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![$a_n ge 0$ and $b_n in mathbbC$ and $sum a_n, sum b_n$ convergent but $sum |a_n b_n|$ divergent? [closed]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
Let $I_1, I_2, cdots$ be any countable collection of intervals, whose union contains some interval $I$, then $sum P(I_j) geq P(I)$.
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I’m going through the construction of Lebesgue measure, which relies on this result. In the following, the $I$ are all intervals defined on $[0,1]$ and $mathbbP(I)$ denotes the length of the interval.
Lemma 1
Let $I_1, I_2, cdots, I_n$ be a finite collection of intervals whose union contain an interval $I$, then $$sum_j=1^n mathbbP(I_j) geq mathbbP(I).$$
Proof:
For any $1leq j leq n$, let $a_j$ be the left end point of $I_j$ and $b_j$ be the right end point of $I_j$. The sets $a_j$ and $b_j$ are finite and thus can be ordered. Let $a_l_i$ and $b_k_i$ be the ordered sets of these end points (from least to greatest).
For the interval $I$ we have
$$bigcup_j=1^n I_j supseteq I.$$
Letting $a$ and $b$ denote the left and right end points of $I$, respectively, without loss of generality we assume
$$a_l_1leq aleq b_l_1, $$
$$a_k_nleq bleq b_k_n. $$
The above simply says that $a$ and $b$ are in the intervals with smallest left end point and largest right end point, respectively (this could of course be the same interval).
Using the ordering of end points, we can pair them up, from smallest to largest, in the sum of interval lengths:
$$sum_j=1^n(b_j - a_j)= sum_i=1^n(b_j_i - a_j_i)= Big[sum_i =2^n-1(b_j_i - a_j_i) +(b_j_1 -a_j_n)Big]+(b_j_n -a_j_1) .$$
In the last expression, we have removed the smallest left end point and largest right end point from the sum. Since we assume some part of the interval $I$ is in every $I_n$, the remaining end points of intervals must overlap, which means, for any $i$,
$$b_j_i-1 geq a_j_i.$$
Thus, we can reorder again to show that
$$sum_i =2^n-1(b_j_i - a_j_i) + (b_j_1-a_j_n)= sum_i=2^n (b_j_i-1- a_j_i) geq0.$$
Returning to our original sum, it follows that
$$sum_j=1^n (b_j - a_j) geq b_k_n - a_l_1 geq b-a,$$
which implies
$$ sum_j=1^n mathbbP(I_j) geq mathbbP(I).$$
$ square$
Lemma 2:
Let $I_1, I_2,cdots$ be a countable collection of open intervals, whos union contains a closed interval, then $$sum_jgeq 1mathbbP(I_j) geq mathbbP(I).$$
Proof:
By the Heine-Borel theorem, for any countable collection of open sets $O_n$ and some closed set $ bigcup_ngeq1 O_n supseteq O$, there exists a finite sub cover
$$O_a_1 cup O_a_2 cup cdots cup O_a_k supseteq O.$$
Thus
$$ bigcup_j=1^infty I_j supseteq I. implies bigcup_a=1^k I_j_a supseteq I$$
and the result follows from lemma 1.
$square$
Lemma 3:
For any countable collection of intervals $I_j$, whose union contains an interval $I$,
$$sum_j geq 1 mathbbP (I_j) geq mathbbP(I).$$
Proof:
Choose some $epsilon >0$ and extend the left and right end points of the $I_j$ by $epsilon 2^-j$ to form an open set
$$I_j_epsilon=left(a_j -epsilon 2^-j , b_j + epsilon 2^-j right).$$
Next, extend the end points of $I$ by $epsilon$ to make the closed set
$$I_epsilon =[a- epsilon , b + epsilon].$$
Since $bigcup_j geq 1 I_j supseteq I,$ there exists some $a_k$ and $b_l$ such that
$$a_k <a leq b < b_l.$$
Further, since $k$ and $l$ are both $geq 1$, we can choose $epsilon$ small enough that
$$a_k - epsilon 2^-k< a_k<a - epsilon quad, quad b+epsilon < b_l < b_l + epsilon 2^-l.$$
Thus, we have constructed a countable collection of open intervals $I_j_epsilon$, whose union contains the closed interval $I_epsilon$. By lemma 2:
$$sum_jgeq1 mathbbP(I_j_epsilon) geq mathbbP(I_epsilon)$$
$$implies sum_j geq 1 Big big( b_j + epsilon 2^-j big) - big(a_j - epsilon 2^-j big)Big geq( b + epsilon )- (a - epsilon).$$
$$implies sum_j geq 1 ( b_j - a_j) + epsilon sum_j geq 1 big (2^1-j big) geq (b-a) + 2epsilon.$$
$$ implies sum_j geq 1 ( b_j - a_j) + 2epsilon geq (b-a) + 2epsilon.$$
$$implies sum_jgeq 1 mathbbP(I_j)=sum_j geq 1 ( b_j - a_j) geq (b-a) = mathbbP(I).$$
$ square$
real-analysis sequences-and-series probability-theory proof-verification
asked Jul 17 at 5:35
Moed Pol Bollo
19518
 |Â
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I’m going through the construction of Lebesgue measure, which relies on this result. In the following, the $I$ are all intervals defined on $[0,1]$ and $mathbbP(I)$ denotes the length of the interval.
