us illustrate this classification of discontinuities by looking at specific examples.
Example 3.2 (Classification of Discontinuities):
a) Consider the function
which is not defined at as illustrated in the next graph.
Apparently, this discontinuity is a removable one since we can simply extend or change the definition of such that .
b) Let us now re-consider the Heavyside Function of Example 2.2, which has a jump discontinuity at . It has and as its left- and right-sided limit. In its graph we can clearly see a jump.
c) An essential or infinite discontinuity is of a very different kind. Consider the following well-known function
and its graph
The limit from the left and the limit from the right are not consistent with each other. Hence, it is an essential or infinite discontinuity.
Note that monotone functions can only have countable many jump discontinuities.
The following theorem outlines the interlinkage between continuous at a point and one-sided continuity at the same point.
Theorem 3.1 (Left- & Right-Continuous & Continuity)
The function is continuous at if and only if it is both right- and left-continuous at .
‘‘ If the function is continuous on its domain , it means that with implies . This holds true for and . Hence, left- and right-sided continuity follows.
‘‘ Assume that is left and right-continuous.
Let further denote a sequence with . Suppose does not converge to , which would contradict continuity. Then there must be an such that for all and corresponding the following holds true:
In other words, we cannot find any such that all terms with are arbitrarily close to the value . This, however, implies that either there exists infinitely many terms of the sequence with , or infinitely many with such that . In either case, there exists a subsequence , with all terms in only one of or , such that for . But such a sequence violates the assumption that is left and right continuous. Hence, the function must be continuous at .
The following is an alternative and much more elegant proof of the reverse direction.
‘‘ Let . By the left and right continuity of at , there are positive numbers such that for all and . Set . Then for all . Therefore, is continuous at .
Let continuous on and let further be a limit point of . If is not closed, then may not be in and so is not defined at . In the following we consider whether can be defined so that is continuous on .
If such an extension exists, then, for any sequence in , which converges to , the corresponding sequence converges to . Thus, for a (not necessarily continuous) function and a limit point , we define
provided that for each convergent sequence in , the corresponding sequence also converges in .
Proposition 3.1 (Neighborhoods and Converging Functions)
The following are equivalent:
(ii) For each neighborhood of in , there is a neighborhood of in such that .
Proof: ‘(i) (ii)’ Suppose that there is a neighborhood of in such that for each neighborhood of in . Consider the sequence of open balls
,, in the complement set . We can chose from to create a sequence, which is in and converges to . However, all terms of are not contained in and thus cannot converge to .
‘(ii) (i)’ Let be a sequence in such that in , and a neighborhood of in . By hypothesis, there is some neighborhood of such that . Since converges to , there is some such that for all . Thus, the image sequence is contained in for all . This means that .
Suppose and with and metric spaces. Assume that is continuous on its domain : for any point and any , there is a corresponding , such that
for any .
In general, depends on the chosen and the point as we can see in the following chart of the function for the domain .
Even though the corresponding tubes with range from to in the illustrative graph above. This in return means, that for a given not an unique would do the job for the entire positive real line. However, if the domain of is bounded, we can find a depending on a given such that a specific continuity is ensured. For more details please refer to Example 4.1 d) and Example 4.2.
In general, we therefore cannot expect that for a fixed the same value of will serve equally well for every point . This might happen, however, and when it does, the function possesses the following properties.
Definition 4.1 (Uniformly Continuous):
Let be a function from one metric space to another . Then is said to be uniformly continuous on a subset of if for every there is a (depending only on ), such that
for all . Note that is not fixed.
Example 4.1 (Uniform Continuity):
a) The function ,
is continuous but not uniformly continuous on its domain.
Since is the restriction of a rational function, it is certainly continuous.
