# Riemann Integrable

#### steven187

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 im 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 couldnt see how you would intergrate such functions

steven

#### shmoe

Homework Helper
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.

#### steven187

I see so if a function is not continuous it can still be riemann integrable, the thing is that i dont get is that, I cant 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 Im just finding it hard to imagine finding the area under the curve of such functions

#### fourier jr

how about the function which =1 for x>=0 and =0 for x<0? that isn't continuous but it's riemann-integrable.

#### steven187

hmmm.. well im still confused I just cant imagine how it is possible to find the area of a open shape, and also I cant see how it is possible to integrate
$$\int_{-1}^{1}f(x)dx$$ if

$$f(x)=\left\{\begin{array}{cc}0,&\mbox{ if } -1\le x\leq 0\\1, & \mbox{ if } 0<x\le 1\end{array}\right$$

#### Hurkyl

Staff Emeritus
Gold Member
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.

#### steven187

hello all

well I cant 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?
$$f(x)=\left\{\begin{array}{cc}0,&\mbox{ if }x\in Q\\1, & \mbox{ if } x\notin Q$$

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

#### matt grime

Homework Helper
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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#### Hurkyl

Staff Emeritus
Gold Member
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.

#### fourier jr

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.

#### Hurkyl

Staff Emeritus
Gold Member
Zero measure sounds better than zero area!

#### rachmaninoff

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"

#### Hurkyl

Staff Emeritus
Gold Member
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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#### fourier jr

edit: whoops never mind

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#### master_coda

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 $$\int_A f d\mu < \infty$$ on a measureable set A wrt a measure $$\mu$$
Not quite. A set can be continuous almost everywhere and still have an uncountable set of discontinuities.

#### steven187

hello all

thanxs guys, you have put things into more perspective now about riemann integrability in relation to discontinuities

#### rachmaninoff

You're right about the Lebesgue integrability - my mistake.

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#### rachmaninoff

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.

#### master_coda

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.

#### rachmaninoff

I see, you're right.

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