Lemma 1
Let $I_1, I_2, cdots, I_n$ be a finite collection of intervals whose union contain an interval $I$, then $$sum_j=1^n mathbbP(I_j) geq mathbbP(I).$$
Proof:
For any $1leq j leq n$, let $a_j$ be the left end point of $I_j$ and $b_j$ be the right end point of $I_j$. The sets $a_j$ and $b_j$ are finite and thus can be ordered. Let $a_l_i$ and $b_k_i$ be the ordered sets of these end points (from least to greatest).
For the interval $I$ we have
$$bigcup_j=1^n I_j supseteq I.$$
Letting $a$ and $b$ denote the left and right end points of $I$, respectively, without loss of generality we assume
$$a_l_1leq aleq b_l_1, $$
$$a_k_nleq bleq b_k_n. $$
The above simply says that $a$ and $b$ are in the intervals with smallest left end point and largest right end point, respectively (this could of course be the same interval).
Using the ordering of end points, we can pair them up, from smallest to largest, in the sum of interval lengths:
$$sum_j=1^n(b_j - a_j)= sum_i=1^n(b_j_i - a_j_i)= Big[sum_i =2^n-1(b_j_i - a_j_i) +(b_j_1 -a_j_n)Big]+(b_j_n -a_j_1) .$$
In the last expression, we have removed the smallest left end point and largest right end point from the sum. Since we assume some part of the interval $I$ is in every $I_n$, the remaining end points of intervals must overlap, which means, for any $i$,
$$b_j_i-1 geq a_j_i.$$
Thus, we can reorder again to show that
$$sum_i =2^n-1(b_j_i - a_j_i) + (b_j_1-a_j_n)= sum_i=2^n (b_j_i-1- a_j_i) geq0.$$
Returning to our original sum, it follows that
$$sum_j=1^n (b_j - a_j) geq b_k_n - a_l_1 geq b-a,$$
which implies
$$ sum_j=1^n mathbbP(I_j) geq mathbbP(I).$$
$ square$
Lemma 2:
Let $I_1, I_2,cdots$ be a countable collection of open intervals, whos union contains a closed interval, then $$sum_jgeq 1mathbbP(I_j) geq mathbbP(I).$$
Proof:
By the Heine-Borel theorem, for any countable collection of open sets $O_n$ and some closed set $ bigcup_ngeq1 O_n supseteq O$, there exists a finite sub cover
$$O_a_1 cup O_a_2 cup cdots cup O_a_k supseteq O.$$
Thus
$$ bigcup_j=1^infty I_j supseteq I. implies bigcup_a=1^k I_j_a supseteq I$$
and the result follows from lemma 1.