Set and suppose we could find a to satisfy the definition of uniform continuity. Taking and , we obtain and
Hence, for these two points we would always have , contradicting the definition of uniform continuity.
b) Let us continue Example 2.1 and re-consider the class of the constant functions. A constant function on a bounded subset is uniformly continuous since we can pick one that works for all as outlined in Example 2.1.
c) Let defined by . This function is continuous and uniformly continuous on the standard metric space . To prove this, let , and set . Then for every there is a (depending only on ), such that
for all . Note that is not fixed.
d) Let defined by . At the beginning of this section, we illustrated that the quadratic function is not uniformly continuous on the entire real line. Now, we formally prove it: Suppose and suppose that is uniformly continuous. For all and , we would find that
for any real . However, this would imply
which is a contradiction since we can choose large.
We will see that the same function is uniformly continuous on specific (bounded) subsets.
Uniform continuity on a set implies continuity on . The converse is also true if the set is compact.
Theorem 4.1 (Heine, Continuous Functions on Compact Sets)
Suppose that is a function from a metric spaces to another . Let be a compact set and assume that is continuous on . Then is uniformly continuous on .
That is, continuous functions on compact sets are uniformly continuous.
Proof: Let be given. Then each point has associated with it a ball , with depending on , such that
Consider the collection of balls for each with radius . These open balls cover and, since is compact, a finite number of them also cover , say
In any ball of twice the radius, , we have
whenever . Let be the smallest of the numbers . We show that this works for the definition of uniform continuity. For this purpose, consider two points of , say and with . By the above discussion there is some ball containing , so . By the triangle inequality we have
Hence, , so we also have . Using the triangle inequality once more we find
In general, the function defined by is not uniformly continuous. However, if we restrict the function to a bounded closed subset the situation is different.
Example 4.2 (Uniform Continuity on Compact Sets):
is continuous and uniformly continuous on its bounded and closed domain . To prove this, observe that
for . Thus, if , then . If is given we only need to take to guarantee that for every pair with . This shows that is uniformly continuous on .
Lipschitz & Hölder Continuity
The next type of continuity is named after the German mathematician Rudolf Lipschitz.
Definition 5.1 (Lipschitz Continuity):
Let be a function from one metric space to another . Then is said to be Lipschitz continuous on if there exists a positive real number (which may depend on ) such that
whenever and .
Think about why the constant needs to be positive?
The inequality contains only absolute values, which is why a change of sign is not possible. Because of a similar reason, the constant cannot be zero since the product would be zero as well.
A Lipschitz continuous function on is (uniformly) continuous: Let be Lipschitz continuous, which means that there is a such that for all . Let and set , then we can imply for any with that . Note that the is not dependent on .
Let us consider two very simple examples.
Example 5.1 (Lipschitz Continuity):
a) The identity defined by is Lipschitz continuous on . To prove this, observe that
which holds true for any constant .
b) The absolute value function is Lipschitz continuous on . To prove this, observe that
due to the reverse triangle inequality. Hence, we can choose as Lipschitz constant. Note, however, that the absolute value function is not differentiable at .
Lipschitz continuity in means that
which is the amount of the secant increase is limited by .
A generalization of Lipschitz continuity is named after another German mathematician Otto Hölder.
Definition 5.2 (Hölder Continuity):
Let be a function from one metric space to another . Then is said to satisfy a Hölder condition of order if there exists a positive real number (which may depend on ) and an exponent such that
Lipschitz continuous functions are Hölder continuous with exponent since Hölder continuity is a generalization of Lipschitz continuity. Hence, Hölder continuous functions are in general not Lipschitz continuous.
A Hölder continuous function is (uniformly) continuous with Lipschitz constant and exponent : let and set (only dependent on ) , then we got for all with
Note that does not depend on .
Example 5.2 (Hölder Continuity):
The root function defined by is Hölder but not Lipschitz continuous on . Let and without loss of generality , then
The second statement that the root function is not Lipschitz continuous can be concluded as follows: Assume there is a Lipschitz constant such that for all . If we then choose we get and thus for all , which contradicts the existence of .