Some questions to understand Riemann integrability criterion

[mcastillo356]

TL;DR Summary

I've got a thirty three pages notes on Riemann integrability criterion, and stucked on page four. Haven`t read it entirely, but my intention is to understand completely all the document (if I can).

Hi, PF

I've began to read the notes, and at the fourth page, there is Theorem 2.4. Quoting it:

Theorem 2.4'. Let $\{P_n\}$ be a sequence of partitions satisfying $\lim_{n\rightarrow\infty}\,||P_n||=0$
1st question: What is the shape of a sequence of partitions?
Attempt:
$\{P_1\}=\{a,b\}$
$\{P_2\}=\{a,\frac{a+b}{2},b\}$
$\{P_3\}=\{a,\frac{2a+b}{3},\frac{a+2b}{3},b\}$
and so on, just to display an easy one.

$\{P_n\}=\frac{a+b}{n}?$

2nd question It's about the expression $0\leq\overline{S}(f,\,P)-\overline{S}(f,\,R)\leq\sum_{j\in{J}}2M\Delta{x_j}\leq{2M\times{m||P||}}$. What is the reason for introducing the number two?
Attempt: There are two partitions going on: $P$, and $R=P\cup{Q}$, that is the reason for including number two.

Best wishes!

PD: The notes are a Google link.

 

[pasmith]

TL;DR Summary: I've got a thirty three pages notes on Riemann integrability criterion, and stucked on page four. Haven`t read it entirely, but my intention is to understand completely all the document (if I can).

Hi, PF

I've began to read the notes, and at the fourth page, there is Theorem 2.4. Quoting it:

Theorem 2.4'. Let $\{P_n\}$ be a sequence of partitions satisfying $\lim_{n\rightarrow\infty}\,||P_n||=0$
1st question: What is the shape of a sequence of partitions?
Attempt:
$\{P_1\}=\{a,b\}$
$\{P_2\}=\{a,\frac{a+b}{2},b\}$
$\{P_3\}=\{a,\frac{2a+b}{3},\frac{a+2b}{3},b\}$
and so on, just to display an easy one.

$\{P_n\}=\frac{a+b}{n}?$

For that case, $P_n = \{ a + (b-a)r/n : r = 0, 1, \dots, n\}$.

2nd question It's about the expression $0\leq\overline{S}(f,\,P)-\overline{S}(f,\,R)\leq\sum_{j\in{J}}2M\Delta{x_j}\leq{2M\times{m||P||}}$. What is the reason for introducing the number two?

$M$ is defined to be $\sup_{ x \in [a,b]} |f(x)|$. For any distinct points $x$ and $y$, $$f(x) - f(y) \leq \sup f - \inf f.$$ Since $\sup f \leq \sup |f|$ and $\inf f \geq -\sup |f|$ we have $$f(x) - f(y) \leq \sup f - \inf f \leq 2\sup |f|,$$ which is essentially where the 2 comes from.

 

[mcastillo356]

Mila esker, @pasmith! (thousand thanks, in Basque)

 

[lekh2003]

It is worth noting that the actual sequence of partitions isn't particularly relevant here. The text makes no effort to specify a sequence because it is a bit of a pain to explicitly write down sometimes as you've seen. Simply specifying that you can pick something like an equal spacing that gets smaller and smaller is usually enough description.

 

[pasmith]

It is worth noting that the actual sequence of partitions isn't particularly relevant here. The text makes no effort to specify a sequence because it is a bit of a pain to explicitly write down sometimes as you've seen. Simply specifying that you can pick something like an equal spacing that gets smaller and smaller is usually enough description.

Starting from an arbitrary partition, one can construct a sequence with $\|P_n\| \to 0$, whereby $P_{n+1}$ is obtained from $P_n$ by adding a point in the middie of its widest interval (for definiteness, if there is more than one widest interval we take the leftmost).

 

[mcastillo356]

Hi, PF

$$0\leq\overline{S}(f,\,P)-\overline{S}(f,\,R)\leq\sum_{j\in{J}}2M\Delta{x_j}\leq{2M\times{m||P||}}$$

where $M=\text{sup}_{[a,b]}|f|$, because the contributions of $\overline{S}(f,\,P)$ and $\overline{S}(f,\,R)$ from the subintervals not in $J$ cancel out.
I still have difficulties; reason: the contributions of $\overline{S}(f,\,P)$ and $\overline{S}(f,\,R)$ from the subintervals not in J cancel out, but the subtraction of two Darboux sums can only lead to a Darboux sum: $M$ cannot be avoided.

