# confused with quantifiers in real analysis problem

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: http://books.google.com.sg/books?id=Wlb-o7HBh2YC&pg=PA337&lpg=PA337&dq=prove+that+the+conclusion+of+theorem+8.1+is+false+if&source=bl&ots=P1wHS88ZLH&sig=UDjp0f9IcGBy6v3jscJlcf-xhW8&hl=en&sa=X&ei=fJsOUbOQEI7KrAfH-oGIDw&redir_esc=y#v=onepage&q=prove%20that%20the%20conclusion%20of%20theorem%208.1%20is%20false%20if&f=false )

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
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