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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.2012-11-05

1 Answers 1

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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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    @emka: Did you read my answer? It handles both directions.2012-11-05