# Proving piecewise function is integrable

• bonfire09
In summary: But you're definitely on the right track.The partition you want to consider is all of the ones that contain the point ##3##. Specifically, what is the contribution of the interval ##[x_k,x_{k+1}]## towards the upper and lower sums when ##x_k=3##?
bonfire09

## Homework Statement

Prove the function ##f(x):[2,4]\rightarrow\mathbb{R}## defined by ##f(x) =\begin{cases} x, & \text{if }2\leq x\leq 3 \\2, & \text{if } 3<x\leq 4 \end{cases}## is Riemann Integrable

## Homework Equations

A function ##f:[a,b]\rightarrow\mathbb{R}## is integrable iff for each positive number ##\epsilon## there is a partition ##P## of the interval ##[a,b]## such that ##U(f,P)-L(f,P)<\epsilon##.

## The Attempt at a Solution

I've been struggling with this problem for a while and I am having a tough time figuring out what my partition should be? I know the function is increasing on ##2\leq x\leq 3## and is constant on ##3<x\leq 4##. And also what is the best way of figuring out the partition?

bonfire09 said:

## Homework Statement

Prove the function ##f(x):[2,4]\rightarrow\mathbb{R}## defined by ##f(x) =\begin{cases} x, & \text{if }2\leq x\leq 3 \\2, & \text{if } 3<x\leq 4 \end{cases}## is Riemann Integrable

## Homework Equations

A function ##f:[a,b]\rightarrow\mathbb{R}## is integrable iff for each positive number ##\epsilon## there is a partition ##P## of the interval ##[a,b]## such that ##U(f,P)-L(f,P)<\epsilon##.

## The Attempt at a Solution

I've been struggling with this problem for a while and I am having a tough time figuring out what my partition should be? I know the function is increasing on ##2\leq x\leq 3## and is constant on ##3<x\leq 4##. And also what is the best way of figuring out the partition?

Are you able to demonstrate that ##f_1:[2,3]\to\mathbb{R},\ x\mapsto x## and and ##f_2:(3,4]\to\mathbb{R},\ x\mapsto 2## are Riemann integrable? What is preventing you from considering a partition that includes the point ##3## so that you can use what you know about ##f_1## and ##f_2## to say something about ##f##? Your only real worry would be considering what happens to the sub interval with ##3## as a left endpoint.

I was considering to let ##P_1=[2,3]## and ##P_2=(3,4]##. And then showing that ##U(f,P_1)-L(f,P_1)<\frac{\epsilon}{2}## and ##U(f,P_2)-L(f,P_2)<\frac{\epsilon}{2}##. And then by combining them we get that ##U(f,P)-L(f,P)<\epsilon## where ##P=P_1\cup P_2##. Yes that 3 is a problem because that is what I am having trouble with.

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Before I can try to help you I really do need to know the answer to the first question; Are you able to demonstrate that ##f_1:[2,3]\to\mathbb{R},\ x\mapsto x## and and ##f_2:(3,4]\to\mathbb{R},\ x\mapsto 2## are Riemann integrable? Specifically, can you prove this from first principles without appealing to any theorems regarding integrability of certain kinds of functions.

I'm not trying to be mean. It's just really hard to know how to help you without knowing if you're able to find the right partitions for the "easier" problems. You really should be able to handle those problems fairly easily, and the way that you figure out how to do those examples will very likely lead to some insight towards how to work through the problem at hand.

If you are comfortable with the easier problem, the I would suggest you think about, given a partition of ##[2,4]## containing ##x_k=3##, what is the contribution of the interval ##[x_k,x_{k+1}]## towards the upper and lower sums?

First show that ##f_1: [2,3]\rightarrow\mathbb{R}## where f(x)=x is R. integrable. We will let ##P_n## be a regular partition of ##[2,3]##. Then we have ##P_n=\{[2,2+\frac{1}{n}], [2+\frac{1}{n},2+\frac{2}{n}],...,[2+\frac{n-1}{n},3]##. Then it follows that

##x_i-x_{i-1}=\frac{1}{n}## and ##m_i=2+\frac{i-1}{n}## and ##M_i=2+\frac{i}{n}##. Then
##U(f,P)-L(f,P)= \sum_{i=1}^{n} M_i(x_i-x_{i-1}) -\sum_{i=1}^{n} m_i(x_i-x_{i-1})

=((2+\frac{1}{n})*\frac{1}{n})+((2+\frac{2}{n})*\frac{1}{n})+...+3( \frac{1}{n}) -((2*\frac{1}{n})-...

-((2-\frac{n-1}{n})*\frac{1}{n})=\frac{1}{n}<\epsilon##.

