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

[Adesh]

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!

 

[Infrared]

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 $\mathbb{R}.$

 

[Adesh]

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 $x_i - X \lt \epsilon$” ?

 

[Infrared]

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?

 

[Adesh]

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 $M_i= x_i-\epsilon$.

 

[Infrared]

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

 

[Adesh]

$M_i$ 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]

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?

 

[Adesh]

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.

 

[Infrared]

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.

 

[Adesh]

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.