Prove the function is Riemann-integrable

[lep11]

Homework Statement

Let $f:[-1,5]→ℝ$ such that $f(x)= \left\{ \begin{array}{l} 3, -1≤x<0 \\ 0 , 0≤x≤2 \\ -2, 2<x≤5 \end{array} \right.$. Prove f is Riemann integrable.

3. The Attempt at a Solution
Obviously f is bounded.
Let $ε>0$ arbitrarily.
Let partition of interval [-1,5] be P={-1,0,2,5}.

Then $U(P,f)=3*(0-(-1))+0*(2-0)+(-2)*(5-2)=-3$
and $L(P,f)=3*(0-(-1))+0*(2-0)+(-2)*(5-2)=-3$

Therefore $U(P,f)-L(P,f)=0<ε$
Here comes the question: Is the partition I chose valid and correct?

 

[FactChecker]

Without knowing what a particular teacher might want, your partition looks very valid and appropriate. I think it must be the answer that was expected.

 

[lep11]

Without knowing what a particular teacher might want, your partition looks very valid and appropriate. I think it must be the answer that was expected.

I am asking because my teacher and one of my classmates said my partition is not valid and I don't get why. May be it has something to do with how the function is defined and/or infimum and supremum of the given function? I am little confused. Or maybe they are just wrong? I chose this particular partition because it seemed to be straightforward and simple. They had included epsilons in their partitions.

 

[FactChecker]

I am asking because my teacher and one of my classmates said my partition is not valid and I don't get why. May be it has something to do with how the function is defined and/or infimum and supremum of the given function? I am little confused. Or maybe they are just wrong? I chose this particular partition because it seemed to be straightforward and simple. They had included epsilons in their partitions.

They must want a partition with equal sized steps of Δx. That is how the Riemann integral is usually defined. You can break up your partition into smaller, Δx=1, steps and keep essentially the same logic. Just sum more terms for the smaller steps.

 

[micromass]

What is the definition of Riemann integrable?

 

[LCKurtz]

Homework Statement

Let $f:[-1,5]→ℝ$ such that $f(x)= \left\{ \begin{array}{l} 3, -1≤x<0 \\ 0 , 0≤x≤2 \\ -2, 2<x≤5 \end{array} \right.$. Prove f is Riemann integrable.

3. The Attempt at a Solution
Obviously f is bounded.
Let $ε>0$ arbitrarily.
Let partition of interval [-1,5] be P={-1,0,2,5}.

Then $U(P,f)=3*(0-(-1))+0*(2-0)+(-2)*(5-2)=-3$

Stop right there. The sup on [0,2] is not 0 and the sup on [2,5] is not -2. There are similar errors for your lower sum.

 

[lep11]

Let $f:[a,b]→ℝ$ be bounded function and $P=${$x_0, x_1, x_2, ..., x_n$} partition of interval $[a,b]$ so $a=x_1<x_2<...<x_n=b$. Let $\Delta x_i=x_i -x_{i-1}$, where $i=1,2,3,...,n$ and let's denote $M_i=sup$ {$f(x)|x∈[x_{i-1},x_i]$}. Then upper sum is $U(P,f)=\sum_{n=1}^{n} M_i \Delta x_i$

 

[lep11]

The sup on [0,2] is not 0

What is it then?

 

[FactChecker]

The sup on [0,2] is 0. The sup on (2,5] is -2, but on [2,5] it is 0. For the formal definition of Riemann integral, you probably should surround the points of discontinuity by arbitrarily small intervals and show that the difference between L(P,f) and U(P,f) can be made arbitrarily small. Since the summation for the Riemann integral is countable, it is unfortunate that the endpoints of the intervals are included in the definition of the M_i. Any countable set of points can be covered by countable many arbitrarily small intervals and will not effect the integral.

 

[geoffrey159]

I don't understand the problem statement. The integral of a step function defined on a segment is a fundamental definition rather than a property. It seems to me that you are asked to prove a definition, which seems to be nonsense to me.

 

[FactChecker]

I don't understand the problem statement. The integral of a step function defined on a segment is a fundamental definition rather than a property. It seems to me that you are asked to prove a definition, which seems to be nonsense to me.

The problem is to prove that the function is Riemann integrable. The property of being Riemann integrable is defined (see http://math.feld.cvut.cz/mt/txtd/1/txe3da1a.htm ) and it is necessary to prove that the given function has that property. Once that property is proven, the Riemann integral is defined.

 

[geoffrey159]

But why would one want to prove integrability of a step function defined on a segment ? Such functions are integrable by definition. It is like if someone said 'Let ABCD a rectangle. Show that angle ABC equals 90 degrees'. I don't get the subtlety.

 

[FactChecker]

But why would one want to prove integrability of a step function defined on a segment ? Such functions are integrable by definition. It is like if someone said 'Let ABCD a rectangle. Show that angle ABC equals 90 degrees'. I don't get the subtlety.

