# How to prove that ##M_i =x_i## in this upper Darboux sum problem?

In summary, the function defined above is between rational and irrational numbers if and only if it is equal to zero on those numbers.
Homework Statement
We have a function ##f:[0,1] \mapsto \mathbb R## such that
$$f(x)= \begin{cases} x& \text{if x is rational} \\ 0 & \text{if x is irrational} \\ \end{cases}$$
Relevant Equations
Upper Darboux sum is
$$U(f,P) = \sum_{i}^{n} M_i (x_i - x_{i-1})$$
We're given a function which is defined as :
$$f:[0,1] \mapsto \mathbb R\\ f(x)= \begin{cases} x& \text{if x is rational} \\ 0 & \text{if x is irrational} \\ \end{cases}$$
Let ##M_i = sup \{f(x) : x \in [x_{i-1}, x_i]\}##. Then for a partition ##P= \{x_0, x_1 , \cdots , x_n\}## we define the upper Darboux sum as
$$U(f,P) = \sum_{i}^{n} M_i (x_i - x_{i-1})$$
Now, I need your help in showing that ##M_i = x_i## for any interval ##[x_{i-1}, x_i]##. Please don't give complete solution, I want to learn it, I'm a beginner in Real Analysis.

Thank you!

You want to show that ##x_i## is the least upper bound for ##f## on the interval ##[x_{i-1},x_i].## This means you should check that:

1) ##x_i## is an upper bound for ##f## on ##[x_{i-1},x_i]##.

2) If ##y<x_i##, then ##y## is not an upper bound for ##f## on ##[x_{i-1},x_i]##.

Item 1 should be apparent from the definition of ##f##. To show 2, you'll want to use the density of ##\mathbb{Q}## in ##\mathbb{R}.##

Infrared said:
show 2, you'll want to use the density of ##\mathbb Q## in ##\mathbb R##
Okay, so we have the Theorem that between any two real numbers we can always find a rational number..

So, from the above Theorem can we conclude that “for any ##\epsilon## we can always find a rational number ##X## such that ##x_i - X \lt \epsilon##” ?

That's true, but probably not what you mean (as written, it would still be true with "integer" instead of "rational number").

And how do you want to use this to solve your question?

Infrared said:
And how do you want to use this to solve your question?
We can conclude that there exist a rational number within any ##\epsilon## of ##x_i## and hence ##M_i= x_i-\epsilon ##.

##M_i## should not depend on ##\varepsilon##. There is no ##\varepsilon## in the definition ##M_i=\sup\{f(x):x\in [x_{i-1},x_i]\}.##

Infrared said:
##M_i## should not depend on ##\varepsilon##. There is no ##\varepsilon## in the definition ##M_i=\sup\{f(x):x\in [x_{i-1},x_i]\}.##
Yes I understand that (I mean I noticed it after you pointed it out). Then how can I write what “I know”, I know that I can find a rational number as close to ##x_i## as I want, but how to write ##M_i## with this extra information?

Infrared said:
2) If ##y<x_i##, then ##y## is not an upper bound for ##f## on ##[x_{i-1},x_i]##.

You want to show this. Showing that ##y## is not an upper bound means finding an element ##x\in[x_{i-1},x_i]## such that ##f(x)>y \ (=x_i)##. How can you do this?

Infrared said:
You want to show this. Showing that ##y## is not an upper bound means finding an element ##x\in[x_{i-1},x_i]## such that ##f(x)>y \ (=x_i)##. How can you do this?
Given that ##y\lt x_i## implies that ##x_i - y \lt \epsilon’## but between any interval of real numbers we can always find a rational number, hence there exist a rational number in the interval ##[y, x_i]## and let that number be ##x##, then we’ve ##f(x) \gt f(y)\implies x \gt y## (I assumed y to be rational). And therefore, ##y## is not an upper bound.

This is better, but I don't know why you have an ##\epsilon'## floating around. Doesn't density of ##\mathbb{Q}## immediately imply that there is a rational number in ##[y,x_i]?## You should also do something like using the interval ##[\text{max}(0,y),x_i]## instead so that you're sure the rational you pick is indeed in ##[0,1]##!

So, ##x_i## is an upper bound, and there is no smaller upper bound. Hence it is the least upper bound.

Infrared said:
This is better, but I don't know why you have an ##\epsilon'## floating around. Doesn't density of ##\mathbb{Q}## immediately imply that there is a rational number in ##[y,x_i]?## You should also do something like using the interval ##[\text{max}(0,y),x_i]## instead so that you're sure the rational you pick is indeed in ##[0,1]##!

So, ##x_i## is an upper bound, and there is no smaller lower bound. Hence it is the least upper bound.
Thank you so much for helping me.

Infrared

## 1. How do I know if ##M_i = x_i## in an upper Darboux sum problem?

The equality ##M_i = x_i## can be proven by showing that the upper and lower Darboux sums for a given partition converge to the same value as the partition becomes finer. This can be done by using the properties of upper and lower sums, such as the fact that the upper sum is always greater than or equal to the lower sum for any partition.

## 2. Can I use a specific example to prove ##M_i = x_i## in an upper Darboux sum problem?

Yes, using a specific example can be a useful way to prove ##M_i = x_i## in an upper Darboux sum problem. By choosing a specific partition and function, you can calculate the upper and lower sums and show that they converge to the same value, thus proving the equality.

## 3. Are there any other methods for proving ##M_i = x_i## in an upper Darboux sum problem?

Yes, there are other methods for proving ##M_i = x_i## in an upper Darboux sum problem. One method is to use the definition of the Darboux sum and show that the supremum of the function on each subinterval is equal to the infimum of the function on that subinterval. Another method is to use the intermediate value theorem to show that the function takes on every value between its supremum and infimum on each subinterval.

## 4. Is it necessary to prove ##M_i = x_i## in an upper Darboux sum problem?

It is not always necessary to prove ##M_i = x_i## in an upper Darboux sum problem. In some cases, it may be enough to show that the upper and lower sums converge to the same value, without explicitly proving the equality. However, in other cases, it may be necessary to prove the equality in order to fully understand the behavior of the function on the interval.

## 5. Can I use calculus to prove ##M_i = x_i## in an upper Darboux sum problem?

Yes, calculus can be a useful tool for proving ##M_i = x_i## in an upper Darboux sum problem. Techniques such as the mean value theorem, the fundamental theorem of calculus, and integration by parts can be used to show that the upper and lower sums converge to the same value, thus proving the equality ##M_i = x_i##.

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