And for showing that ##f_2:(3,4]\rightarrow\mathbb{R}## defined by ##f(x)=2## is integrable. Consider ##P=[3+\frac{\epsilon}{20},4]##. It also follows that ##m=M=2##. Thus ##U(f_2,P)-L(f_2,P)=2(4-\frac{\epsilon}{20})-2(4-\frac{\epsilon}{20})=0<\epsilon##.

Just zoom out it seems to be spilling over. And as for my partition do I just consider ##P=P_n\cup [3+\frac{\epsilon}{20},4]##?,

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bonfire09 said:

## Homework Statement

Prove the function ##f(x):[2,4]\rightarrow\mathbb{R}## defined by ##f(x) =\begin{cases} x, & \text{if }2\leq x\leq 3 \\2, & \text{if } 3<x\leq 4 \end{cases}## is Riemann Integrable

## Homework Equations

A function ##f:[a,b]\rightarrow\mathbb{R}## is integrable iff for each positive number ##\epsilon## there is a partition ##P## of the interval ##[a,b]## such that ##U(f,P)-L(f,P)<\epsilon##.

## The Attempt at a Solution

I've been struggling with this problem for a while and I am having a tough time figuring out what my partition should be? I know the function is increasing on ##2\leq x\leq 3## and is constant on ##3<x\leq 4##. And also what is the best way of figuring out the partition?

What makes you think it matters? In fact, integrability requires verification of something for all partitions whose mesh sizes → 0. So, restricting yourself to some 'special' partitions really will not do the complete job.

I guess you are worried about a partition in which the breakpoint x = 3 occurs somewhere in the middle of a partition interval. Well, if you look at the definition of Riemann integrability, you should be able to see why that does matter in the end.

bonfire09 said:
First show that ##f_1: [2,3]\rightarrow\mathbb{R}## where f(x)=x is R. integrable. We will let ##P_n## be a regular partition of ##[2,3]##. Then we have ##P_n=\{[2,2+\frac{1}{n}], [2+\frac{1}{n},2+\frac{2}{n}],...,[2+\frac{n-1}{n},3]##. Then it follows that

##x_i-x_{i-1}=\frac{1}{n}## and ##m_i=2+\frac{i-1}{n}## and ##M_i=2+\frac{i}{n}##. Then
##U(f,P)-L(f,P)= \sum_{i=1}^{n} M_i(x_i-x_{i-1}) -\sum_{i=1}^{n} m_i(x_i-x_{i-1})

=((2+\frac{1}{n})*\frac{1}{n})+((2+\frac{2}{n})*\frac{1}{n})+...+3( \frac{1}{n}) -((2*\frac{1}{n})-...

-((2-\frac{n-1}{n})*\frac{1}{n})=\frac{1}{n}<\epsilon##.

This mostly looks ok. If you are intending for this to be a full-blown proof, you should somewhere toward the beginning (a) say something along the lines of "fix ##\epsilon>0##" and (b) state the relationship between the chosen ##n## and fixed ##\epsilon##.

Also, while your computation regarding the difference between the upper and lower sums is perfectly correct, I would encourage you consider thinking about it as $$U(f,P)-L(f,P)= \sum_{i=1}^{n} [M_i(x_i-x_{i-1}) - m_i(x_i-x_{i-1})]= \sum_{i=1}^{n} [(M_i - m_i)(x_i-x_{i-1})]$$ rather than (or in addition to) $$U(f,P)-L(f,P)= \sum_{i=1}^{n} M_i(x_i-x_{i-1}) -\sum_{i=1}^{n} m_i(x_i-x_{i-1})$$ For one thing, it'll make for a slightly easier (and more readable) write-up. And it's just a slightly different way of thinking about the difference that might prove useful in more general or more difficult situations.

Again, though, what you wrote is fine.

And for showing that ##f_2:(3,4]\rightarrow\mathbb{R}## defined by ##f(x)=2## is integrable. Consider ##P=[3+\frac{\epsilon}{20},4]##. It also follows that ##m=M=2##. Thus ##U(f_2,P)-L(f_2,P)=2(4-\frac{\epsilon}{20})-2(4-\frac{\epsilon}{20})=0<\epsilon##.

Does your text/prof/class define definite integrals over (half-)open integrals as limits of integrals over closed intervals? Why ##\frac{\epsilon}{20}## specifically?

Regardless of how an integral/partition over ##(3,4]## is defined, is there any partition here that wouldn't work? Would choosing a partition similar to the one you chose in the first part make your job here more difficult? Would it maybe make your job later on easier?

And as for my partition do I just consider ##P=P_n\cup [3+\frac{\epsilon}{20},4]##?,

##P=P_n\cup [3+\frac{\epsilon}{20},4]## does not include our "problem interval". If we were to add that interval to this collection, it would be ##[3,3+\frac{\epsilon}{20}]##. What is the max of ##f## on that interval? The min? What is the contribution to ##U(f,P)-L(f,P)## that comes from that sub interval (where the contribution of the ##i##th sub interval in the first case above was ##(M_i−m_i)(x_i−x_{i−1})##)?

