Upper Derivative and Increasing Function on a Compact IntervalVitali Covering Lemma ProofQuestion about Vitali Covering (from a Lemma in Royden and Fitzpatrick's book)A question regarding Vitali Covering Covering LemmaConsequence Vitali covering lemmaOn why the Vitali Covering Lemma does not apply when the covering collection contains degenerate closed intervalsProof of an Analogue to Vitali Covering LemmaLebesgue outer measure with open ballsFailure of the Vitali Covering Lemma for open coveringsQuestion about Vitali Covering (from a Lemma in Royden and Fitzpatrick's book)Using the limit definition to Prove that a Set has a Vitali CoveringA question regarding Vitali Covering Covering LemmaImproving the constant in Hardy-Littlewood maximal inequality from $3^d$ to $2^d$Show that the function $x mapsto dim(mathrmIm(D_xf))$ is locally increasing and upper semicontinuous
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Upper Derivative and Increasing Function on a Compact Interval
Vitali Covering Lemma ProofQuestion about Vitali Covering (from a Lemma in Royden and Fitzpatrick's book)A question regarding Vitali Covering Covering LemmaConsequence Vitali covering lemmaOn why the Vitali Covering Lemma does not apply when the covering collection contains degenerate closed intervalsProof of an Analogue to Vitali Covering LemmaLebesgue outer measure with open ballsFailure of the Vitali Covering Lemma for open coveringsQuestion about Vitali Covering (from a Lemma in Royden and Fitzpatrick's book)Using the limit definition to Prove that a Set has a Vitali CoveringA question regarding Vitali Covering Covering LemmaImproving the constant in Hardy-Littlewood maximal inequality from $3^d$ to $2^d$Show that the function $x mapsto dim(mathrmIm(D_xf))$ is locally increasing and upper semicontinuous
$begingroup$
Definition. For a real valued function $f$ and an interior point $x$ of its domain, the uppper derivative of $f$ at $x$ denoted by $overlineDf(x)$ is defined as follows: $$overlineDf(x)=lim_hrightarrow0left[ sup left leq h right right]$$
I am working through Royden and Fitzpatrick's proof of the following lemma:
Lemma. Let $f$ be an increasing function on the closed, bounded interval $[a,b]$. Then for each $alpha>0$, $$m^*xin (a,b) : overlineDf(x) geq
alpha leq frac1alpha[f(b)-f(a)].$$
Here is the relevant part of the proof giving me trouble.
Let $alpha>0$. Define $E_alpha:=xin (a,b): overlineDf(x)geqalpha $. Choose $alpha' in (0,alpha)$. Let $mathscrF$ be the collection of closed, bounded intervals $[c,d]$ contained in $(a,b)$ for which $f(d)-f(c)geq alpha ' (d-c)$. Since $overlineDfgeq alpha$ on $E_alpha$, $mathscrF$ is a Vitali covering for $E_alpha$.
I realize that the questions am I about to pose have been asked here (indeed, I copied-&-pasted the relevant parts from that post), but астон вілла олоф мэллбэрг's answer is much too brief for my liking and the long string of comments, which appear to contain part of the answer to Kurome's question, are way to jumbled to get anything out of them. I have the same question as the OP in the link: why is $scrF neq emptyset$ and why is $scrF$ a Vitali covering of $E_alpha$. Specifically, I don't understand the implication
$$t<delta impliesfracf(x+t)-f(x)tgeqalpha'$$
holds (I tried unpacking the definition of the upper derivative, but I couldn't see it); nor do I understand how астон вілла олоф мэллбэрг is able to choose $d in (a,b)$ such that $d-x < delta$. And why does does $[x,d]$ having a length less than $delta$ imply $scrF$ is a Vitali covering of $E_alpha$? That $delta$ wasn't arbitrary. For ease of references, here is астон вілла олоф мэллбэрг's answer:
[астон вілла олоф мэллбэрг's answer]: Take any $x in E_alpha$. Now, since $overlineDf(x)geqalpha$, it follows that for some small $delta$, $t<delta impliesfracf(x+t)-f(x)tgeqalpha'$.
