Re: Possible proof of Gabriel's Theorem?

From: Jason (logamath_at_yahoo.com)
Date: 03/07/05 

Date: 7 Mar 2005 06:39:40 -0800

> Definition: Suppose that f:[a,b] -> R is bounded. We say f is
> (Riemann) integrable if for every epsilon > 0 there exists
> delta > 0 such that if a = t_0 < ... < t_n = b,
> I_j = [t_{j-1}, t_j], M_j is the sup of f on I_j and m_j is
> the inf of f on I_j then the sum of (M_j - m_j)*(length(I_j))
> is less than epsilon.
> You really didn't know that? If you don't know what the word
> "integrable" means then the fact that you've been replying
> to posts where people have been discussing the concept seems
> very curious.

Actually I know exactly what it is. I am questioning the relevance of
his post. I am not sure he knows exactly what he is talking about.

See, this is an example of you skirting the issues. Forget about all
the other bozos on here! I am trying to discourage anyone who does not
know what they are talking about from posting to this thread.

As for the definition you provided, although I know it, I do not fully
agree with it. As I stated earlier, I don't like to talk about constant
functions such as f(x) = c for c some constant as being differentiable.
They are not differentiable in my opinion. Now the above definition
covers your ass for such functions but in truth, the ftoc fails
miserably for any such function. i.e. f' = 0. The indefinite integral
of 0 is some constant and its evaluation is 0 since f(x+w)-f(x) = 0 for
any such constant function. So here you have a constant function which
is differentiable but not *integrable*. Wow, makes a lot of sense for a
learner, doesn't it?
The epsilon-delta definition is a load of hogwash in my opinion. In
this case of a constant functioin it is stating that f(x) = c is *not*
integrable since no epsilon > 0 exists. I hinted at this earlier but
seeing you were all so stupid, I simply avoided it.

So why don't you try to answer the questions and stop being so arrogant!

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