$square$
Lemma 3:
For any countable collection of intervals $I_j$, whose union contains an interval $I$,
$$sum_j geq 1 mathbbP (I_j) geq mathbbP(I).$$
Proof:
Choose some $epsilon >0$ and extend the left and right end points of the $I_j$ by $epsilon 2^-j$ to form an open set
$$I_j_epsilon=left(a_j -epsilon 2^-j , b_j + epsilon 2^-j right).$$
Next, extend the end points of $I$ by $epsilon$ to make the closed set
$$I_epsilon =[a- epsilon , b + epsilon].$$
Since $bigcup_j geq 1 I_j supseteq I,$ there exists some $a_k$ and $b_l$ such that
$$a_k <a leq b < b_l.$$
Further, since $k$ and $l$ are both $geq 1$, we can choose $epsilon$ small enough that
$$a_k - epsilon 2^-k< a_k<a - epsilon quad, quad b+epsilon < b_l < b_l + epsilon 2^-l.$$
Thus, we have constructed a countable collection of open intervals $I_j_epsilon$, whose union contains the closed interval $I_epsilon$. By lemma 2:
$$sum_jgeq1 mathbbP(I_j_epsilon) geq mathbbP(I_epsilon)$$
$$implies sum_j geq 1 Big big( b_j + epsilon 2^-j big) - big(a_j - epsilon 2^-j big)Big geq( b + epsilon )- (a - epsilon).$$
$$implies sum_j geq 1 ( b_j - a_j) + epsilon sum_j geq 1 big (2^1-j big) geq (b-a) + 2epsilon.$$
$$ implies sum_j geq 1 ( b_j - a_j) + 2epsilon geq (b-a) + 2epsilon.$$
$$implies sum_jgeq 1 mathbbP(I_j)=sum_j geq 1 ( b_j - a_j) geq (b-a) = mathbbP(I).$$
$ square$
real-analysis sequences-and-series probability-theory proof-verification
asked Jul 17 at 5:35
Moed Pol Bollo
19518
What's the question?
– Lord Shark the Unknown
Jul 17 at 5:37
@LordSharktheUnknown I just want to make sure the proof is correct, as it is necessary for the construction of Lebesgue measure on $[0,1]$.
– Moed Pol Bollo
Jul 17 at 5:39
What does "Choose $a_k$ and $b_k$ such that: $a_kle a_j$, $b_jle b_k$" mean?
– Lord Shark the Unknown
Jul 17 at 5:42
@LordSharktheUnknown That’s a poorly worded way of saying $a_k$ is the smallest element of the set of left end points and $b_k$ is the largest element of the set of right end points.
– Moed Pol Bollo
Jul 17 at 5:45
Then that's an error straight away; the least left endpoint may not belong to the same interval $I_k$ as the largest right endpoint.
– Lord Shark the Unknown
Jul 17 at 5:46
 |Â
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I’m going through the construction of Lebesgue measure, which relies on this result. In the following, the $I$ are all intervals defined on $[0,1]$ and $mathbbP(I)$ denotes the length of the interval.
Lemma 1
Let $I_1, I_2, cdots, I_n$ be a finite collection of intervals whose union contain an interval $I$, then $$sum_j=1^n mathbbP(I_j) geq mathbbP(I).$$
Proof:
For any $1leq j leq n$, let $a_j$ be the left end point of $I_j$ and $b_j$ be the right end point of $I_j$. The sets $a_j$ and $b_j$ are finite and thus can be ordered. Let $a_l_i$ and $b_k_i$ be the ordered sets of these end points (from least to greatest).
For the interval $I$ we have
$$bigcup_j=1^n I_j supseteq I.$$
Letting $a$ and $b$ denote the left and right end points of $I$, respectively, without loss of generality we assume
$$a_l_1leq aleq b_l_1, $$
$$a_k_nleq bleq b_k_n. $$
The above simply says that $a$ and $b$ are in the intervals with smallest left end point and largest right end point, respectively (this could of course be the same interval).
Using the ordering of end points, we can pair them up, from smallest to largest, in the sum of interval lengths:
$$sum_j=1^n(b_j - a_j)= sum_i=1^n(b_j_i - a_j_i)= Big[sum_i =2^n-1(b_j_i - a_j_i) +(b_j_1 -a_j_n)Big]+(b_j_n -a_j_1) .$$
In the last expression, we have removed the smallest left end point and largest right end point from the sum. Since we assume some part of the interval $I$ is in every $I_n$, the remaining end points of intervals must overlap, which means, for any $i$,
$$b_j_i-1 geq a_j_i.$$
Thus, we can reorder again to show that
$$sum_i =2^n-1(b_j_i - a_j_i) + (b_j_1-a_j_n)= sum_i=2^n (b_j_i-1- a_j_i) geq0.$$
Returning to our original sum, it follows that
$$sum_j=1^n (b_j - a_j) geq b_k_n - a_l_1 geq b-a,$$
which implies
$$ sum_j=1^n mathbbP(I_j) geq mathbbP(I).$$
$ square$
Lemma 2:
Let $I_1, I_2,cdots$ be a countable collection of open intervals, whos union contains a closed interval, then $$sum_jgeq 1mathbbP(I_j) geq mathbbP(I).$$
Proof:
By the Heine-Borel theorem, for any countable collection of open sets $O_n$ and some closed set $ bigcup_ngeq1 O_n supseteq O$, there exists a finite sub cover
$$O_a_1 cup O_a_2 cup cdots cup O_a_k supseteq O.$$
Thus
$$ bigcup_j=1^infty I_j supseteq I. implies bigcup_a=1^k I_j_a supseteq I$$
and the result follows from lemma 1.