(The definition for the upper derivative above is slightly wrong, I will edit it)
Putting $t=d-x$, this means that $t<delta implies f(d)-f(x) geq alpha'(d-x)$. The interval $[d,x]$ is in $mathscrF$ for every $d$ close enough to $x$,for arbitrary $x$ in $E_alpha$. This makes $mathscrF$ a Vitali covering for $E_alpha$.
real-analysis measure-theory proof-explanation
$endgroup$
add a comment |
$begingroup$
Definition. For a real valued function $f$ and an interior point $x$ of its domain, the uppper derivative of $f$ at $x$ denoted by $overlineDf(x)$ is defined as follows: $$overlineDf(x)=lim_hrightarrow0left[ sup left leq h right right]$$
I am working through Royden and Fitzpatrick's proof of the following lemma:
Lemma. Let $f$ be an increasing function on the closed, bounded interval $[a,b]$. Then for each $alpha>0$, $$m^*xin (a,b) : overlineDf(x) geq
alpha leq frac1alpha[f(b)-f(a)].$$
Here is the relevant part of the proof giving me trouble.
Let $alpha>0$. Define $E_alpha:=xin (a,b): overlineDf(x)geqalpha $. Choose $alpha' in (0,alpha)$. Let $mathscrF$ be the collection of closed, bounded intervals $[c,d]$ contained in $(a,b)$ for which $f(d)-f(c)geq alpha ' (d-c)$. Since $overlineDfgeq alpha$ on $E_alpha$, $mathscrF$ is a Vitali covering for $E_alpha$.
I realize that the questions am I about to pose have been asked here (indeed, I copied-&-pasted the relevant parts from that post), but астон вілла олоф мэллбэрг's answer is much too brief for my liking and the long string of comments, which appear to contain part of the answer to Kurome's question, are way to jumbled to get anything out of them. I have the same question as the OP in the link: why is $scrF neq emptyset$ and why is $scrF$ a Vitali covering of $E_alpha$. Specifically, I don't understand the implication
$$t<delta impliesfracf(x+t)-f(x)tgeqalpha'$$
holds (I tried unpacking the definition of the upper derivative, but I couldn't see it); nor do I understand how астон вілла олоф мэллбэрг is able to choose $d in (a,b)$ such that $d-x < delta$. And why does does $[x,d]$ having a length less than $delta$ imply $scrF$ is a Vitali covering of $E_alpha$? That $delta$ wasn't arbitrary. For ease of references, here is астон вілла олоф мэллбэрг's answer:
[астон вілла олоф мэллбэрг's answer]: Take any $x in E_alpha$. Now, since $overlineDf(x)geqalpha$, it follows that for some small $delta$, $t<delta impliesfracf(x+t)-f(x)tgeqalpha'$.
(The definition for the upper derivative above is slightly wrong, I will edit it)
Putting $t=d-x$, this means that $t<delta implies f(d)-f(x) geq alpha'(d-x)$. The interval $[d,x]$ is in $mathscrF$ for every $d$ close enough to $x$,for arbitrary $x$ in $E_alpha$. This makes $mathscrF$ a Vitali covering for $E_alpha$.
real-analysis measure-theory proof-explanation
$endgroup$
1
$begingroup$
Side note: I was flipping through the 3rd edition (one of the editions produced without Fitpatrick), and it appears that this result doesn't even figure into it (please correct me if I'm wrong). I suppose this is another bad proof purely ascribable to Fitzpatrick, and therefore more evidence that Fitzpatrick soiled Royden's book.
$endgroup$
– user193319
Apr 19 '18 at 14:47
$begingroup$
When I say "another", I am referring to this, which still hasn't received a full answer yet (any takers!?!?).