PD: I hate to quote myself, but ain't got time.

 

[mcastillo356]

The aim of the whole sentence is to introduce $\overline{S}(f)$. I recall the proof:

Theorem 2.4'. Let $\{P_n\}$ be a sequence of partitions satisfying $\lim_{n\rightarrow\infty}{||P_n||=0}$. Then

$$\displaystyle\lim_{n\to\infty}{\overline{S}(f,\,P_n)}=\overline{S}(f)$$

On the other hand, the supremum of a function on any given interval is the supremum of its range, this is, $\overline{S}(f)$; in other words,

$$M=\text{sup}_{[a,b]}|f|$$

 

[mcastillo356]

Nope, $\overline{S}(f)$ is the Riemann integral, not $M=\text{sup}_{[a,b]}|f|$

 

[mcastillo356]

Hi, PF

I will write down the paragraph that keeps me wondering, then the doubts, and eventually mi attempt. There it goes:

" Theorem 2.4. For every $\epsilon>0$, there exists some $\delta$ such that
$$0\leq{\overline{S}(f,\,P)-\overline{S}(f)}<\epsilon$$

and

$$0\leq{\underline{S}(f)-\underline{S}(f,\,P)}<\epsilon$$

for any partition $P$, $||P||<\delta$

This proposition implies that by simply taking any sequence of partitions whose lengths tend to zero, the limit of the corresponding Darboux upper and lower sums always exist and give you the Riemann upper and lower integrals respectively. Alternatively, we state

Theorem 2.4'. Let $\{P_n\}$ be a sequence of partitions satisfying $\lim_{n\rightarrow\infty}{||P_n||=0}$.
Then

$$\displaystyle\lim_{n\rightarrow{\infty}}{\overline{S}(f,\,P_n)=\overline{S}(f)}$$

$$\displaystyle\lim_{n\rightarrow{\infty}}{\underline{S}(f,\,P_n)=\underline{S}(f)}$$

Proof. Let $m$ be the number of partition points of $Q$ (excluding the endpoints). Consider any partition $P$ and let $R$ be the partition by puting together $P$ and $Q$. Note that the number of subintervals in $P$ which contain some partition points of $Q$ in its interior must be less than or equal to $m$. Denote the indices of the collection of these subintervals in $P$ by $J$. We have

$$0\leq{\overline{S}(f,\,P)-\overline{S}(f,\,R)}\leq{\displaystyle\sum_{j\in J}{2M\Delta{x_j}}}\leq{2M\times{m||P||}}$$.

Where $M=\text{sup}_{[a,b]}|f|$, because the contributions of $\overline{S}(f,\,P)$ and $\overline{S}(f,\,R)$ from the subintervals not in $J$ cancel out.

Doubt: What means "because the contributions of $\overline{S}(f,\,P)$ and $\overline{S}(f,\,R)$ from the subintervals not in $J cancel out$" This is, why is it that "where $M=\text{sup}_{[a,b]}|f|$, because the contributions of $\overline{S}(f,\,P)$ and $\overline{S}(f,\,R)$ from the subintervals not in $J$ cancel out".

Attempt: I've drawn three pictures, $\overline{S}(f,\,P)$, $\overline{S}(f,\,R)$, and $\overline{S}(f,\,Q)$, and tried to figure out why the contributions of $\overline{S}(f,\,P)$ and $\overline{S}(f,\,R)$ not in $J$ cancel out. There I go:

,

Conclusion: $\overline{S}(f,\,P)$ overlaps $\overline{S}(f,\,R)$, and leaves just the function itself. Therefore it only rests a Darboux sum for which only counts $2M\times{m||P||}$. But still amoused why renders two times $M$.

Best whises!

PD: The function is $f(x)=\cos{(x)}+2,\quad{1\leq{x}\leq{5}}$

 

[pasmith]

Start with the simplest case.