$square$
Lemma 3:
For any countable collection of intervals $I_j$, whose union contains an interval $I$,
$$sum_j geq 1 mathbbP (I_j) geq mathbbP(I).$$
Proof:
Choose some $epsilon >0$ and extend the left and right end points of the $I_j$ by $epsilon 2^-j$ to form an open set
$$I_j_epsilon=left(a_j -epsilon 2^-j , b_j + epsilon 2^-j right).$$
Next, extend the end points of $I$ by $epsilon$ to make the closed set
$$I_epsilon =[a- epsilon , b + epsilon].$$
Since $bigcup_j geq 1 I_j supseteq I,$ there exists some $a_k$ and $b_l$ such that
$$a_k <a leq b < b_l.$$
Further, since $k$ and $l$ are both $geq 1$, we can choose $epsilon$ small enough that
$$a_k - epsilon 2^-k< a_k<a - epsilon quad, quad b+epsilon < b_l < b_l + epsilon 2^-l.$$
Thus, we have constructed a countable collection of open intervals $I_j_epsilon$, whose union contains the closed interval $I_epsilon$. By lemma 2:
$$sum_jgeq1 mathbbP(I_j_epsilon) geq mathbbP(I_epsilon)$$
$$implies sum_j geq 1 Big big( b_j + epsilon 2^-j big) - big(a_j - epsilon 2^-j big)Big geq( b + epsilon )- (a - epsilon).$$
$$implies sum_j geq 1 ( b_j - a_j) + epsilon sum_j geq 1 big (2^1-j big) geq (b-a) + 2epsilon.$$
$$ implies sum_j geq 1 ( b_j - a_j) + 2epsilon geq (b-a) + 2epsilon.$$
$$implies sum_jgeq 1 mathbbP(I_j)=sum_j geq 1 ( b_j - a_j) geq (b-a) = mathbbP(I).$$
$ square$
real-analysis sequences-and-series probability-theory proof-verification
asked Jul 17 at 5:35
Moed Pol Bollo
19518
I’m going through the construction of Lebesgue measure, which relies on this result. In the following, the $I$ are all intervals defined on $[0,1]$ and $mathbbP(I)$ denotes the length of the interval.
Lemma 1
Let $I_1, I_2, cdots, I_n$ be a finite collection of intervals whose union contain an interval $I$, then $$sum_j=1^n mathbbP(I_j) geq mathbbP(I).$$
Proof:
For any $1leq j leq n$, let $a_j$ be the left end point of $I_j$ and $b_j$ be the right end point of $I_j$. The sets $a_j$ and $b_j$ are finite and thus can be ordered. Let $a_l_i$ and $b_k_i$ be the ordered sets of these end points (from least to greatest).
For the interval $I$ we have
$$bigcup_j=1^n I_j supseteq I.$$
Letting $a$ and $b$ denote the left and right end points of $I$, respectively, without loss of generality we assume
$$a_l_1leq aleq b_l_1, $$
$$a_k_nleq bleq b_k_n. $$
The above simply says that $a$ and $b$ are in the intervals with smallest left end point and largest right end point, respectively (this could of course be the same interval).
