Intuition behind $m(b-a) \leq L(P,f) \leq U(P,f) \leq M(b-a)$

Question

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?

Answer 1

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 $M_i \le M$ $\forall i$

Answer 2

I try to think of it less rigorously. The maximum possible sum, is the maximum value of the function times the length of the interval. Or, if you think visually, the box with the biggest area on the interval is that where the height is the biggest possible (i.e the max/sup). A similar analogy can be used for the inf.