Riemann Integrability: Bounded Functions on [a,b]

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In summary: I was mixing up Lebesgue integrability with something else. Sorry about that!In summary, the conversation discusses Riemann's Criterion, which states that a bounded function on a closed interval is Riemann integrable if and only if the upper and lower integrals exist and are equal. This leads to a question about whether a function needs to be continuous to be Riemann integrable, to which the answer is no. Examples of discontinuous functions that are Riemann integrable are given, and the conversation also touches on Lebesgue integrability and the role of discontinuities in determining Riemann integrability.
  • #1
steven187
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hello all

well I was working through Riemanns Criterion :

let f be a bounded function on the closed interval [a,b]. then f is riemann integrable on [a,b] if and only if , given any epsilon>0, there exist a partition P of [a,b] such that U(f,P)-L(f,P)<epsilon

but there is one thing that I am confused about, riemanns integrability only requires a function to be bounded on a closed interval, if that is the case a piecewise function or a function that is discontinuous at a point which are bounded on a closed interval should be riemann integrable would that be be correct? i just couldn't see how you would intergrate such functions

steven
 
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  • #2
The bounded requirement guarantees that the upper and lower integrals exist. This isn't enough to declare it riemann integrable, you also need the upper and lower integrals to be equal (or your equivalent statement if you prefer).

A function does not have to be continuous to be riemann integrable.
 
  • #3
I see so if a function is not continuous it can still be riemann integrable, the thing is that i don't get is that, I can't see how it is is possible to integrate something which is discontinuous or piecewise and bounded in both cases, do you have any examples of a function being riemann integrable and not continuous at the same time I am just finding it hard to imagine finding the area under the curve of such functions
 
  • #4
how about the function which =1 for x>=0 and =0 for x<0? that isn't continuous but it's riemann-integrable.
 
  • #5
hmmm.. well I am still confused I just can't imagine how it is possible to find the area of a open shape, and also I can't see how it is possible to integrate
[tex]\int_{-1}^{1}f(x)dx[/tex] if

[tex]f(x)=\left\{\begin{array}{cc}0,&\mbox{ if }
-1\le x\leq 0\\1, & \mbox{ if } 0<x\le 1\end{array}\right[/tex]
 
  • #6
Try splitting your partition into two parts -- the intervals on which f is continuous, and the intervals on which f is not continuous, and analyze their behavior separately as the norm of the partition goes to zero.
 
  • #7
hello all

well I can't get out how you can find an interval where f is not continuous because it is only discontinuous at one point x=0

now would this function be riemann integrable?
[tex]f(x)=\left\{\begin{array}{cc}0,&\mbox{ if }x\in Q\\1, & \mbox{ if } x\notin Q[/tex]

and what type of functions which are riemann integrable but discontinuous at the same time? I think that would help me understand what riemann integrability actually applies for.

thank you
 
  • #8
that function is definitely not riemann integrable. you should prove it; it is the first thing yuo should ever be asked to prove about non-integrable functions.


Now the function that is 0 from -1 to 0 then 1 from 0 to 1
Hurkyl's hint was quite sepcific, that fuction is discontinuous ate exactly one point, zero. the upper and lower riemann sums differ by exactly the width of the interval of the partition containing 0, so obviuosly as the max length of an interval tends to zero the upper and lower sums must converge (to 1)

note a function is discontinuois on a set if it has a discontinuity at any point of that set, not if it is discontinuous at all points of the set. discontinuous is the negation of continuous after all.
 
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  • #9
If f is discontinuous at x = 0, then f is not continuous on any open interval containing zero. :tongue2:


If I remember correctly...

A bounded function is Riemann integrable if and only if its set of discontinuities has zero area.
 
  • #10
Hurkyl said:
If f is discontinuous at x = 0, then f is not continuous on any open interval containing zero. :tongue2:


If I remember correctly...

A bounded function is Riemann integrable if and only if its set of discontinuities has zero area.

i think that's how it goes. a function is riemann integrable if & only if it is continuous almost everywhere, meaning the set of discontinuities has measure zero.
 
  • #11
Zero measure sounds better than zero area!
 
  • #12
Wait a minute... isn't that the criterion for Lebesgue integrability? Riemann integrability requires a countable set of discontinuities (I think)...

edit: replaced word "definition" with "criterion"
 
  • #13
Nah, you can Lesbegue integrate things that are everywhere discontinuous. Every bounded, measurable function is Lesbegue integrable over any set of finite measure. (And some unbounded functions!)

For example, the Lesbegue integral of the salt-and-pepper function over [0, 1] is 1.
 
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  • #14
edit: whoops never mind
 
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  • #15
fourier jr said:
that what it means to be continuous "almost everywhere". countable (finite & infinite) sets of real numbers have measure zero. a function is lebesgue integrable if [tex]\int_A f d\mu < \infty[/tex] on a measureable set A wrt a measure [tex]\mu[/tex]

Not quite. A set can be continuous almost everywhere and still have an uncountable set of discontinuities.
 
  • #16
hello all

thanxs guys, you have put things into more perspective now about riemann integrability in relation to discontinuities
 
  • #17
You're right about the Lebesgue integrability - my mistake.
 
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  • #18
Not quite. A set can be continuous almost everywhere and still have an uncountable set of discontinuities.

Wouldn't that mean a function with measure zero of discontinuities could be not Riemann integrable?

edit: That's what I originally meant to say, before bringing Lebesgue into the mess.
 
  • #19
rachmaninoff said:
Wouldn't that mean a function with measure zero of discontinuities could be not Riemann integrable?

edit: That's what I originally meant to say, before bringing Lebesgue into the mess.

No -- because the set of discontinuities doesn't have to be countable, it just has to have measure zero.

I believe the exact condition is that a function on [a,b] is Riemann integrable if and only if the function is bounded and the set of discontinuities has measure zero.
 
  • #20
I see, you're right.
 

1. What is Riemann integrability?

Riemann integrability is a mathematical concept that refers to the ability to calculate the area under a curve using a method called Riemann sums. It is used to determine the integral of a function on a specific interval.

2. How do you know if a function is Riemann integrable?

A function is Riemann integrable if it is continuous on the given interval and its upper and lower Riemann sums converge to the same value as the partition of the interval approaches zero.

3. Can a function be Riemann integrable if it is unbounded?

No, a function must be bounded on the given interval to be Riemann integrable. If a function is unbounded, the upper or lower Riemann sums will not converge and the integral cannot be calculated.

4. What is the importance of Riemann integrability?

Riemann integrability is important in mathematics because it allows us to calculate the area under a curve, which has many real-world applications. It is also a fundamental concept in calculus and is used to find the antiderivative of a function.

5. Can any function be Riemann integrable?

No, not all functions are Riemann integrable. A function must meet certain criteria, such as being continuous and bounded, to be Riemann integrable. Functions that have discontinuities, such as jump or infinite discontinuities, are not Riemann integrable.

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