# General questions about integrals (definitions)

• I
• JulienB
In summary: I'm sorry, I don't remember well. It seems that the condition is that the integral exists and is positive, but I'm not sure about the details.
JulienB
Hi everybody! I'm currently studying integrals, and I would like to clarify a few definitions, especially about the criterions of convergence/divergence of an integral. Basically if that's okay for you guys I'm going to list and number a few statements and I'd like to know if they are true or not.

1. If a definite integralab f(x) dx has both its domain of integration and integrand bounded, but its set of locations where it is not continuous is an uncountable set, then f(x) is not integrable.
(or do we have to run other tests to find out if the function is integrable or not?)

2. If a definite integralab f(x) dx has both its domain of integration and integrand unbounded, then it is called an improper integral. While the term improper integral normally designates the limit of an unbounded integral, this "ambiguity is resolved as both the proper and improper integral will coincide in value" (Wikipedia, improper integral).

3. The improper integral of a function f(x) exists, if there exists a function g(x) so that ∫ab g(x) dx converges and |f(x)| ≤ g(x) ∀ x ∈ [a,b). If one of those criterions is not met, then we cannot conclude anything about the existence of the improper integral of f(x) and we must run other tests. (Majorant criterion)

4. The improper integral of a function f(x) doesn't exist, if there exists a function g(x) so that ∫ab g(x) dx diverges and f(x) ≥ g(x) ∀ x ∈ [a,b). If one of those criterions is not met, then we cannot conclude anything about the non-existence of the improper integral of f(x) and we must run other tests. (Minorant criterion)I stop here, it would already mean a lot to me if those simple assumptions would become facts. :)

For 1), consider the function ##f: \mathbb R \to \mathbb R## defined by:
##\left\{\begin{array}{l}
0 \ on \ \mathbb Q\\
1 \ on \ \mathbb R \setminus \mathbb Q
\end{array}
\right.##

For 4), consider ##g(x)=-\frac{1}{x}## and ##f(x)=0## on ##]0,1[##

JulienB

Samy_A said:
For 1), consider the function ##f: \mathbb R \to \mathbb R## defined by:
##\left\{\begin{array}{l}
0 \ on \ \mathbb Q\\
1 \ on \ \mathbb R \setminus \mathbb Q
\end{array}
\right.##

It seems to me that this function is integrable though its set of non continuous locations is uncountable. Is that right? Then I would conclude that my statement was false and that this criterion only let's us know if a function is integrable but not if it is not integrable.

Samy_A said:
For 4), consider ##g(x)=-\frac{1}{x}## and ##f(x)=0## on ##]0,1[##

I see, f(x) converges though it is bigger as g(x).. Would that problem be solved if I had stated |f(x)| ≥ g(x)?

Oh, and also are 2. and 3. correct statements then?

Thank you so much Samy.

Julien.

Another question related to 1. would then be: if f(x) is bounded on both its domain of integrability and integrand but its set of non continuous locations is uncountable, what criterion(s) am I left with to check if f(x) is integrable or not? Julien.

JulienB said:
It seems to me that this function is integrable though its set of non continuous locations is uncountable. Is that right? Then I would conclude that my statement was false and that this criterion only let's us know if a function is integrable but not if it is not integrable.
Statement 1) is indeed false. I don't exactly understand what criterion for being integrable or not it could yield.
JulienB said:
I see, f(x) converges though it is bigger as g(x).. Would that problem be solved if I had stated |f(x)| ≥ g(x)?
No, see the same counterexample.
JulienB said:
Oh, and also are 2. and 3. correct statements then?
3) is correct (provided f is a measurable function). I'm not sure about what 2) is supposed to mean.
JulienB said:
Another question related to 1. would then be: if f(x) is bounded on both its domain of integrability and integrand but its set of non continuous locations is uncountable, what criterion(s) am I left with to check if f(x) is integrable or not?
That's a difficult question to answer. It depends on how you studied integration: Riemann integral, Lebesgue integral?

JulienB

Samy_A said:
Statement 1) is indeed false. I don't exactly understand what criterion for being integrable or not it could yield.

Mmm that comes from the script of my maths teacher. It's in German but I try to translate it:

Bounded functions are integrable, if they are "almost everywhere" continuous: a bounded function f: [a,b] → ℂ is integrable, if the set of its discontinuities is countable.

Samy_A said:
No, see the same counterexample.

Oh yes indeed. Maybe f(x) ≥ |g(x)| then? Or is it that the minorant criterion does not exist at all for integration?

Samy_A said:
3) is correct (provided f is a measurable function). I'm not sure about what 2) is supposed to mean.

About 2: I was under the impression that what we called an improper integral is actually the limit of a definite integral as one of the boundaries tends towards ±∞ or as the integrand tends towards ±∞. Actually I just realized I wrote "both" instead of "either" in my original statement and that's not how I meant it, sorry. And what I wished to check was if that is the correct definition of an improper integral.

Samy_A said:
That's a difficult question to answer. It depends on how you studied integration: Riemann integral, Lebesgue integral?

I study Riemann integrals, as far as I know. Wikipedia seems to say that statement number 1 (or a corrected version of it) is called "Lebesgue's integrability condition", but I still believe it applies to Riemann integrals right?Julien.

JulienB said:
Mmm that comes from the script of my maths teacher. It's in German but I try to translate it:

Bounded functions are integrable, if they are "almost everywhere" continuous: a bounded function f: [a,b] → ℂ is integrable, if the set of its discontinuities is countable.
Ah yes, that is correct.

