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    The title of this thread is misleading.
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    (Original post by Simplicity)
    The title of this thread is misleading.
    Not anymore.

    Another question:
    Let f:[a,b] \to \mathbb{R} be continuous, and consider the partition P_n = \{a + k(b-a)/n : k = 0,...,n\}. Then:
    \lim_{n \to \infty} L(P_n,f) = \lim_{n \to \infty} U(P_n,f) = \int^b_af
    Where U and L are the upper and lower sums as I defined previously.

    Using this, I'm trying to show that for g:[0,1] \to \mathbb{R}, that
    \int^1_0g = \lim_{n\to\infty}\frac{1}{n}\sum  ^n_{k=1}g(\frac{k}{n})

    Now, this after calculating the lower sum for such a function g, I need to show that g(\frac{1}{n}) + g(\frac{2}{n}) + ... + g(\frac{n-1}{n}) + g(1) = min\{g(x) : 0 \le x \le \frac{1}{n}\} + min\{g(x) : \frac{1}{n} \le x \le \frac{2}{n}\} + ... + min\{g(x) : \frac{n-1}{n} \le x \le 1\}

    I can't see how this is true in general. If g is increasing or decreasing then I can see one of the sums might amount to this, but if g is something like an appropriate scaled sine function then it won't be true all the time. Any hints?
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    You're not showing the sums are equal (as you surmise, they won't be). What you need to do is show that the difference can be made arbitrarily small.

    As I said about 7 posts ago, you should use the fact (or prove the fact) that a continuous function on a closed interval is uniformly continuous.
 
 
 
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