Using the ordering of end points, we can pair them up, from smallest to largest, in the sum of interval lengths:
$$sum_j=1^n(b_j - a_j)= sum_i=1^n(b_j_i - a_j_i)= Big[sum_i =2^n-1(b_j_i - a_j_i) +(b_j_1 -a_j_n)Big]+(b_j_n -a_j_1) .$$
In the last expression, we have removed the smallest left end point and largest right end point from the sum. Since we assume some part of the interval $I$ is in every $I_n$, the remaining end points of intervals must overlap, which means, for any $i$,
$$b_j_i-1 geq a_j_i.$$
Thus, we can reorder again to show that
$$sum_i =2^n-1(b_j_i - a_j_i) + (b_j_1-a_j_n)= sum_i=2^n (b_j_i-1- a_j_i) geq0.$$
Returning to our original sum, it follows that
$$sum_j=1^n (b_j - a_j) geq b_k_n - a_l_1 geq b-a,$$
which implies
$$ sum_j=1^n mathbbP(I_j) geq mathbbP(I).$$
$ square$
Lemma 2:
Let $I_1, I_2,cdots$ be a countable collection of open intervals, whos union contains a closed interval, then $$sum_jgeq 1mathbbP(I_j) geq mathbbP(I).$$
Proof:
By the Heine-Borel theorem, for any countable collection of open sets $O_n$ and some closed set $ bigcup_ngeq1 O_n supseteq O$, there exists a finite sub cover
$$O_a_1 cup O_a_2 cup cdots cup O_a_k supseteq O.$$
Thus
$$ bigcup_j=1^infty I_j supseteq I. implies bigcup_a=1^k I_j_a supseteq I$$
and the result follows from lemma 1.
$square$
Lemma 3:
For any countable collection of intervals $I_j$, whose union contains an interval $I$,
$$sum_j geq 1 mathbbP (I_j) geq mathbbP(I).$$
Proof:
Choose some $epsilon >0$ and extend the left and right end points of the $I_j$ by $epsilon 2^-j$ to form an open set
$$I_j_epsilon=left(a_j -epsilon 2^-j , b_j + epsilon 2^-j right).$$
Next, extend the end points of $I$ by $epsilon$ to make the closed set
$$I_epsilon =[a- epsilon , b + epsilon].$$
Since $bigcup_j geq 1 I_j supseteq I,$ there exists some $a_k$ and $b_l$ such that
$$a_k <a leq b < b_l.$$
Further, since $k$ and $l$ are both $geq 1$, we can choose $epsilon$ small enough that
$$a_k - epsilon 2^-k< a_k<a - epsilon quad, quad b+epsilon < b_l < b_l + epsilon 2^-l.$$
Thus, we have constructed a countable collection of open intervals $I_j_epsilon$, whose union contains the closed interval $I_epsilon$. By lemma 2:
$$sum_jgeq1 mathbbP(I_j_epsilon) geq mathbbP(I_epsilon)$$
$$implies sum_j geq 1 Big big( b_j + epsilon 2^-j big) - big(a_j - epsilon 2^-j big)Big geq( b + epsilon )- (a - epsilon).$$
$$implies sum_j geq 1 ( b_j - a_j) + epsilon sum_j geq 1 big (2^1-j big) geq (b-a) + 2epsilon.$$
$$ implies sum_j geq 1 ( b_j - a_j) + 2epsilon geq (b-a) + 2epsilon.$$
$$implies sum_jgeq 1 mathbbP(I_j)=sum_j geq 1 ( b_j - a_j) geq (b-a) = mathbbP(I).$$
$ square$
real-analysis sequences-and-series probability-theory proof-verification
asked Jul 17 at 5:35
Moed Pol Bollo
19518
edited Jul 21 at 10:21
asked Jul 17 at 5:35
Moed Pol Bollo
19518
asked Jul 17 at 5:35
Moed Pol Bollo
19518
asked Jul 17 at 5:35
Moed Pol Bollo
19518
19518
What's the question?
– Lord Shark the Unknown
Jul 17 at 5:37
@LordSharktheUnknown I just want to make sure the proof is correct, as it is necessary for the construction of Lebesgue measure on $[0,1]$.
– Moed Pol Bollo
Jul 17 at 5:39
What does "Choose $a_k$ and $b_k$ such that: $a_kle a_j$, $b_jle b_k$" mean?
– Lord Shark the Unknown
Jul 17 at 5:42
@LordSharktheUnknown That’s a poorly worded way of saying $a_k$ is the smallest element of the set of left end points and $b_k$ is the largest element of the set of right end points.
– Moed Pol Bollo
Jul 17 at 5:45
Then that's an error straight away; the least left endpoint may not belong to the same interval $I_k$ as the largest right endpoint.
