1. ## Riemann Integrable

Prove the following: Let $\displaystyle I$ be a generalized interval, and let $\displaystyle f$ be a function which is uniformly continuous on $\displaystyle I$. Then $\displaystyle f$ is Riemann integrable.

We know that $\displaystyle f$ is bounded. We need to show that $\displaystyle \int\limits_{-I} f = \int\limits^{-}_{I} f$ (e.g upper and lower integrals are equal). Let $\displaystyle \varepsilon > 0$. Then there exists a $\displaystyle \delta > 0$ such that $\displaystyle |f(x)-f(y)| < \varepsilon$ whenever $\displaystyle |x-y| < \delta$. Skipping some steps, we have $\displaystyle f(x) < f(y) + \varepsilon$ for all $\displaystyle x,y \in J_{k}$ where $\displaystyle J_{k}$ are the subintervals of $\displaystyle I$ with length $\displaystyle |J_{k}| = (b-a)/N$. Then $\displaystyle \sup_{x \in J_{k}} f(x) \leq \inf_{y \in J_{k}} f(y) + \varepsilon$. And so $\displaystyle \int\limits^{-}_{I} f - \int\limits_{-I} f \leq \sum_{k=1}^{N} \varepsilon |J_{k}|$ or $\displaystyle \int\limits^{-}_{I} f - \int\limits_{-I} f \leq \sum_{k=1}^{N} \varepsilon(b-a)$.

From here, how would I conclude that $\displaystyle \int\limits^{-}_{I} f = \int\limits_{-I} f$?

2. Originally Posted by particlejohn

or $\displaystyle \int\limits^{-}_{I} f - \int\limits_{-I} f \leq \sum_{k=1}^{N} \varepsilon(b-a)$.
the RHS of your inequality is incorrect. the correct one is $\displaystyle \varepsilon (b-a),$ because $\displaystyle \varepsilon \sum_{k=1}^N |J_k|=\varepsilon(b-a).$ so $\displaystyle \forall \epsilon > 0: \ 0 \leq \int_{I}^{-} f - \int_{- \ I} f \leq \varepsilon (b-a).$ letting $\displaystyle \varepsilon \rightarrow 0$ gives us: $\displaystyle \int_{I}^{-}f=\int_{- \ I} f. \ \ \ \square$

3. I meant $\displaystyle \int\limits^{-}_{I} f - \int\limits_{-I} f \leq \sum_{k=1}^{N} \varepsilon|J_{k}|$.