$endgroup$
– user193319
Apr 19 '18 at 14:49
add a comment |
$begingroup$
Definition. For a real valued function $f$ and an interior point $x$ of its domain, the uppper derivative of $f$ at $x$ denoted by $overlineDf(x)$ is defined as follows: $$overlineDf(x)=lim_hrightarrow0left[ sup left leq h right right]$$
I am working through Royden and Fitzpatrick's proof of the following lemma:
Lemma. Let $f$ be an increasing function on the closed, bounded interval $[a,b]$. Then for each $alpha>0$, $$m^*xin (a,b) : overlineDf(x) geq
alpha leq frac1alpha[f(b)-f(a)].$$
Here is the relevant part of the proof giving me trouble.
Let $alpha>0$. Define $E_alpha:=xin (a,b): overlineDf(x)geqalpha $. Choose $alpha' in (0,alpha)$. Let $mathscrF$ be the collection of closed, bounded intervals $[c,d]$ contained in $(a,b)$ for which $f(d)-f(c)geq alpha ' (d-c)$. Since $overlineDfgeq alpha$ on $E_alpha$, $mathscrF$ is a Vitali covering for $E_alpha$.
I realize that the questions am I about to pose have been asked here (indeed, I copied-&-pasted the relevant parts from that post), but астон вілла олоф мэллбэрг's answer is much too brief for my liking and the long string of comments, which appear to contain part of the answer to Kurome's question, are way to jumbled to get anything out of them. I have the same question as the OP in the link: why is $scrF neq emptyset$ and why is $scrF$ a Vitali covering of $E_alpha$. Specifically, I don't understand the implication
$$t<delta impliesfracf(x+t)-f(x)tgeqalpha'$$
holds (I tried unpacking the definition of the upper derivative, but I couldn't see it); nor do I understand how астон вілла олоф мэллбэрг is able to choose $d in (a,b)$ such that $d-x < delta$. And why does does $[x,d]$ having a length less than $delta$ imply $scrF$ is a Vitali covering of $E_alpha$? That $delta$ wasn't arbitrary. For ease of references, here is астон вілла олоф мэллбэрг's answer:
[астон вілла олоф мэллбэрг's answer]: Take any $x in E_alpha$. Now, since $overlineDf(x)geqalpha$, it follows that for some small $delta$, $t<delta impliesfracf(x+t)-f(x)tgeqalpha'$.
(The definition for the upper derivative above is slightly wrong, I will edit it)
Putting $t=d-x$, this means that $t<delta implies f(d)-f(x) geq alpha'(d-x)$. The interval $[d,x]$ is in $mathscrF$ for every $d$ close enough to $x$,for arbitrary $x$ in $E_alpha$. This makes $mathscrF$ a Vitali covering for $E_alpha$.
real-analysis measure-theory proof-explanation
$endgroup$
Definition. For a real valued function $f$ and an interior point $x$ of its domain, the uppper derivative of $f$ at $x$ denoted by $overlineDf(x)$ is defined as follows: $$overlineDf(x)=lim_hrightarrow0left[ sup left leq h right right]$$
I am working through Royden and Fitzpatrick's proof of the following lemma:
Lemma. Let $f$ be an increasing function on the closed, bounded interval $[a,b]$. Then for each $alpha>0$, $$m^*xin (a,b) : overlineDf(x) geq
alpha leq frac1alpha[f(b)-f(a)].$$
Here is the relevant part of the proof giving me trouble.
Let $alpha>0$. Define $E_alpha:=xin (a,b): overlineDf(x)geqalpha $. Choose $alpha' in (0,alpha)$. Let $mathscrF$ be the collection of closed, bounded intervals $[c,d]$ contained in $(a,b)$ for which $f(d)-f(c)geq alpha ' (d-c)$. Since $overlineDfgeq alpha$ on $E_alpha$, $mathscrF$ is a Vitali covering for $E_alpha$.
