Fix a closed interval $[a, b]$. For a bounded function $f$ on $[a, b]$ and a dissection $\mathcal{D}$ of $[a, b]$, how are the lower sum $s(f, \mathcal{D})$ and upper sum $S(f, \mathcal{D})$ defined? Show that $s(f, \mathcal{D}) \leqslant S(f, \mathcal{D})$.

Suppose $\mathcal{D}^{\prime}$ is a dissection of $[a, b]$ such that $\mathcal{D} \subseteq \mathcal{D}^{\prime}$. Show that

$$s(f, \mathcal{D}) \leqslant s\left(f, \mathcal{D}^{\prime}\right) \text { and } S\left(f, \mathcal{D}^{\prime}\right) \leqslant S(f, \mathcal{D})$$

By using the above inequalities or otherwise, show that if $\mathcal{D}_{1}$ and $\mathcal{D}_{2}$ are two dissections of $[a, b]$ then

$$s\left(f, \mathcal{D}_{1}\right) \leqslant S\left(f, \mathcal{D}_{2}\right)$$

For a function $f$ and dissection $\mathcal{D}=\left\{x_{0}, \ldots, x_{n}\right\}$ let

$$p(f, \mathcal{D})=\prod_{k=1}^{n}\left[1+\left(x_{k}-x_{k-1}\right) \inf _{x \in\left[x_{k-1}, x_{k}\right]} f(x)\right]$$

If $f$ is non-negative and Riemann integrable, show that

$$p(f, \mathcal{D}) \leqslant e^{\int_{a}^{b} f(x) d x} .$$

[You may use without proof the inequality $e^{t} \geqslant t+1$ for all $t$.]