Apparently they are not integrable by definition because that is not how the Riemann integral has been defined in that class. The link I gave above tells what it takes to prove that a Riemann integral exists in terms the OP is using . That is a common definition of the Riemann integral. It is necessary to prove at least once that a step function satisfies the conditions. As far as the subtlety goes, this example is a good exercise on how to do such proofs. It's nice to practice on easy examples because there may be much harder examples later.

 

[micromass]

Why is everybody guessing? Shouldn't we wait for the OP to first tell us what his definition of Riemann integrable is?

 

[FactChecker]

Why is everybody guessing? Shouldn't we wait for the OP to first tell us what his definition of Riemann integrable is?

It's a standard definition; he is using the standard terminology; the teacher's objection is absolutely correct.

 

[micromass]

It's a standard definition

It's not. I know at least 4 different definitions of Riemann integrability. All are equivalent of course. I just want to hear which one he adopts.

 

[FactChecker]

It's not. I know at least 4 different definitions of Riemann integrability. All are equivalent of course. I just want to hear which one he adopts.

In my opinion, posts 1 and 7 leave no room for doubt. And the inclusion of both endpoints in the M_i of post 7 causes the problem that the original approach in post 1 refers to.

 

[micromass]

In my opinion, posts 1 and 7 leave no room for doubt. And the inclusion of both endpoints in the M_i of post 7 causes the problem that the original approach in post 1 refers to.

Maybe not. But since this thread is about proving Riemann integrability, I don't see it as a bad thing that the OP actually states what this means.

 

[FactChecker]

Maybe not. But since this thread is about proving Riemann integrability, I don't see it as a bad thing that the OP actually states what this means.

Agree. You make a good point.

 

[LCKurtz]

Since we are speculating here, I'm guessing the OP may be taking a course from little Rudin where, if my memory from long ago serves me, whatever particular definition he gave, he proves a theorem that $f$ is integrable on $[a,b]$ if you can show that given $\epsilon > 0$ you can find a partition $P$ of the interval such that$$
U(P,f) - L(P,f) < \epsilon$$It's pretty clear that is what the OP is trying to show. Under ideal circumstances the OP would have stated the definition of integrability and given the theorem in the home work template.

 

[lep11]

First of all, I appreciate your inputs, gents. I should have stated the definition of Riemann integrability in the very beginning, but I thought many of you (or at least some of you) are familiar with it.

Since we are speculating here, I'm guessing the OP may be taking a course from little Rudin where, if my memory from long ago serves me, whatever particular definition he gave, he proves a theorem that $f$ is integrable on $[a,b]$ if you can show that given $\epsilon > 0$ you can find a partition $P$ of the interval such that$$
U(P,f) - L(P,f) < \epsilon$$It's pretty clear that is what the OP is trying to show. Under ideal circumstances the OP would have stated the definition of integrability and given the theorem in the home work template.

That's correct and $U(P,f)$, $L(P,f)$ are defined as in http://math.feld.cvut.cz/mt/txtd/1/txe3da1a.htm
Now you don't need to speculate and keep guessing any more. :)

So to wrap this up, my partition is not correct and it leads to problems.

By the way, my answers are always wrong in math class which sucks. Or that's how I feel. No matter how hard I try. Maybe math and especially mathematical analysis are too abstract for me. I am thinking if studying math is really worth it after all. :/

 

[LCKurtz]

First of all, I appreciate your inputs, gents. I should have stated the definition of Riemann integrability in the very beginning, but I thought many of you (or at least some of you) are familiar with it.

That's correct and $U(P,f)$, $L(P,f)$ are defined as in http://math.feld.cvut.cz/mt/txtd/1/txe3da1a.htm
Now you don't need to speculate and keep guessing any more. :)

So to wrap this up, my partition is not correct and it leads to problems.

By the way, my answers are always wrong in math class which sucks. Or that's how I feel. No matter how hard I try. Maybe math and especially mathematical analysis are too abstract for me. I am thinking if studying math is really worth it after all. :/

Don't be too discouraged. Sometimes it literally takes time to pass before the more abstract ideas sink in. I didn't appreciate my advanced calculus course until a year later.

Getting back to your original problem, try a partition that includes short intervals containing the points where the function has a jump discontinuity in their interior.

 

[FactChecker]

First of all, I appreciate your inputs, gents. I should have stated the definition of Riemann integrability in the very beginning, but I thought many of you (or at least some of you) are familiar with it.

That's correct and $U(P,f)$, $L(P,f)$ are defined as in http://math.feld.cvut.cz/mt/txtd/1/txe3da1a.htm
Now you don't need to speculate and keep guessing any more. :)

So to wrap this up, my partition is not correct and it leads to problems.

By the way, my answers are always wrong in math class which sucks. Or that's how I feel. No matter how hard I try. Maybe math and especially mathematical analysis are too abstract for me. I am thinking if studying math is really worth it after all. :/

Your problem here was not abstraction, it was just in the details. Mathematical proofs can get very picky about details. It looks like you were blind-sided by the detail of the jump discontinuities forcing different U and L values. That requires something like @LCKurtz's suggestion above. Your intuition about the meaning of the Riemann integral is essentially correct.