Test your Skill Ceiling in Integration

Calculus Level 4

$\large \int _{ 0 }^{ 2 }{ \lceil x \rceil \, d \lfloor x \rfloor }$

Evaluate the Riemann-Stieltjes integral above, using the definition given in the link.

Enter 666 if you come to the conclusion that this integral fails to exist.

Bonus What if we use the definition of the Riemann-Stieltjes integral found here (Definition 6.7.2)

Notations : $\lfloor \cdot \rfloor$ denotes the floor function and $\lceil \cdot \rceil$ denotes the ceiling function .

Inspiration .

The answer is 666.

1 solution

Andrew Tawfeek
May 23, 2016

Not exactly a solution, but wouldn't putting dx into a floor function cause the integral to become the infinite sum of the product of the function and zero?

Any approximating sum will have at most two non-zero summands; we need to think about those.

Otto Bretscher - 5 years ago

Otto, I think you already knew I was going to reply... Ok, my definition of a real bounded function $f$ respect to a real increasing (not necessarialy strictly) $g$ , i.e, my definition about $f$ is Riemann-Stieljes integrable respect to $g$ is like this ,i.e, if there is a number L so that for all ε > 0, there is a partition Pε so that, for all partitions P with P ⊇ Pε and all evaluation sequences Z for P, we have | Ig(f,P,Z) − L | < ε. With this definition I can prove (Stament1) that if $f$ is R-S integrable respect to $g$ in [a, c] and in [c, b] (a < c < b), then $f$ is R-S integrable respect to $g$ in [a, b] and $\int_a^b f \space dg = \int_a^c f \space dg + \int_c^b f \space dg$ . For example, in this case, $\int_0^2 \lceil x \rceil \space d\lfloor x \rfloor = \int_0^1 \lceil x \rceil \space d\lfloor x \rfloor + \int_1^2 \lceil x \rceil \space d\lfloor x \rfloor = 1 + 2 = 3$ . But, I think also your proof is based in the characterization which I consider a sufficient condition, but not a necessary condition here ( at the end of the first page), am I wrong? Do you want me to prove my statement (Statement1)? my proof is very similar to that used when we deal with the Riemann integral.. Anyway, I'm looking forward a proof of this exercise, which must be due to the characterization... I don't see other possibility.

Guillermo Templado - 5 years ago

It's great to hear from you, Compañero! You are giving a very lucid solution of the Bonus question; thank you! (+1)

I will be quite busy today as my "summer travel season" is about to start; lots of packing to do! Like many aliens in the USA, I'm a migrant worker; I move down to Boston to teach Calculus at the Harvard Summer Program, as I have done for 30 years.

One of my old Soviet texts explains these Riemann-Stieltjes issues quite clearly. They define the useful concept of the variance of a partition (with respect to given $f$ and $g$ on a given interval $[a,b]$ ) as the supremum of the difference of any two approximating sums. The two definitions of the R-S integrability can then be states succinctly as follows:

(I) Wolfram For every $\epsilon >0$ there exists a $\delta>0$ such that the variance of any partition with mesh size $<\delta$ is less than $\epsilon$ .

(II) For every $\epsilon >0$ there exists a partition with variance $< \epsilon$ . (This is equivalent to Rudin's definition and the definition given here , but they use much more convoluted language.)

Obviously, (I) implies (II), but our example here shows that they are not equivalent.

In this problem we use Definition I and in the Bonus we use Definition II.

The key to showing that our integral here is not R-S integrable in the sense of Definition (I) is to consider partitions where 1 is not a mesh point. The variance of any such a partition will be at least 1 since the integrand $f(x)=\lceil x\rceil$ attaind the values 1 and 2 on the interval containing 1.

Otto Bretscher - 5 years ago

Wouldn't everything being summed up with this integral be zero though? Just be an infinite sum of 0+0+0+...?

Andrew Tawfeek - 5 years ago

We have to think about the interval(s) where the "integrator" $g(x)=\lfloor x \rfloor$ makes a step. See my comments to @Guillermo Templado 's post

Otto Bretscher - 5 years ago

Our professor told us that this integral DOES exist if we use Rudin's definition, because floor function is continuous from the right and ceiling function is continuous from the left, so ceiling function is continuous at every discontinuity of floor function.

Tianxia Jia - 4 years, 1 month ago

Test your Skill Ceiling in Integration

The answer is 666.

1 solution

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