– Lord Shark the Unknown
Jul 17 at 5:46
 |Â
show 4 more comments
What's the question?
– Lord Shark the Unknown
Jul 17 at 5:37
@LordSharktheUnknown I just want to make sure the proof is correct, as it is necessary for the construction of Lebesgue measure on $[0,1]$.
– Moed Pol Bollo
Jul 17 at 5:39
What does "Choose $a_k$ and $b_k$ such that: $a_kle a_j$, $b_jle b_k$" mean?
– Lord Shark the Unknown
Jul 17 at 5:42
@LordSharktheUnknown That’s a poorly worded way of saying $a_k$ is the smallest element of the set of left end points and $b_k$ is the largest element of the set of right end points.
– Moed Pol Bollo
Jul 17 at 5:45
Then that's an error straight away; the least left endpoint may not belong to the same interval $I_k$ as the largest right endpoint.
– Lord Shark the Unknown
Jul 17 at 5:46
What's the question?
– Lord Shark the Unknown
Jul 17 at 5:37
What's the question?
– Lord Shark the Unknown
Jul 17 at 5:37
@LordSharktheUnknown I just want to make sure the proof is correct, as it is necessary for the construction of Lebesgue measure on $[0,1]$.
– Moed Pol Bollo
Jul 17 at 5:39
@LordSharktheUnknown I just want to make sure the proof is correct, as it is necessary for the construction of Lebesgue measure on $[0,1]$.
– Moed Pol Bollo
Jul 17 at 5:39
What does "Choose $a_k$ and $b_k$ such that: $a_kle a_j$, $b_jle b_k$" mean?
– Lord Shark the Unknown
Jul 17 at 5:42
What does "Choose $a_k$ and $b_k$ such that: $a_kle a_j$, $b_jle b_k$" mean?
– Lord Shark the Unknown
Jul 17 at 5:42
@LordSharktheUnknown That’s a poorly worded way of saying $a_k$ is the smallest element of the set of left end points and $b_k$ is the largest element of the set of right end points.
– Moed Pol Bollo
Jul 17 at 5:45
@LordSharktheUnknown That’s a poorly worded way of saying $a_k$ is the smallest element of the set of left end points and $b_k$ is the largest element of the set of right end points.
– Moed Pol Bollo
Jul 17 at 5:45
Then that's an error straight away; the least left endpoint may not belong to the same interval $I_k$ as the largest right endpoint.
– Lord Shark the Unknown
Jul 17 at 5:46
Then that's an error straight away; the least left endpoint may not belong to the same interval $I_k$ as the largest right endpoint.
– Lord Shark the Unknown
Jul 17 at 5:46
 |Â
show 4 more comments
1 Answer
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Prove Lemma 2 directly, using a finite recursion, and you will not need Lemma 1. As follows: Let $I=[a,b]$ with $a<b$ and let $C=c_j:1leq jleq n$ be a family of open intervals that covers $[a,b].$
Let $f(1)$ be the least $m$ such that $0in c_m.$
For $jgeq 1,$ if $sup c_f(j)leq b,$ let $f(j+1)$ be the least $m$ such that $sup c_f(j)in c_m.$ If $sup c_f(j)>b$ then $f(j')$ is not defined for $j'>j$.
By induction on $j:$ If $f(j)$ exists then
$(alpha).;sup (cup_ileq jc_f(i))=sup c_f(j),$ and
$(beta).; cup_ileq jc_f(i)supset [0,sup c_f(j)), $ and
$(gamma).;$ If $1leq i<j$ then $f(j)ne f(i) $ and $c_f(i)ne c_f(j)$.
Now dom($f)=1,..,K$ for some $Kleq n.$ And $sup c_f(K)>b.$ For convenience define $U(0)=a$ and $U(j)=sup c_f(j)$ for $1leq jleq K.$
For $1leq jleq K $ we have $U(j-1)in c_f(j)$ so we have $P(c_f(j))>(sup c_f(j))-U(j-1)=U(j)-U(j-1).$
Therefore, (especially by $(gamma);$), we have $sum_i=1^nP(c_i)geq$ $ sum_j=1^KP(c_f(j))>$ $sum_j=1^K U(j)-U(j-1)=$ $=U(K)-U(0)=U(K)-a>b-a=P(I).$
answered Jul 20 at 12:03
DanielWainfleet
31.7k31644
To the proposer: You seem to be having difficulties with Lemma 1. But your main result (Lemma 3) uses Lemma 2 only, so I decided to post a direct proof of Lemma 2 .