I realize that the questions am I about to pose have been asked here (indeed, I copied-&-pasted the relevant parts from that post), but астон вілла олоф мэллбэрг's answer is much too brief for my liking and the long string of comments, which appear to contain part of the answer to Kurome's question, are way to jumbled to get anything out of them. I have the same question as the OP in the link: why is $scrF neq emptyset$ and why is $scrF$ a Vitali covering of $E_alpha$. Specifically, I don't understand the implication
$$t<delta impliesfracf(x+t)-f(x)tgeqalpha'$$
holds (I tried unpacking the definition of the upper derivative, but I couldn't see it); nor do I understand how астон вілла олоф мэллбэрг is able to choose $d in (a,b)$ such that $d-x < delta$. And why does does $[x,d]$ having a length less than $delta$ imply $scrF$ is a Vitali covering of $E_alpha$? That $delta$ wasn't arbitrary. For ease of references, here is астон вілла олоф мэллбэрг's answer:
[астон вілла олоф мэллбэрг's answer]: Take any $x in E_alpha$. Now, since $overlineDf(x)geqalpha$, it follows that for some small $delta$, $t<delta impliesfracf(x+t)-f(x)tgeqalpha'$.
(The definition for the upper derivative above is slightly wrong, I will edit it)
Putting $t=d-x$, this means that $t<delta implies f(d)-f(x) geq alpha'(d-x)$. The interval $[d,x]$ is in $mathscrF$ for every $d$ close enough to $x$,for arbitrary $x$ in $E_alpha$. This makes $mathscrF$ a Vitali covering for $E_alpha$.
real-analysis measure-theory proof-explanation
real-analysis measure-theory proof-explanation
asked Apr 19 '18 at 14:24
user193319user193319
2,4552927
2,4552927
1
$begingroup$
Side note: I was flipping through the 3rd edition (one of the editions produced without Fitpatrick), and it appears that this result doesn't even figure into it (please correct me if I'm wrong). I suppose this is another bad proof purely ascribable to Fitzpatrick, and therefore more evidence that Fitzpatrick soiled Royden's book.
$endgroup$
– user193319
Apr 19 '18 at 14:47
$begingroup$
When I say "another", I am referring to this, which still hasn't received a full answer yet (any takers!?!?).
$endgroup$
– user193319
Apr 19 '18 at 14:49
add a comment |
1
$begingroup$
Side note: I was flipping through the 3rd edition (one of the editions produced without Fitpatrick), and it appears that this result doesn't even figure into it (please correct me if I'm wrong). I suppose this is another bad proof purely ascribable to Fitzpatrick, and therefore more evidence that Fitzpatrick soiled Royden's book.
$endgroup$
– user193319
Apr 19 '18 at 14:47
$begingroup$
When I say "another", I am referring to this, which still hasn't received a full answer yet (any takers!?!?).
$endgroup$
– user193319
Apr 19 '18 at 14:49
1
1
$begingroup$
Side note: I was flipping through the 3rd edition (one of the editions produced without Fitpatrick), and it appears that this result doesn't even figure into it (please correct me if I'm wrong). I suppose this is another bad proof purely ascribable to Fitzpatrick, and therefore more evidence that Fitzpatrick soiled Royden's book.
$endgroup$
– user193319
Apr 19 '18 at 14:47
$begingroup$
Side note: I was flipping through the 3rd edition (one of the editions produced without Fitpatrick), and it appears that this result doesn't even figure into it (please correct me if I'm wrong). I suppose this is another bad proof purely ascribable to Fitzpatrick, and therefore more evidence that Fitzpatrick soiled Royden's book.
$endgroup$
– user193319
Apr 19 '18 at 14:47
$begingroup$
When I say "another", I am referring to this, which still hasn't received a full answer yet (any takers!?!?).
$endgroup$
– user193319
Apr 19 '18 at 14:49
$begingroup$
When I say "another", I am referring to this, which still hasn't received a full answer yet (any takers!?!?).
$endgroup$
– user193319
Apr 19 '18 at 14:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Based on definition of upper derivative the following implication is wrong:
$$exists delta>0,0<|t|<deltaimplies fracf(x+t) - f(x) t geqalpha'tag1$$ The correct implication is that $$exists delta>0,0<h<deltaimpliessupleft<hright>alpha'tag2$$ This is obtained by using $epsilon=alpha-alpha'$ in the limit definition for $overlineD f(x) $.
