Prove that if f(x) is continuous on a closed interval [a,b] then it is integrable there.

I need some help with this one guys. Thanx.

Is Anyone Up to the challenge?

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- Apr 25th 2009, 10:32 AMVonNemo19Integrability of a function
Prove that if f(x) is continuous on a closed interval [a,b] then it is integrable there.

I need some help with this one guys. Thanx.

Is Anyone Up to the challenge? - Apr 25th 2009, 10:50 AMTheEmptySet

Here is the basic Idea

Let P be any partion of [a,b]

$\displaystyle U(P,f)-L(P,f)=\sum_{i=0}^{N}[f(x_M)-f(x_m)][x_{i+1}-x_{i}]$

Where $\displaystyle x_M$ is the max on each sub interval

$\displaystyle x_m $ is the min on each sub interval

They exsist by the extreem value theorem.

Then on each subinterval by the MVT there exists a $\displaystyle c \in (x_{i},x_{i+1})$ such that

$\displaystyle f(x_m)(x_{i+1}-x_i)=F(c_i) \mbox{ and } f(x_M)(x_{i+1}-x_i)=F(c_{i+1}$

so we get

$\displaystyle U(P,f)-L(P,f)=\sum_{i=0}^{N}[f(x_M)-f(x_m)][x_{i+1}-x_{i}]=\sum_{i=0}^{N}F(c_{i+1})-F(c_i)$

$\displaystyle =F(c_{N+1})-F(C_0)=F(b)-F(a)$

This is just a sketch you will need to justify and fill in the details

TES - Apr 25th 2009, 12:11 PMVonNemo19HUH?
How did you derive this from the MVT?:

Then on each subinterval by the MVT there exists a http://www.mathhelpforum.com/math-he...a3eb3899-1.gif such that

http://www.mathhelpforum.com/math-he...900ab120-1.gif - Apr 25th 2009, 02:46 PMTheEmptySet
I will leave the above proof here but it is NOT what you asked for. I should have read the question more carefully. That is my mistake.

To show that this is integrable we need to show that there exists a partition of [a,b] such that $\displaystyle U(f,p)-L(f,p)< \epsilon$

let $\displaystyle \epsilon > 0$ be given

Since f is continous of a closed bounded (i.e compact ) interval f is uniformly continous.

Now by the uniform cont. of f there exists a $\displaystyle \delta > 0$ such that for $\displaystyle |x-y|< \delta \implies |f(x)-f(y)| < \frac{\epsilon}{b-a}$

Choose a partition P of [a,b] such that the mesh (each subdivsion) is less that $\displaystyle \delta$ $\displaystyle P=\{ a=x_0,x_1,...x_{N+1}=b \}$

Now by the extreem value theeorem and the continuity of f in each subinterval $\displaystyle [x_i,x_{i+1}]$ there exists points

$\displaystyle x_m,x_M$ such that $\displaystyle f(x_m)$ is a min and $\displaystyle f(x_M)$ is a max on each sub interval

Then for this partition note that since the mesh is less that delta $\displaystyle f(x_M)-f(x_m) < \frac{\epsilon}{b-a}$

$\displaystyle U(f,p) -L(f,p)=\sum_{i=0}^{N}(f(x_M)-f(x_m))(x_{i+1}-x_1)< \frac{\epsilon}{b-a}\sum_{i=0}^{N}(x_{i+1}-x_i)=$

$\displaystyle \frac{\epsilon}{b-a}(b-a)=\epsilon$ - Apr 25th 2009, 04:06 PMKrizalid
- Apr 25th 2009, 04:11 PMVonNemo19Now for an Example
Now, can I apply this (by epsilon-delta proof) to the function, let's say... 2x, and show that this function is integrable?

I'm kind of at a loss, because in most cases of the epsilon-delta proofs of ordinary limits of functions ( as in taking the derivative for instance), the limit L is either known, or can be found through algebraic methods. But, in the case of integration, The value of the integral must remain a mystery until the function has, in fact, been integrated.

Now, if I choose an epsilon, how can I show that the Riemann Sum is approaching some value A without having previous knowlege of what that value may be?

Why does this stuff make my head spin?(Headbang)