Let $f:[a,b] \rightarrow \mathbb{R}$ be called a step function if there is a partition $P=\{a=t_0,t_1,…,t_n=b\}$ when $f$ restricted to $(t_{i-1},t_i)$ is equal to a constant $c_i$ for any $1\leq i \leq n$.
Prove that if $f$ is a step function then
$(i) \overline{\int_a^b} f(x) \, dx = \sum_{i=1}^n c_i(t_i-t_{i-1})$
$(ii) \underline{\int_a^b} f(x) \, dx = \sum_{i=1}^n c_i(t_i-t_{i-1})$
$(iii) f$ is riemann integrable and $\int_a^b f(x) \, dx = \sum_{i=1}^n c_i(t_i-t_{i-1})$
$(i)$ and $(ii) \implies (iii)$, so I only need to prove $(i)$ and $(ii)$
I know that if I prove that for any $1\leq i \leq n$, $\overline{\int_a^b}\left.f\right|_{[t_{i-1},t_i]}\, dx = c_i(t_i-t_{i-1})$ then $\overline{\int_a^b} f(x) \, dx = \sum_{i=1}^n c_i(t_i-t_{i-1})$. And it is the same for the lower integral.
So how can i prove that $1\leq i \leq n$, $\underline{\int_a^b}\left.f\right|_{[t_{i-1},t_i]}\, dx = c_i(t_i-t_{i-1})$ and $1\leq i \leq n$, $\overline{\int_a^b}\left.f\right|_{[t_{i-1},t_i]}\, dx = c_i(t_i-t_{i-1})$?
Best Answer
$\textbf{Hint:}$ For partition $\mathbb{P}_i=\{I_1,I_2,\cdots,I_n\}$ of $[t_{i-1},t_i]$ $$\displaystyle\underline{\int_a^b}\left.f\right|_{[t_{i-1},t_i]}\, dx =\displaystyle\underline{\int_{t_i-1}^{t_i}}f dx=\sup_{\mathbb{P}_i}\left\{\sum_{i=0}^nm_i\Delta x_i\right\}=\cdots$$
$\displaystyle m_i=\inf_{I_k\::\:I_k\in\mathbb{P}_i} f(x)$ and $\Delta x_k=|x_k-x_{k-1}|$ where $(x_k,x_{k-1})= I_k\in\mathbb{P}_i$