Actually I am learning this on my own. I've taken an analysis class already but we did not cover integration at all. My book bases all their theorems on compact sets and I used ##\frac{\epsilon}{20}## because I chose it since my book likes to use these types of epsilons. As for my partition what I had said earlier would not work. I'm thinking along of the lines of ##P=\{[2, 2+\frac{1}{n}],...,[2+\frac{n-1}{n},3],[3,3+\frac{1}{n}],...[3+\frac{n-1}{n},4]\}## But I still can't seem to figure out a partition that avoids that ##x=3## problem.

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bonfire09 said:
As for my partition what I had said earlier would not work. I'm thinking along of the lines of ##P=\{[2, 2+\frac{1}{n}],...,[2+\frac{n-1}{n},3],[3,3+\frac{1}{n}],...[3+\frac{n-1}{n},4]\}## But I still can't seem to figure out a partition that avoids that ##x=3## problem.

I think this is a fabulous partition. You know how it behaves on ##[2,3]##; the total difference between the upper and lower sums is ##\frac{1}{n}## on that interval. On the interval ##[3,3+\frac{1}{n}]##, the max of ##f## is ##3## (at ##x=3##) and the min is ##2## (everywhere else). So the contribution of this interval to the total difference in the upper and lower sums is ##(3-2)\frac{1}{n}=\frac{1}{n}##. On every sub interval of ##[3+\frac{1}{n},4]##, the max of ##f## and the min of ##f## are the same, so there is no contribution to the total difference between the upper and lower sums there. I reckon you can take it from there?

So we aren't going to find a partition that avoids ##3##, because ##3## is in the interval of integration. But it turns out that this one little discontinuity doesn't give us too much trouble since we can make the partition as small as we need to virtually eliminate any contribution it could have made to the difference between the upper and lower sums. We didn't even really need to choose a partition that included ##3## as an endpoint of a sub interval. It just happened in this particular problem that it was really easy to do so.

Oh thanks a lot for you help. I honestly did not realize this whole time that I was thinking that f(x) takes on two different values at ##x=3## instead it takes only one value. I can take it on from here since its quite easy to do now. I'll make some more functions to practice with so I get a good hang of it.

No problem.

For what it's worth, your ##\frac{\epsilon}{2}## idea in the second post is solid and may prove very useful in other exercises, especially when you start using facts about the Riemann integrability of continuous functions.

Also, once you feel like you're getting the hang of this argument in specific cases, you may want to try and tackle the general case:

Prove the function ##f:[a,c]\rightarrow\mathbb{R}## defined by ##f(x) =\begin{cases} f_1(x), & \text{if }a\leq x\leq b \\f_2(x), & \text{if } b<x\leq c \end{cases}## is Riemann Integrable whenever ##f_1:[a,b]\rightarrow\mathbb{R}## and ##f_2:(b,c]\rightarrow\mathbb{R}## are.

And if that is too easy, you can try the piecewise defined by finitely many (instead of just two) Riemann integrable functions. It's basically the same argument. Or you can try countably many, but that's a bit tricky. It's a similar argument, but you need to be a bit clever.

## 1. What is a piecewise function?

A piecewise function is a mathematical function that is defined by different rules or equations on different parts of its domain. This means that the function may have different formulas for different intervals or sets of numbers.

## 2. How can I prove that a piecewise function is integrable?

In order to prove that a piecewise function is integrable, you can use the definition of integrability which states that if the function is bounded and the set of discontinuities has measure zero, then the function is integrable. You can also use the Riemann integrability criterion, which states that a function is integrable if and only if the upper and lower sums converge to the same value as the partition of the domain approaches zero.

## 3. Can a piecewise function be non-integrable?

Yes, a piecewise function can be non-integrable if it does not meet the criteria for integrability. For example, if the function is unbounded or if the set of discontinuities has a non-zero measure, then the function would not be considered integrable.

## 4. How does the concept of integrability apply to piecewise functions?

The concept of integrability applies to piecewise functions in the same way it applies to other functions. The goal is to determine if the function can be integrated over a certain interval or set of numbers. For piecewise functions, this involves considering the different equations or rules that define the function on different parts of its domain.

## 5. Can I use different integration techniques for different parts of a piecewise function?

Yes, it is possible to use different integration techniques for different parts of a piecewise function. This is because each part of the function may have different properties and requirements for integration. For example, one part of the function may be continuous and can be integrated using the fundamental theorem of calculus, while another part may be discontinuous and require a different method of integration.

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