# Showing Riemann integrability

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.

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

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