# Homework Help: Test question we don't know who is right

1. Apr 7, 2009

### 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.

BACKGROUND

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.

Last edited: Apr 8, 2009
2. Apr 8, 2009

### futurebird

*Bump*

This is driving me crazy.

3. Apr 8, 2009

5. Apr 8, 2009

### futurebird

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

6. Apr 8, 2009

### 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.

7. Apr 8, 2009

### 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.

8. Apr 8, 2009

### Billy Bob

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 $$c$$.

9. Apr 8, 2009

### 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.)

10. Apr 8, 2009

### Billy Bob

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.

11. Apr 8, 2009

### 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.

12. Apr 8, 2009

### 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!

13. Apr 8, 2009

### futurebird

Thanks so much for all of your help!!

14. Apr 8, 2009

### Hurkyl

Staff Emeritus
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.