infimum and supremum of riemann sum

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We'll see this in words, and then translate it into mathematical language. Then the supremum/infimum thing will fall out.

In every interval, we know that there is at least one rational and one irrational. $f(x) = 0$ if $x$ is rational and $f(x)=1$ if $x$ is irrational.

We have a dissection of $[2,3]$.This will give the intervals $(x_{i-1},x_i)$.

$\sup_{(x_{i-1},x_i)} f(x)$ is the largest value that $f$ takes in this interval. However, this interval contains at least one irrational, call that irrational $r$ , then $f(r) =1$, where $r \in (x_{i-1},x_i)$. Now, the maximum value $f$ takes anywhere is $1$ (from definition), so the supremum of $f$ over every interval is $1$, since in every interval, $f$ takes the value $1$ at least once.

Similarly, $\inf_{(x_{i-1},x_i)} f(x)$ is the smallest value that $f$ takes in this interval. However, this interval contains at least one rational, call that rational $s$ , then $f(s) =0$, where $s \in (x_{i-1},x_i)$. Now, the minimum value $f$ takes anywhere is $0$ (from definition), so the infimum of $f$ over every interval is $0$, since in every interval, $f$ takes the value $0$ at least once.

Now, we have from the definition of $\mathcal U(f,D)$ (for all dissections $D$):

$$ \mathcal U (f,D) = \sum_{i=0}^{n-1} (x_{i} - x_{i-1}) \left(\sup_{(x_{i-1},x_i)} f(x)\right) = \sum_{i=0}^{n-1} (x_{i} - x_{i-1}) \times 1 = 3-2 = 1 $$

Now, we have from the definition of $\mathcal L(f,D)$ (once again, for all dissections $D$): $$ \mathcal L(f,D) = \sum_{i=0}^{n-1} (x_{i} - x_{i-1}) \left(\inf_{(x_{i-1},x_i)} f(x)\right) = \inf_{i=0}^{n-1} (x_{i} - x_{i-1}) \times 0 = 0 $$

This is essentially what is happening in the proof.

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shahrina ismail
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shahrina ismail

I am a PhD student doing elliptic curves and I love every single thing about my research.

Updated on April 23, 2020

Comments

  • shahrina ismail
    shahrina ismail over 3 years

    I saw this example on internet as I was studying for my exam. I am a little bit confused as why the supremum is 1 and infimum is 0? Also at the end it has a conclusion that f is not integrable in [2,3], is it because L and U have different answers? Please help me to clarify this. Thank you.

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    • Doug M
      Doug M over 6 years
      If it is Riemann integrable then as the partition gets sufficiently fine, then the upper sum should approach the lower sum. But in this case, they don't.
    • Sarvesh Ravichandran Iyer
      Sarvesh Ravichandran Iyer over 6 years
      Let me confirm: You are not sure why the supremum of $f$ over any such interval contained in $[2,3]$ is $1$, and why the infimum is zero? For the second question, you are right. $1 \neq 0$, so the upper and lower Riemann sums don't coincide, hence the function isn't integrable.
    • shahrina ismail
      shahrina ismail over 6 years
      @астонвіллаолофмэллбэрг Yes I am not sure why the infimum and supremum are as such given in the example.. as for the second question it is clear now to me that 1 does not equal 0, thus function f is not integrable. It is just teh infimum and the supremum part confusing me. Thank you for your help.
    • Sarvesh Ravichandran Iyer
      Sarvesh Ravichandran Iyer over 6 years
      Ok then. I will answer the question.
  • shahrina ismail
    shahrina ismail over 6 years
    @actoh this is brilliant. Thank you so much..
  • Sarvesh Ravichandran Iyer
    Sarvesh Ravichandran Iyer over 6 years
    You are welcome!