infimum and supremum of riemann sum
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 $(x_{i1},x_i)$.
$\sup_{(x_{i1},x_i)} f(x)$ 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 $r \in (x_{i1},x_i)$. 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, $\inf_{(x_{i1},x_i)} f(x)$ 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 $s \in (x_{i1},x_i)$. 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 $D$):
$$ \mathcal U (f,D) = \sum_{i=0}^{n1} (x_{i}  x_{i1}) \left(\sup_{(x_{i1},x_i)} f(x)\right) = \sum_{i=0}^{n1} (x_{i}  x_{i1}) \times 1 = 32 = 1 $$
Now, we have from the definition of $\mathcal L(f,D)$ (once again, for all dissections $D$): $$ \mathcal L(f,D) = \sum_{i=0}^{n1} (x_{i}  x_{i1}) \left(\inf_{(x_{i1},x_i)} f(x)\right) = \inf_{i=0}^{n1} (x_{i}  x_{i1}) \times 0 = 0 $$
This is essentially what is happening in the proof.
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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, 2020Comments

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.

Doug M over 6 yearsIf 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 over 6 yearsLet 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 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 over 6 yearsOk then. I will answer the question.


shahrina ismail over 6 years@actoh this is brilliant. Thank you so much..

Sarvesh Ravichandran Iyer over 6 yearsYou are welcome!