Intuition behind $m(b-a) \leq L(P,f) \leq U(P,f) \leq M(b-a)$ In the text "An Introduction to Measure and Integration by Rana" I'm having trouble gaining intuition behind the following Proposition in $(1)$
$(1)$
$$\text{Proposition}$$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,  \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \, \, \, \,\,\,\,\,\,\,$For every partition $P$ of $[a,b]$:
$$m(b-a) \leq L(P,f) \leq U(P,f) \leq M(b-a)$$ 
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,$where $m:=\inf\{f(x)|a\le x\le b\} \, \text{and} \, M:=\sup\{f(x)|a\le x\le b\}$
$\text{Remark}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$Substituting one can observe the following:
$$\inf\big\{f(x)| a \leq x\leq b \big\}(b-a) \leq \sum_{i=1}^{n}\inf \big\{f(x)|x_{i-1} \leq x \leq x_{i} \big\}(x_{i}-x_{i-1}) \leq \sum_{i=1}^{n}\sup \big\{f(x)|x_{i-1} \leq x \leq x_{i} \big\}(x_{i}-x_{i-1}) \leq \sup\big\{f(x)| a \leq x\leq b \big\}$$
$\text{Remark}$
Looking at the proportion I managed to break $(1)$ in to various cases:
Case $(1)$:
$$\inf\big\{f(x)| a \leq x\leq b \big\}(b-a) \leq \sum_{i=1}^{n}\inf \big\{f(x)|x_{i-1} \leq x \leq x_{i} \big\}(x_{i}-x_{i-1})$$
Case $(2)$:
$$\sum_{i=1}^{n}\inf \big\{f(x)|x_{i-1} \leq x \leq x_{i} \big\}(x_{i}-x_{i-1}) \leq \sum_{i=1}^{n}\sup \big\{f(x)|x_{i-1} \leq x \leq x_{i} \big\}(x_{i}-x_{i-1})$$
Case $(3)$:
$$\sum_{i=1}^{n}\sup \big\{f(x)|x_{i-1} \leq x \leq x_{i} \big\}(x_{i}-x_{i-1}) \leq \sup\big\{f(x)| a \leq x\leq b \big\}(b-a)$$
For each of the distinct cases I'm having trouble putting things together and creating a picture?

Let $L(P, f) = \sum{m_i\Delta x_i}$ and $U(P, f) = \sum{M_i\Delta x_i}$
Case $(1)$:
$m(b-a) = \sum{m\Delta x_i} \le \sum{m_i\Delta x_i}$, because $m$ is minimum of $f$ on $[a,b]$, and each $m_i$ is a local minimum for each $\Delta x_i$, so $m_i \ge m$ $\forall i$
Case $(2)$:
$\sum{m_i\Delta x_i} \le \sum{M_i\Delta x_i}$, because $m_i$ and $M_i$ are local $min$ and $max$ for each $\Delta x_i$, so $m_i \le M_i$
Case $(3)$:
$\sum{M_i\Delta x_i} \le \sum{M \Delta x_i} = M(b-a)$, because $M$ is maximum of $f$ on $[a,b]$, and each $M_i$ is a local maximum for each $\Delta x_i$, so $M_i \le M$ $\forall i$
