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so I'm working on this problem in real analysis. the exercise in the book says that "prove that the conclusion of the theorem is false."

I am rather confused because the conclusion of the theorem is not standard for me. The theorem proceeds as follows:

$f$ is continuous on $[a,b]$. Then there exists $I$ s.t. for all $\epsilon >0$ there exists $\delta>0$ s.t. $|\text{Riemann sum} - I| < \epsilon$ where $I$ is the definite integral.

I'd appreciate if someone could help me identify what I should show so that the conclusion of the above theorem is false. Should I show that no $I$ exists? If so, should I still assume the validity of the other given data in the theorem?

Help appreciated!

(Sorry if it's vague. The complete theorem can be found here on page 333. it's theorem 8.1: )

EDIT (Chris Eagle)

The actual statement of the theorem in question is

Let $f$ be a continuous function on an interval $[a,b]$. Then there is a number $I$, called the definite integral of $f$ on $[a,b]$, such that for each $\varepsilon>0$ there is a $\delta>0$ so that $$\left|\sum\limits_{k=1}^n f(\xi_k)(x_k-x_{k-1})-I\right |<\varepsilon$$ whenever $[x_0,x_1],[x_1,x_2],\ldots,[x_{n-1},x_n]$ is a partition of the interval $[a,b]$ into subintervals of length less than $\delta$ and each $\xi_k$ is a point in the interval $[x_{k-1},x_k]$.

The exercise is

Prove that the conclusion of the theorem is false if $f$ is discontinuous at any point in the interval and is not bounded.

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Have you given the exact, complete wording of the exercise and theorem as in your book? I can't make any sense of it as it stands. – Chris Eagle Feb 3 '13 at 16:59
It doesn't look like $I$ is really used. It's mentioned, but not used. – goblin Feb 3 '13 at 17:08

My guess is that they've given you some function that somehow fails to satisfy the hypotheses of the theorem and then said "prove that the conclusion of the theorem is false." If it's a theorem, then its conclusion is true when its hypotheses are satisfied. In order to say more, I'd need to know exactly what comes just before the words "prove that the conclusion of the theorem is false."

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Hi michael, yes they specify that the function in consideration is discountinuous and unbounded. Using this information, I have to prove that the conclusion of the theorem is false... – Tomas Jorovic Feb 3 '13 at 17:27
@user1237300 Please edit your question so that this essential detail is included. The first line is literally quoting the question out of context. – Erick Wong Feb 3 '13 at 17:35

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