# Antiderivative as Area under Curve

1. Aug 28, 2009

### Gear300

The proof I'm familiar with for relating the antiderivative to the area under a curve involves usage of the mean value theorem, which for that particular case, implies continuity for the curve. Thus, integration as a process for finding the area under a curve should be valid under the conditions that the curve is continuous for the given domain, right?

2. Aug 28, 2009

### elemental09

The definition for the area under a bounded curve y=f(x) over some interval [a,b] is the value of the integral:
$$\int_a^b \! f(x) \, dx$$
Of course f must be nonnegative on [a,b] to yield the area - otherwise the integral calculates the signed area.
The question remaining is of the existence of this integral. Continuity of f on [a,b] is certainly sufficient for existence, however it is not necessary. More generally, integrability is guaranteed if the set S of discontinuities of f on [a,b] has zero content; that is, if, for every e>0, S can be covered by finite collection of closed intervals whose total length is less than e. In particular, if f is continuous on [a,b] then S is the empty set, which has zero content.

3. Aug 28, 2009

### Gear300

This may be a bit obvious...but what is the difference in the 2 claims?

4. Aug 28, 2009

### HallsofIvy

I'm not sure what your question is. Yes, if a function is continuous on a given domain, and bounded (which, if the domain is closed and bounded, follows from continuity, then it is certainly integrable. You seem to asking if the other way is true: does integrability imply continuity. The answer to that is "no". If the function has isolated points of discontinuity, apply the proof you are familiar with to the intervals between the points of discontinuity. Whatever partition of the interval you are using, you can always add the points of discontinuity as endpoints, getting a "refinement" of your partition. Now whether you can do that in a way that guarentees integrability depends on the function.

Note that if a function is continuous on a closed and bounded interval, then it is integrable there. If the function is not continuous then it MAY be integrable.

5. Aug 28, 2009

### HallsofIvy

"zero content" means "measure 0" as defined in Lebesque measure. In addition to the empty set any finite or countable set has measure 0. And there exist uncountable sets that have measure 0.

6. Aug 28, 2009

### Gear300

I see...so essentially, points of discontinuity may be taken as endpoints of continuous partitions of the domain, right?

By this, do you mean that every set contains the null set?

7. Aug 28, 2009

### Tac-Tics

"Null set" has two meanings.

THE null set, $$\varnothing$$ is the set that has no elements.

A null set, on the other hand, is a set that has no "width" to it. It's also called a "set of measure zero."

A null set is roughly a set that doesn't contain some real interval, [a, b]. A single point has no width to it. Neither does two points. Even if you have several million points, together they have no width. Even if you choose a countably infinite number of points (such as the set of all integers or the set of all rationals), it STILL doesn't have any width to it.

THE null set is just one example of A null set. Be careful to distinguish between them!

8. Aug 28, 2009

### elemental09

The statement that any set contains the empty set is, by convention, true. This is an example of a vacuous truth.

@HallsofIvy: I don't believe content zero (as I've defined it) is equivalent to Lebesgue measure zero. The latter does not require a finite cover, merely a countable one.

9. Aug 28, 2009

### Gear300

I suppose this falls under measure theory...interesting, though I'm not really aware of its content. The only null set I knew of is THE null set. What branch of mathematics does measure theory fall under?

What meaning does a vacuous truth have (mathematically)?

10. Aug 28, 2009

### elemental09

Not sure how to precisely phrase it, but, roughly, a vacuous truth is a statement which only applies to elements of the empty set. For example, "every element of the empty set is an even number" is vacuously true. Wikipedia has a decent article on this.
My statement that any set S contains the empty set is vacuously true since, turning to the definition of set containment, every element of the empty set is in S. This is true because there are no elements in the empty set.

11. Aug 28, 2009

### Gear300

Heheh...the definition is slightly odd to me. Its sort of like saying that everything that has something also has nothing.

12. Aug 28, 2009

### snipez90

It's not so much a definition as it is something that has to be true. By the definition of containment, the only way that the empty set cannot be contained in another set A is if there exists an element of the empty set that is in A, and this is clearly absurd. It's not so strange that the empty set is part of every set when you realize that you obtain the empty set by applying the axiom of specification to an arbitrary set A with the condition "x =/= x".

13. Aug 28, 2009

### Elucidus

Just to toss this in...

I believe this is correct:

A function defined on an interval [a, b] is Riemann integrable (on [a, b]) if and only if it of bounded variation.

The definition of bounded variation is a bit deep and any book on real analysis should contain it (lots of lim sups and lim infs).

--Elucidus

14. Aug 28, 2009

### Gear300

I see...I'm getting it bit by bit...but for clarity, what is in bold was supposed to be "is not in A," right?

Interesting...I guess I'll have to look that up...but if I were to take a guess from a bit of what I just read, it would have something to do with asymptotic behavior?

15. Aug 29, 2009

### Elucidus

Just to give a glimpse.

Let $P=\{x_0,x_1,x_2, \dots,x_n\}[/tex] be a set of points so that [itex]a=x_0<x_1<x_2< \dots <x_n=b$.

Given a function f defined on [a, b] we say the variation of f with respect to P is

$$var (f,P)=\sum_{k=1}^{n}\left| f(x_k)-f(x_{k-1}) \right|.$$​

If a different list of points (say Q) is used then var(f, Q) may differ from var(f, P) (either more or less).

But if there exists a bound B so that var(f, P*) <= B for any list of points P* then f is said to be of bounded variation.

I believe that all functions that are piecewise continuous (and just continuous) are of bounded variation.

Now, there is a small chance I am slightly mistaken in my original claim (my analysis book is not handy so I can't look it up).

--Elucidus

16. Aug 29, 2009

### wofsy

The characteristic function of the complement of the Cantor set on [0,1] is Riemann integrable and its set of discontinuities is uncountable. Measure zero is the correct criterion.

With Lebesgue integrals the set of discontinuities can have positive measure. e.g.the characteristic function of the irrationals on [0,1].

When the function is continuous, it integral is differentiable. Otherwise its integral is only continuous - I don't event think that is is piece wise differentiable in the case of the Cantor set example.

Interestingly, if you take the characteristc function of the complement of a Cantor set of positive measure then it can not be integrated. I wish I had an intuition for this.

Last edited: Aug 29, 2009
17. Aug 29, 2009

### wofsy

A bounded function that is continuous almost everywhere is the condition you want. There are bounded continuous functions that are not of bounded variation.

Last edited: Aug 29, 2009