By choosing arbitrarily small values of $h$ this gives you infinitely many values of $t$ of arbitrarily small modulus such that ratio $(f(x+t) - f(x)) /t>alpha' $ (why? If $sup S>s$ then there are some members in $S$ definitely greater than $s$). Thus the collection $mathscrF $ contains intervals of the form $[x, x+t] $ or $[x+t, x] $ depending on whether $t>0$ or $t<0$ and $mathscrF $ is non-empty. Since there is an infinite supply of such $t$ with arbitrarily small modulus, it follows that $mathscrF $ is a Vitali covering for $E_alpha $.
I have to admit that the definition of upper derivative is somewhat difficult to decode and the proof in your book is very terse ("since $overlineD f(x) geq alpha$ on $E_alpha$, $mathscrF $ is a Vitali covering for $E_alpha $"). Also the answer you have linked contains a wrong implication $(1)$. It also appears that there are many people who are facing problems with this particular proof.
On a closer inspection the definition of upper derivative as given in your post matches that of a limit superior $$overlineD f(x) =limsup_hto 0fracf(x+h)-f(x)h=limsup_yto x frac f(y) - f(x) y-x $$ and then the conclusion is obvious from the properties of $limsup$.
Remember the two fundamental / defining properties of $limsup$. If $M=limsup_xto a f(x) $ then
- for any number $A<M$ there are infinitely many values of $x$ arbitrarily close to $a$ such that $f(x) > A$. In order to make this even more confusing a number $A<M$ is written as $A=M-epsilon$ where $epsilon>0$.
- for any number $B>M$ we have $f(x) <B$ for all values of $x$ sufficiently close to $a$. A number $B>M$ is usually written as $B=M+epsilon $ where $epsilon>0$.
Some books try to replace "arbitrarily" and "sufficiently" using another number $delta>0$.
For the current question one needs only the first property of $limsup$.
$endgroup$
add a comment |
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$begingroup$
Based on definition of upper derivative the following implication is wrong:
$$exists delta>0,0<|t|<deltaimplies fracf(x+t) - f(x) t geqalpha'tag1$$ The correct implication is that $$exists delta>0,0<h<deltaimpliessupleft<hright>alpha'tag2$$ This is obtained by using $epsilon=alpha-alpha'$ in the limit definition for $overlineD f(x) $.
By choosing arbitrarily small values of $h$ this gives you infinitely many values of $t$ of arbitrarily small modulus such that ratio $(f(x+t) - f(x)) /t>alpha' $ (why? If $sup S>s$ then there are some members in $S$ definitely greater than $s$). Thus the collection $mathscrF $ contains intervals of the form $[x, x+t] $ or $[x+t, x] $ depending on whether $t>0$ or $t<0$ and $mathscrF $ is non-empty. Since there is an infinite supply of such $t$ with arbitrarily small modulus, it follows that $mathscrF $ is a Vitali covering for $E_alpha $.
I have to admit that the definition of upper derivative is somewhat difficult to decode and the proof in your book is very terse ("since $overlineD f(x) geq alpha$ on $E_alpha$, $mathscrF $ is a Vitali covering for $E_alpha $"). Also the answer you have linked contains a wrong implication $(1)$. It also appears that there are many people who are facing problems with this particular proof.
On a closer inspection the definition of upper derivative as given in your post matches that of a limit superior $$overlineD f(x) =limsup_hto 0fracf(x+h)-f(x)h=limsup_yto x frac f(y) - f(x) y-x $$ and then the conclusion is obvious from the properties of $limsup$.
Remember the two fundamental / defining properties of $limsup$. If $M=limsup_xto a f(x) $ then
- for any number $A<M$ there are infinitely many values of $x$ arbitrarily close to $a$ such that $f(x) > A$. In order to make this even more confusing a number $A<M$ is written as $A=M-epsilon$ where $epsilon>0$.
