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Suppose that $f$ is continuous on $[a,b]$. Prove that given $\epsilon>0$, there exist points $x_0=a such that if

$E_k=\{y: f(x)=y\ for\ some\ x \in [x_{k-1}, x_k]\}$,

then sup$E_k -$inf$E_k<\epsilon$ for $k=1,2,...,n$.

Just from looking at the question, I suspect that I need to eventually the Bolzano Weierstrass Theorem and Extreme Value Theorem to complete this proof. However, I am having some difficulty setting up the problem so that I can use theorems as needed. Please help! I really appreciate any guidance that I can get.

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    @LJym89,not everyone speaks English as their first language.2012-10-28

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Notice that what the question asks is that continuous functions are Riemann-integrable.

If you have the theorem that states that continuous functions defined on compact sets are uniformly continuous, then it suffices you to choose $ a = x_0 < x_1 < \dots < x_n = b $ such that $ |x_{i+1} - x_i| < \delta, i = 0, 1, \dots, n-1 $ where $ \forall \varepsilon > 0,\quad \exists \delta > 0,\quad \forall x,y \in [a,b], \quad |x-y| < \delta \quad \Rightarrow \quad |f(x) - f(y)| < \varepsilon, $ i.e. the definition of uniform continuity. I believe this is a sufficiently good hint, if you want more details you can ask me again.

Hope that helps,

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    The problem with your argument is that the last line works if the $\delta_k$ for uniform continuity is such that |x_M - x_m| < \delta_k, which nothing ensures you that that will happen. You need to do the exact same reasoning from the start, but begin with the interval $[a,b]$ instead ; if you choose your partition with x_{k-1} - x_k < \delta for all $k$, then uniform continuity over $[a,b]$ will ensure you that |\mathrm{sup} - \mathrm{inf}| < \varepsilon/(b-a), so that when you compute the Riemann sum, it is smaller than $(b-a) \varepsilon/(b-a) = \varepsilon$.2012-11-11