Suppose $P = \{x_0, \dots, x_n\}$ and $R$ is $P$ with just one additional point, $\zeta \notin P$, so that $R = P \cup \{\zeta\}$ and $Q = R$. Then there exists $1 \leq i \leq n$ such that $\zeta \in (x_{i-1}, x_i)$. To save typing I set $A = \sup_{[x_{i-1},x_i]} f$, $A' = \sup_{[x_{i-1},\zeta]} f$, $A'' = \sup_{[\zeta,x_i]} f$, $\delta' = \zeta - x_{i-1}$ and $\delta'' = x_i - \zeta$. Then, since the contributions to the upper sums from all other subintervals are equal, we have $$\bar{S}(f,P) - \bar{S}(f,R) = A\delta_i - A'\delta' - A''\delta''.$$ We can bound the right hand side above by replacing $A\delta_i$ with a greater (or equal) quantity and $A'\delta' + A'' \delta''$ with a lesser (or equal) quantity. Here, that is done by taking $$A \leq \sup_{[a,b]}f \leq \sup_{[a,b]}|f| = M$$ and $$\begin{split} A' &\geq \inf_{[x_{i-1},\zeta]} f \geq \inf_{[a,b]}f \geq -M, \\ A'' &\geq \inf_{[\zeta,x_i]} f \geq \inf_{[a,b]}f \geq -M. \end{split}$$ (It is obvious that $$-\sup |f| \leq \inf f \leq \sup f \leq \sup |f|.)$$ Hence $$\begin{split} \bar{S}(f,P) - \bar{S}(f,R) &\leq M\delta_i - (-M)(\delta' + \delta'') \\ &= M\delta_i + M\delta_i \\ &= 2M\delta_i \\ &\leq 2M\|P\|. \end{split}$$

 

[mcastillo356]

Fine. Let see if I've understood

Start with the simplest case.

Fine

Suppose $P = \{x_0, \dots, x_n\}$ and $R$ is $P$ with just one additional point, $\zeta \notin P$, so that $R = P \cup \{\zeta\}$ and $Q = R$.

Fine. $Q$ left aside

Then there exists $1 \leq i \leq n$ such that $\zeta \in (x_{i-1}, x_i)$. To save typing I set $A = \sup_{[x_{i-1},x_i]} f$, $A' = \sup_{[x_{i-1},\zeta]} f$, $A'' = \sup_{[\zeta,x_i]} f$, $\delta' = \zeta - x_{i-1}$ and $\delta'' = x_i - \zeta$.

Simple: $${\{A',A''\}}\subset{\{A\}}$$, and $${\{\delta',\delta''\}}\subset{\{\delta_i\}}$$

Then, since the contributions to the upper sums from all other subintervals are equal,

Due to the simplicity of $R$, constrained to contain only one additional point: $\zeta$

we have $$\bar{S}(f,P) - \bar{S}(f,R) = A\delta_i - A'\delta' - A''\delta''.$$ We can bound the right hand side above by replacing $A\delta_i$ with a greater (or equal) quantity and $A'\delta' + A'' \delta''$ with a lesser (or equal) quantity. Here, that is done by taking $$A \leq \sup_{[a,b]}f \leq \sup_{[a,b]}|f| = M$$ and $$\begin{split} A' &\geq \inf_{[x_{i-1},\zeta]} f \geq \inf_{[a,b]}f \geq -M, \\ A'' &\geq \inf_{[\zeta,x_i]} f \geq \inf_{[a,b]}f \geq -M. \end{split}$$ (It is obvious that $$-\sup |f| \leq \inf f \leq \sup f \leq \sup |f|.)$$ Hence $$\begin{split} \bar{S}(f,P) - \bar{S}(f,R) &\leq M\delta_i - (-M)(\delta' + \delta'') \\ &= M\delta_i + M\delta_i \\ &= 2M\delta_i \\ &\leq 2M\|P\|. \end{split}$$

This last quote doesn't need anything; from my point of view, fine, perfect. It is the first, the second and the third quotes I am not sure if I've solved right

 

[mcastillo356]

Hi, PF

Continuing forward, I've got one question. I've been thinking how to post, and I've decided to write down a rather large quote, with the intention to contextualize it.

There goes the quote, then my doubts, and finally my attempt

"Theorem 2.8. Let $f$ and $g$ be integrable on $[a,b]$. We have (...)
(b) $fg$ is integrable on $[a,b]$
Suppose $f$, $g$ are Riemann integrable on $[a,b]$. Then they are bounded functions, so there exists $M_1, M_2>0$ such that $|f(x)|\leq{M_1}$ and $|g(x)|\leq{M_2}$ for all $x\in{[a,b]}$. It follows that $fg$ is also a bounded funtion (...)

