Test question we don't know who is right

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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.
 
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*Bump*

This is driving me crazy.
 
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, \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.
 
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?
 
Billy Bob said:
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...
 
<br /> m_i = \inf \{ f(x) : x_{i-1} \leq x \leq x_i \}<br />

so

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

Then

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

It seems like the same idea as the sum I described above.
 
But now it looks like:

<br /> <br /> \inf \{ f(x) : a \leq x \leq b \} = \sup \{ f(x) : a \leq x \leq b \}<br /> <br />

making f(x) a constant function. I'm so lost.
 
futurebird said:
<br /> L(f, P) = \inf \{ f(x) : x_{i-1} \leq x \leq x_i \ni c \in (x_{i-1},x_i) \}<br />

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.
 
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
futurebird said:
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.
 
  • #11
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
Right! In fact, why not take P_j to be the simple partition a&lt;c-\frac1j&lt;c+\frac1j&lt;b?

Just three subintervals: two fat and one skinny!
 
  • #13
Thanks so much for all of your help!
 
  • #14
futurebird said:
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.
 
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