Test question we don't know who is right

[futurebird]

Suppose $$\alpha(x) = [x]$$ is the floor function, then what is the value of $$\int_{0}^{n}f d \alpha$$
for $$f(x) \in R_{\alpha}[0,n]$$

where n is an integer?This was a question on my exam. I want to know if I got it right. Some say the answer is zero, but I think it is:

$$\sum_{i=1}^{n-1} \max \{ f(x) : i < x \leq i+1\}$$

Because $$M_i = \sup \{ f(x) | x_{i-1} \leq x \leq x_i \}$$ gets multiplied with the difference of the endpoints of every possible partition and, with a fine enough partition, we will get the value 0 most of the time and we will get the vale 1 n-1 times.BACKGROUNDLet me add some more info:

To take the Stietjes integral you write $$P = \{a= x_0 < x_1 < \cdots < x_n = b \}$$ a partition of the interval [a ,b]. Then $$\Delta \alpha_i = \alpha( x_i) - \alpha (x_{i-1})$$ for i= 1, ..., n. Next for each i = 1, ..., n we define:

$$m_i = \inf \{ f(x) : x_{i-1} \leq x \leq x_i \}$$
$$M_i = \sup \{ f(x) : x_{i-1} \leq x \leq x_i \}$$

Now we can define the lower and upper Stietjes sums:

$$L(f, P) = \sum_{i=1}^{n}m_i\Delta \alpha_i$$
$$U(f, P) = \sum_{i=1}^{n}M_i\Delta \alpha_i$$

Now we can define the lower and upper Stietjes integrals, which are equal for any Stietjes integrable function over a given $$\alpha$$.

$$\bar{\int_{a}^{b}}f d \alpha = \inf_P U(f,P)$$

$$\int_{a}^{b}f d \alpha = \sup_P L(f,P)$$

So that's what we are talking about with this problem.

 

[futurebird]

*Bump*

This is driving me crazy.

 

[HallsofIvy]

What you have is basically right. The best way to get the details straight is to look at simple examples. $\alpha(x)$ is constant except where x is an integer so $d\alpha$ is 0 except at integers. If n= 1, [tex]\int_0^1 f(x)d\alpha= 0[/itex]. If n= 2, the integral is f(0)+ f(1). If n= 3, the integral is f(0)+ f(1)+ f(3), etc.

 

[Billy Bob]

futurebird, the n in the Stieltjes sum is not the same n as in the statement of your problem.

Your intuition should be able to guide you quickly the correct answer. Review the simpler case where $$\alpha(x)=0$$ for $$x<c$$ and $$\alpha(x)=1$$ for $$x\ge c$$. What then is $$\int_a^b f(x)\,d\alpha(x)$$ when $$a<c<b$$?

 

[futurebird]

futurebird, the n in the Stieltjes sum is not the same n as in the statement of your problem.

Your intuition should be able to guide you quickly the correct answer. Review the simpler case where $$\alpha(x)=0$$ for $$x<c$$ and $$\alpha(x)=1$$ for $$x\ge c$$. What then is $$\int_a^b f(x)\,d\alpha(x)$$ when $$a<c<b$$?

Well for $$\int_a^b f(x)\,d\alpha(x)$$ the change in alpha is 1 at least once for all partitions...

 

[futurebird]

$$m_i = \inf \{ f(x) : x_{i-1} \leq x \leq x_i \}$$

so

$$L(f, P) = \inf \{ f(x) : x_{i-1} \leq x \leq x_i \ni c \in (x_{i-1},x_i) \}$$

Then

$$\int_{a}^{b}f d \alpha = \sup_P L(f,P) = \inf \{ f(x) : a \leq x \leq b \}$$

It seems like the same idea as the sum I described above.

 

[futurebird]

But now it looks like:

$$\inf \{ f(x) : a \leq x \leq b \} = \sup \{ f(x) : a \leq x \leq b \}$$

making f(x) a constant function. I'm so lost.

 

[Billy Bob]

$$L(f, P) = \inf \{ f(x) : x_{i-1} \leq x \leq x_i \ni c \in (x_{i-1},x_i) \}$$

This is $$L(f, P)$$ where $$P$$ depends on the choice of $$x_{i-1}$$ and $$x_i$$, with the only requirement being that $$c \in (x_{i-1},x_i)$$.

Take $$x_{i-1}$$ and $$x_i$$ very close to $$c$$.

 

[futurebird]

Why would I want to make the partition so snug around c? I mean I know that can be done... but why? I want to say that the vale is f(c).

But I don't think this is justified. (or I don't understand why it is justified.)

 

[Billy Bob]

Why would I want to make the partition so snug around c? I mean I know that can be done... but why? I want to say that the vale is f(c).

But I don't think this is justified. (or I don't understand why it is justified.)

You have to find $$\sup_P L(f,P)$$, and you get closer and closer to the sup when the partition is more and more snug.

 

[futurebird]

OK. I just thought "sup" meant whatever partition gave the greatest value and had nothing to do with "snugness" -- so if a big messy partition gives the largest value it's the "sup" -- but now I'm thinking that this might not ever happen I think we had a theorem that said refinements are bigger... OK...

This is making more sense now.

Thanks.

 

[Billy Bob]

Right! In fact, why not take $$P_j$$ to be the simple partition $$a<c-\frac1j<c+\frac1j<b$$?

Just three subintervals: two fat and one skinny!

 

[futurebird]

Thanks so much for all of your help!

 

[Hurkyl]

OK. I just thought "sup" meant whatever partition gave the greatest value and had nothing to do with "snugness" -- so if a big messy partition gives the largest value it's the "sup" -- but now I'm thinking that this might not ever happen I think we had a theorem that said refinements are bigger... OK...

This is making more sense now.

Thanks.

I've lost your train of thought, but I think you may have overlooked the fact that the lower integral is the supremum of an infimum -- and you forgot about the infimum part.