# 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?

• The point is that $L(P,f)$ is non decreasing, and the finer the partition $P$ the better is the approximation of the area by $L(P,f)$. On the other hand $b-a$ comes from the coarsest partition of $I$, namely $z_0=a,z_1=b.$ Similarly for $U(P,f)$. Jul 12, 2017 at 1:11
• I think you should draw a picture. Jul 12, 2017 at 2:02
• @user 1987 what do you mean by the "coarsest partition" ? Jul 12, 2017 at 16:24
• @Zophikel Coarsest means the smallest number of subintervals. And the smallest number is obviously 1. Jul 12, 2017 at 19:24
• @Zophikel $U(P,f)$ is non-increasing. Also, the ultimate goal is not to approximate upper/lower sum. The goal is to make the gap $U(P,f)-L(P,f)$ smaller and smaller so that whats in between, namely, the definite integral, gets squeezed in. Jul 26, 2017 at 13:48

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 $M_i \le M$ $\forall i$

• Can you include a visual in your answer please Jul 26, 2017 at 17:01
• @Zophikel what do you mean by visual?
– user466918
Jul 26, 2017 at 17:03
• A picture basically Jul 26, 2017 at 17:04
• Is my intution correct:the goal is to make the gap U(P,f)−L(P,f) smaller and smaller so that whats in between, namely, the definite integral, gets squeezed in. Jul 26, 2017 at 17:09
• @Zophikel Maybe this will help? imgur.com/a/Lvkge
– user466918
Jul 26, 2017 at 17:12

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.