Absolute continuity, function of partition bound

[SqueeSpleen]

Given [a,b] a bounded interval, and f \in L^{p} ([a,b]) 1 < p < \infty, we define:
F(x) = \displaystyle \int_{a}^{x} f(t) dt, x \in [a,b]
Prove that exists K \in R such that for every partition:
a_{0} = x_{0} < x_{1} < ... < x_{n} = b :
\displaystyle \sum_{i=0}^{n-1} \frac{| F(x_{i+1}) - F(x_{i}) |^{p}}{|x_{i+1}-x_{i} |^{p-1}} \leq K

I know that F(x) is absolutely continuous and of bounded variation.

\frac{|F(x_{i+1}) - F(x_{i})|}{|x_{i+1} - x_{i}|} = F(\xi) for some \xi \in [x_{i}, x_{i+1}] (Lagrange Intermediate Value Theorem)
Then
\frac{|F(x_{i+1}) - F(x_{i})|^{p}}{|x_{i+1} - x_{i}|^{p-1}} = f(\xi)^{p} (x_{i+1}-x_{i})
As f \in L^{p} when x_{i+1} \to x_{i} f(x_{i+1}-x_{i}) when grows slower in some neighborhood of x_{i} than \frac{1}{x_{i+1}-x_{i}} because that function doesn't isn't L^{1}.
I was trying prove that the closenes of the x_{n} can't make the sum diverges, and that neither the number of terms but I don't know how to approach all the issues at the same time.
When the partition norm tends to zero we got \displaystyle \int_{a}^{b} | f (t) |^{p} dt
And when it doesn't I think that (2M)^{p} (b-a) where M is the maximum of F(x) with x \in [a,b] (which exists because F(x) is continuous in that closed set).
So I guess K = (2M)^{p} (b-a) + \displaystyle \int_{a}^{b} | f (t) |^{p} dt works
But I'm not sure how to prove it.

 

[brmath]

I got hung up when you wrote:

$\frac {F(x_{i+1} - F(x_i)|}{|x_{i+1} - x_i} = F(\zeta)$... surely you mean $F'(\zeta)$ which would be $f(\zeta)$.. From the next line I see you did mean that.

Sum up your $f^p(\zeta)(x_{i+1} - x_i).$ As the width of the partition goes to 0 you will have the Riemann integral of $f^p$.

I think you have to show (or state) that if it is Riemann integrable then it is Lebesgue integrable.
Can you take it from there?