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Why does choosing $\varepsilon$ to be arbitrarily small mean equality?

Discussion in 'Mathematics' started by angelo086, Aug 1, 2020.

  1. angelo086

    angelo086 Guest

    Suppose that $\int_a^bf(x)dx$ exists and there is a number A such that, for every $\varepsilon> 0$ and $\delta > 0$ , there is a partition $P$ of $[a,b]$ with $||P||<\delta$ and the Riemann sum of $f$ over $P$ that satisfies the inequality $|\sigma -A|$ . Show that $\int_a^bf(x)dx=A$.

    In the last part of this proof it follows that,

    $|A-\int_a^bf(x)dx| \leq |A-\sigma|+|\sigma-\int_a^bf(x)dx| \leq 2\varepsilon$

    Then it says we can choose $\varepsilon$ to be arbitrarily small so that $A=\int_a^bf(x)dx$.

    If we choose $\varepsilon$ to be very small how does $A=\int_a^bf(x)dx$ follow?

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