- for any number $B>M$ we have $f(x) <B$ for all values of $x$ sufficiently close to $a$. A number $B>M$ is usually written as $B=M+epsilon $ where $epsilon>0$.
Some books try to replace "arbitrarily" and "sufficiently" using another number $delta>0$.
For the current question one needs only the first property of $limsup$.
$endgroup$
add a comment |
$begingroup$
Based on definition of upper derivative the following implication is wrong:
$$exists delta>0,0<|t|<deltaimplies fracf(x+t) - f(x) t geqalpha'tag1$$ The correct implication is that $$exists delta>0,0<h<deltaimpliessupleft<hright>alpha'tag2$$ This is obtained by using $epsilon=alpha-alpha'$ in the limit definition for $overlineD f(x) $.
By choosing arbitrarily small values of $h$ this gives you infinitely many values of $t$ of arbitrarily small modulus such that ratio $(f(x+t) - f(x)) /t>alpha' $ (why? If $sup S>s$ then there are some members in $S$ definitely greater than $s$). Thus the collection $mathscrF $ contains intervals of the form $[x, x+t] $ or $[x+t, x] $ depending on whether $t>0$ or $t<0$ and $mathscrF $ is non-empty. Since there is an infinite supply of such $t$ with arbitrarily small modulus, it follows that $mathscrF $ is a Vitali covering for $E_alpha $.
I have to admit that the definition of upper derivative is somewhat difficult to decode and the proof in your book is very terse ("since $overlineD f(x) geq alpha$ on $E_alpha$, $mathscrF $ is a Vitali covering for $E_alpha $"). Also the answer you have linked contains a wrong implication $(1)$. It also appears that there are many people who are facing problems with this particular proof.
On a closer inspection the definition of upper derivative as given in your post matches that of a limit superior $$overlineD f(x) =limsup_hto 0fracf(x+h)-f(x)h=limsup_yto x frac f(y) - f(x) y-x $$ and then the conclusion is obvious from the properties of $limsup$.
Remember the two fundamental / defining properties of $limsup$. If $M=limsup_xto a f(x) $ then
- for any number $A<M$ there are infinitely many values of $x$ arbitrarily close to $a$ such that $f(x) > A$. In order to make this even more confusing a number $A<M$ is written as $A=M-epsilon$ where $epsilon>0$.
- for any number $B>M$ we have $f(x) <B$ for all values of $x$ sufficiently close to $a$. A number $B>M$ is usually written as $B=M+epsilon $ where $epsilon>0$.
Some books try to replace "arbitrarily" and "sufficiently" using another number $delta>0$.
For the current question one needs only the first property of $limsup$.
$endgroup$
add a comment |
$begingroup$
Based on definition of upper derivative the following implication is wrong:
$$exists delta>0,0<|t|<deltaimplies fracf(x+t) - f(x) t geqalpha'tag1$$ The correct implication is that $$exists delta>0,0<h<deltaimpliessupleft<hright>alpha'tag2$$ This is obtained by using $epsilon=alpha-alpha'$ in the limit definition for $overlineD f(x) $.
By choosing arbitrarily small values of $h$ this gives you infinitely many values of $t$ of arbitrarily small modulus such that ratio $(f(x+t) - f(x)) /t>alpha' $ (why? If $sup S>s$ then there are some members in $S$ definitely greater than $s$). Thus the collection $mathscrF $ contains intervals of the form $[x, x+t] $ or $[x+t, x] $ depending on whether $t>0$ or $t<0$ and $mathscrF $ is non-empty. Since there is an infinite supply of such $t$ with arbitrarily small modulus, it follows that $mathscrF $ is a Vitali covering for $E_alpha $.
I have to admit that the definition of upper derivative is somewhat difficult to decode and the proof in your book is very terse ("since $overlineD f(x) geq alpha$ on $E_alpha$, $mathscrF $ is a Vitali covering for $E_alpha $"). Also the answer you have linked contains a wrong implication $(1)$. It also appears that there are many people who are facing problems with this particular proof.