JulienB said:
Oh yes indeed. Maybe f(x) ≥ |g(x)| then? Or is it that the minorant criterion does not exist at all for integration?
Yes, I think that one is correct.
JulienB said:
About 2: I was under the impression that what we called an improper integral is actually the limit of a definite integral as one of the boundaries tends towards ±∞ or as the integrand tends towards ±∞. Actually I just realized I wrote "both" instead of "either" in my original statement and that's not how I meant it, sorry. And what I wished to check was if that is the correct definition of an improper integral.
That is indeed the definition of an improper integral.
JulienB said:
I study Riemann integrals, as far as I know. Wikipedia seems to say that statement number 1 (or a corrected version of it) is called "Lebesgue's integrability condition", but I still believe it applies to Riemann integrals right?
Yes.

JulienB
@Samy_A Thank you for your answers, that was very helpful. I like to be sure I understand definitions well before I go further :)

Regarding 1 ("bounded functions are integrable, if they are "almost everywhere" continuous"), do you have a suggestion of another criterion to check integrability if the set of discontinuities of the function is uncountable?

I have another one that says:

Functions that are "almost everywhere" equal have the same integrals: If f: [a,b] → ℂ is integrable and g: [a,b] → ℂ bounded, and if the set of all x ∈ [a,b] with f(x) ≠ g(x) is countable, then g is also integrable and ∫ab f(x)dx = ∫ab g(x)dx.

Not sure if that helps in many cases though. What's your opinion about that?

Julien.

JulienB said:
I have another one that says:

Functions that are "almost everywhere" equal have the same integrals: If f: [a,b] → ℂ is integrable and g: [a,b] → ℂ bounded, and if the set of all x ∈ [a,b] with f(x) ≠ g(x) is countable, then g is also integrable and ∫ab f(x)dx = ∫ab g(x)dx.

Not sure if that helps in many cases though. What's your opinion about that?
Is that true for Riemann integrals? It is for Lebesgue integral, but I doubt it is for the Riemann integral.
The Lebesgue integrability condition you mentioned states that for a bounded f: [a,b] → ℂ, f is Riemann integrable if and only if the set of discontinuities of f has Lebesgue measure 0.
But one can easily take a nice continuous function f, change it's values on the rational numbers in such a way that the resulting function g is nowhere continuous.
g will be Lebesgue integrable, but not Riemann integrable.

Samy_A said:
Is that true for Riemann integrals? It is for Lebesgue integral, but I doubt it is for the Riemann integral.

Well what I quoted in my previous post is absolutely all there is in my script about that criterion. But to be honest, the script my teacher made is very unclear and I doubt it is complete. He has proved in the past to have a certain talent for misleading us :(

I hear what you're saying with your example, but not even is "Lebesgue" mentioned once in my script, while Riemann is at many occasions. I will ask other students and search a bit on the internet too.

Thanks a lot again!Julien.

Samy_A said:
Is that true for Riemann integrals?

It is if you assume both are integrable. So if ##f## and ##g## are Riemann-integrable and ##f=g## a.e. then the integrals coincide.

micromass said:
It is if you assume both are integrable. So if ##f## and ##g## are Riemann-integrable and ##f=g## a.e. then the integrals coincide.

Unfortunately that's not what is stated in my script...It says that if:
1. f: [a,b] → ℂ is integrable,
2. g: [a,b] → ℂ is bounded,
3. the set of all x ∈ [a,b] for which f(x) g(x) is countable,
then g(x) is integrable and ab f(x)dx = ∫ab g(x)dx.

It does look wrong to me, especially with the example Samy submitted.Julien.

JulienB said:
Unfortunately that's not what is stated in my script...It says that if:
1. f: [a,b] → ℂ is integrable,
2. g: [a,b] → ℂ is bounded,
3. the set of all x ∈ [a,b] for which f(x) g(x) is countable,
then g(x) is integrable and ab f(x)dx = ∫ab g(x)dx.

It does look wrong to me, especially with the example Samy submitted.Julien.

Yes, that is incorrect. It becomes correct for the Lebesgue integral though. It is also correct if we demand finite in (3).

micromass said:
Yes, that is incorrect. It becomes correct for the Lebesgue integral though. It is also correct if we demand finite in (3).

Alright, thank you for your answer. I reformulate what you said with my own words to make sure I get it right:

1. f: [a,b] → ℂ is integrable,
2. g: [a,b] → ℂ is bounded,
3. the set of all x ∈ [a,b] for which f(x) g(x) is finite,
then g(x) is integrable and ab f(x)dx = ∫ab g(x)dx.

About Lebesgue integrals, we didn't do that yet so I guess that criterion for integrability was meant for Riemann integrals.

Thx a lot,Julien.

## 1. What is an integral?

An integral is a mathematical concept that represents the area under a curve on a graph. It is used to calculate the total value or accumulated quantity of a function over a given interval.

## 2. What is the difference between a definite and indefinite integral?

A definite integral has specific limits of integration, meaning it calculates the area under a curve between two specific points on the x-axis. An indefinite integral does not have specific limits, and instead represents a general antiderivative of a function.

## 3. How do you calculate an integral?

To calculate an integral, you can use various integration techniques such as substitution, integration by parts, or partial fractions. You can also use software or calculators to numerically approximate the integral.

## 4. What are the applications of integrals?

Integrals have many real-world applications, such as calculating the area under a curve in physics, finding the volume of a solid in engineering, and determining the average value of a function in economics. They are also used in optimization problems and probability calculations.

## 5. Can any function be integrated?

Not all functions can be integrated analytically, meaning with a closed-form solution. However, most functions can be numerically integrated using numerical methods such as the trapezoidal rule or Simpson's rule. Some functions, such as those with infinite discontinuities, are not integrable at all.

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