Spoiler

$$\left.\begin{matrix}|f(x)|\leq{M_1}\\|g(x)|\leq{M_2}\end{matrix}\right\}\Rightarrow{|f(x)g(x)|\leq{M_1M_2}}$$

Now, if $P$ is a partition of $[a,b]$, we compute $\overline{S}(fg,\,P)-\underline{S}(fg,\,P)$. Let $P$ be the partition given by $a=x_0<x_1<\cdots{<x_n=b}$. Observe that for any $x,y\in{[x_{i-1},x_i]}$, we have

$$|f(x)g(x)-f(y)g(y)|\leq{|f(x)||g(x)-g(y)|+|g(y)||f(x)-f(y)|}\leq{M_1|g(x)-g(y)|+M_2|f(x)-f(y)|}$$

Spoiler

$$|f(x)g(x)-f(y)g(y)|=|f(x)g(x)-f(x)g(y)+f(x)g(y)-f(y)g(y)|$$

https://proofwiki.org/wiki/Supremum..._between_Supremum_and_Infimum#google_vignette

Hence

$$\sup\limits_{x\in{[x_{i-1},x_i]}}{|f(x)g(x)|}-\inf\limits_{x\in{[x_{i-1},x_i]}}{|f(x)g(x)|}$$

$$\leq{M_1\Big(\sup\limits_{x\in{[x_{i-1},x_i]}}{g(x)}
-\inf\limits_{x\in{[x_{i-1},x_i]}}{g(x)}
\Big)
+M_2\Big(
\sup\limits_{x\in{[x_{i-1},x_i]}}{f(x)}
-\inf\limits_{x\in{[x_{i-1},x_i]}}{f(x)}
\Big)}$$

which implies that

$$\overline{S}(fg,\,P)-\underline{S}(fg,\,P)\leq{M_1(\overline{S}(g,\,P)-\underline{S}(g,\,P))+M_2(\overline{S}(f,\,P)-\underline{S}(f,\,P))}$$

Now given $\epsilon>0$, there exist partitions $P_1, P_2$ of $[a,b]$ such that

$$\overline{S}(f,\,P_1)-\underline{S}(f,\,P_1)<\displaystyle\frac{\epsilon}{M_1+M_2},\quad{\overline{S}(f,\,P_2)-\underline{S}(f,\,P_2)<\displaystyle\frac{\epsilon}{M_1+M_2}}\,(\ast)$$

Let $P$ be the partition of $[a,b]$ formed by putting together $P_1$ and $P_2$. Then we also have

$$\overline{S}(f;\,P)-\underline{S}(f,\,P)<\displaystyle\frac{\epsilon}{M_1+M_2},\quad{\overline{S}(f;\,P)-\underline{S}(f,\,P)<\displaystyle\frac{\epsilon}{M_1+M_2}}$$

so the above estimate of $\overline{S}(fg,\,P)-\underline{S}(fg,\,P)$ gives

$\overline{S}(fg,\,P)-\underline{S}(fg,\,P)<M_1\displaystyle\frac{\epsilon}{M_1+M_2}+M_2\displaystyle\frac{\epsilon}{M_1+M_2}=\epsilon$

This proves that $fg$ is Riemann integrable by Integrability Criterion II."

Spoiler: Integrability Criterion II

Theorem 2.7. Let $f$ be a bounded function on $[a,b]$. Then $f$ is Riemann integrable on $[a,b]$ if and only if for every $\epsilon>0$, there exists a partition $P$ such that

$$\overline{S}(f,\,P)-\underline{S}(f,\,P)<\epsilon$$

Question: Is there a typo at $(\ast)$, where the document begins writing the indicated formula this way: $\overline{S}(f,\,P_1)-\underline{S}(f,\,P_2)<$ etc? Shouldn't be $\overline{S}(f,\,P_1)-\underline{S}(f,\,P_1)<$ etc?

Reason: $P=P_1\cup{P_2}$. But this is a very weak reasoning.

Best wishes!

PD: What means "you can't use macroparameters in mathmode?

 

[mcastillo356]

Hi, PF

Continuing forward, or better said, backwards

$M$ is defined to be $\sup_{ x \in [a,b]} |f(x)|$. For any distinct points $x$ and $y$, $$f(x) - f(y) \leq \sup f - \inf f.$$ Since $\sup f \leq \sup |f|$ and $\inf f \geq -\sup |f|$ we have $$f(x) - f(y) \leq \sup f - \inf f \leq 2\sup |f|,$$ which is essentially where the 2 comes from.

Doubt: Why do we pick two distinct points?

Attempt:

The supremum of a subset $S$ of a partially ordered set $P$ is the least element in $P$ that is greater than or equal to each element of $S$, if such an element exists. A partial order on a set is an arrangement such that, for certain pairs of elements, one precedes the other. A closed interval is an interval that includes all its endpoints. For example

$$[a,b]=\{x\in{\mathbb{R}}\,|\,a\leq{x}\leq{b}\}$$

So... we must consider a pair of elements, because the context is a partially ordered set.

Best.