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Let $f: [a,b] \to \mathbb{R}$ be bounded. Show that $f$ is Riemann intergrable iff $$\bar{\int_{a}^{b}} f = -\left[\bar{\int_{a}^{b}} -f\right]$$

My attempt is as follows.

"$\Leftarrow$" $$\bar{\int_{a}^{b}} -f= \inf\{U(-f;P)\}=\inf\{-L(f;P)\}$$

So,

$$-[\bar{\int_{a}^{b}} -f = -[-\sup\{L(f;P)\}]=\sup\{f;P\}$$

I'm stuck on making sure I'm pushing definitions through properly and the $\to$ direction of the proof.

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I think your reasoning is right. –  Martin Argerami Nov 5 '12 at 1:37

1 Answer 1

up vote 0 down vote accepted

Expanded version:

\begin{align} \bar{\int_{a}^{b}} f &= -\bar{\int_{a}^{b}} -f \\ &= -\inf\{\sum (x_i-x_{i-1}) \sup_{x \in [x_{i-1},x_i]} (-f(x)) \} \\ &= -\inf\{-\sum (x_i-x_{i-1}) \inf_{x \in [x_{i-1},x_i]} f(x) \} \\ &= \sup\{\sum (x_i-x_{i-1}) \inf_{x \in [x_{i-1},x_i]} f(x) \} \\ &= \mbox{lower integral} \\ \end{align}

So that equation is equivalent to Riemann integrability.

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I'm trying to prove this identity: $\bar{\int_{a}^{b}} f = -\bar{\int_{a}^{b}} -f$. I'm stuck on messing with definitions of upper integrals. –  emka Nov 5 '12 at 1:50
    
@emka: Did you read my answer? It handles both directions. –  wj32 Nov 5 '12 at 1:53

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