Proof of Riemann condition on $\mathbb{R^n}$

Question

Let $Q$ be a rectangle; let $f:Q\to\mathbb{R}$ be a bounded function. Then

$$\underline{\int_{Q}} f \leq \overline{\int_{Q}} f$$

equality holds iff given $\epsilon>0$, there exists a corresponding partition $P$ of $Q$ for which

$$U(f,P)-L(f,P)<\epsilon.$$

Proof. Case 1: Assume $\underline{\int_{Q}} f = \overline{\int_{Q}} f.$ Chooese a partition $P$ so that $\int_{Q} f-L(f, P)<\frac{\varepsilon}{2}$ and a partition $P'$ so that $\int_{Q} f-U(f, P')<\frac{\varepsilon}{2}.$ Let $P''$ be common refinement of both $P_1$ and $P_2$. Since

$$L(f, P) \leq L\left(f, P^{\prime \prime}\right) \leq \int_{Q} f \leq U\left(f, P^{\prime \prime}\right) \leq U\left(f, P^{\prime}\right),$$

then lower and upper sums for $f$ determined by $P''$ are within $\epsilon$ of each other.

Case 2: Now assume $\underline{\int_{Q}} f <\overline{\int_{Q}} f.$ Show $U(f,P)-L(f,P)<\epsilon.$

My question is: 1) I coulnd't show Case 2, can you help?

2) Can you explain that why these inequality $\int_{Q} f-L(f, P)<\frac{\varepsilon}{2}$ and $\int_{Q} f-U(f, P')<\frac{\varepsilon}{2}$ are true?

Thanks...

Answer 1

Given $\epsilon > 0$, let $P$ be such that $U(f, P) - L(f, P) < \frac{\epsilon}{2}$. Then we have $$L(f, P) \leq \underline{\int_Q}f \leq \overline{\int_Q}f \leq U(f, P)$$ So by the triangle inequality, we have \begin{align*}\overline{\int_Q}f - \underline{\int_Q}f &\leq \Big(U(f, P) - \overline{\int_Q}f\Big) + \Big(\underline{\int_Q}f - L(f, P)\Big) \\ &\leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\end{align*} The desired conclusion follows.