On a closer inspection the definition of upper derivative as given in your post matches that of a limit superior $$overlineD f(x) =limsup_hto 0fracf(x+h)-f(x)h=limsup_yto x frac f(y) - f(x) y-x $$ and then the conclusion is obvious from the properties of $limsup$.
Remember the two fundamental / defining properties of $limsup$. If $M=limsup_xto a f(x) $ then
- for any number $A<M$ there are infinitely many values of $x$ arbitrarily close to $a$ such that $f(x) > A$. In order to make this even more confusing a number $A<M$ is written as $A=M-epsilon$ where $epsilon>0$.
- for any number $B>M$ we have $f(x) <B$ for all values of $x$ sufficiently close to $a$. A number $B>M$ is usually written as $B=M+epsilon $ where $epsilon>0$.
Some books try to replace "arbitrarily" and "sufficiently" using another number $delta>0$.
For the current question one needs only the first property of $limsup$.
$endgroup$
Based on definition of upper derivative the following implication is wrong:
$$exists delta>0,0<|t|<deltaimplies fracf(x+t) - f(x) t geqalpha'tag1$$ The correct implication is that $$exists delta>0,0<h<deltaimpliessupleft<hright>alpha'tag2$$ This is obtained by using $epsilon=alpha-alpha'$ in the limit definition for $overlineD f(x) $.
By choosing arbitrarily small values of $h$ this gives you infinitely many values of $t$ of arbitrarily small modulus such that ratio $(f(x+t) - f(x)) /t>alpha' $ (why? If $sup S>s$ then there are some members in $S$ definitely greater than $s$). Thus the collection $mathscrF $ contains intervals of the form $[x, x+t] $ or $[x+t, x] $ depending on whether $t>0$ or $t<0$ and $mathscrF $ is non-empty. Since there is an infinite supply of such $t$ with arbitrarily small modulus, it follows that $mathscrF $ is a Vitali covering for $E_alpha $.
I have to admit that the definition of upper derivative is somewhat difficult to decode and the proof in your book is very terse ("since $overlineD f(x) geq alpha$ on $E_alpha$, $mathscrF $ is a Vitali covering for $E_alpha $"). Also the answer you have linked contains a wrong implication $(1)$. It also appears that there are many people who are facing problems with this particular proof.
On a closer inspection the definition of upper derivative as given in your post matches that of a limit superior $$overlineD f(x) =limsup_hto 0fracf(x+h)-f(x)h=limsup_yto x frac f(y) - f(x) y-x $$ and then the conclusion is obvious from the properties of $limsup$.
Remember the two fundamental / defining properties of $limsup$. If $M=limsup_xto a f(x) $ then
- for any number $A<M$ there are infinitely many values of $x$ arbitrarily close to $a$ such that $f(x) > A$. In order to make this even more confusing a number $A<M$ is written as $A=M-epsilon$ where $epsilon>0$.
- for any number $B>M$ we have $f(x) <B$ for all values of $x$ sufficiently close to $a$. A number $B>M$ is usually written as $B=M+epsilon $ where $epsilon>0$.
Some books try to replace "arbitrarily" and "sufficiently" using another number $delta>0$.
For the current question one needs only the first property of $limsup$.
edited Mar 21 at 3:19
answered Mar 21 at 1:00
Paramanand SinghParamanand Singh
51.1k558170
51.1k558170
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1
$begingroup$
Side note: I was flipping through the 3rd edition (one of the editions produced without Fitpatrick), and it appears that this result doesn't even figure into it (please correct me if I'm wrong). I suppose this is another bad proof purely ascribable to Fitzpatrick, and therefore more evidence that Fitzpatrick soiled Royden's book.
$endgroup$
– user193319
Apr 19 '18 at 14:47
$begingroup$
When I say "another", I am referring to this, which still hasn't received a full answer yet (any takers!?!?).
$endgroup$
– user193319
Apr 19 '18 at 14:49