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    The title of this thread is misleading.
    • Thread Starter

    (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?

    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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Updated: August 10, 2009


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