– DanielWainfleet
Jul 20 at 12:06
Thanks for the alternative proof! I am still trying to understand all of it. A few questions for clarification: (1) Is your definition of $C$ not that of a finite collection of open intervals? (2) By “let $f(j+1)$ be the least $m$ such that $sup c_f(j) in c_m$†are you essentially specifying an overlap of intervals similar to that in my proof of Lemma 1? Thanks again.
– Moed Pol Bollo
Jul 21 at 7:52
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Prove Lemma 2 directly, using a finite recursion, and you will not need Lemma 1. As follows: Let $I=[a,b]$ with $a<b$ and let $C=c_j:1leq jleq n$ be a family of open intervals that covers $[a,b].$
Let $f(1)$ be the least $m$ such that $0in c_m.$
For $jgeq 1,$ if $sup c_f(j)leq b,$ let $f(j+1)$ be the least $m$ such that $sup c_f(j)in c_m.$ If $sup c_f(j)>b$ then $f(j')$ is not defined for $j'>j$.
By induction on $j:$ If $f(j)$ exists then
$(alpha).;sup (cup_ileq jc_f(i))=sup c_f(j),$ and
$(beta).; cup_ileq jc_f(i)supset [0,sup c_f(j)), $ and
$(gamma).;$ If $1leq i<j$ then $f(j)ne f(i) $ and $c_f(i)ne c_f(j)$.
Now dom($f)=1,..,K$ for some $Kleq n.$ And $sup c_f(K)>b.$ For convenience define $U(0)=a$ and $U(j)=sup c_f(j)$ for $1leq jleq K.$
For $1leq jleq K $ we have $U(j-1)in c_f(j)$ so we have $P(c_f(j))>(sup c_f(j))-U(j-1)=U(j)-U(j-1).$
Therefore, (especially by $(gamma);$), we have $sum_i=1^nP(c_i)geq$ $ sum_j=1^KP(c_f(j))>$ $sum_j=1^K U(j)-U(j-1)=$ $=U(K)-U(0)=U(K)-a>b-a=P(I).$
answered Jul 20 at 12:03
DanielWainfleet
31.7k31644
To the proposer: You seem to be having difficulties with Lemma 1. But your main result (Lemma 3) uses Lemma 2 only, so I decided to post a direct proof of Lemma 2 .
– DanielWainfleet
Jul 20 at 12:06
Thanks for the alternative proof! I am still trying to understand all of it. A few questions for clarification: (1) Is your definition of $C$ not that of a finite collection of open intervals? (2) By “let $f(j+1)$ be the least $m$ such that $sup c_f(j) in c_m$†are you essentially specifying an overlap of intervals similar to that in my proof of Lemma 1? Thanks again.
– Moed Pol Bollo
Jul 21 at 7:52
add a comment |Â
up vote
1
down vote
Prove Lemma 2 directly, using a finite recursion, and you will not need Lemma 1. As follows: Let $I=[a,b]$ with $a<b$ and let $C=c_j:1leq jleq n$ be a family of open intervals that covers $[a,b].$
Let $f(1)$ be the least $m$ such that $0in c_m.$
For $jgeq 1,$ if $sup c_f(j)leq b,$ let $f(j+1)$ be the least $m$ such that $sup c_f(j)in c_m.$ If $sup c_f(j)>b$ then $f(j')$ is not defined for $j'>j$.
By induction on $j:$ If $f(j)$ exists then
$(alpha).;sup (cup_ileq jc_f(i))=sup c_f(j),$ and
$(beta).; cup_ileq jc_f(i)supset [0,sup c_f(j)), $ and
$(gamma).;$ If $1leq i<j$ then $f(j)ne f(i) $ and $c_f(i)ne c_f(j)$.
Now dom($f)=1,..,K$ for some $Kleq n.$ And $sup c_f(K)>b.$ For convenience define $U(0)=a$ and $U(j)=sup c_f(j)$ for $1leq jleq K.$
For $1leq jleq K $ we have $U(j-1)in c_f(j)$ so we have $P(c_f(j))>(sup c_f(j))-U(j-1)=U(j)-U(j-1).$
Therefore, (especially by $(gamma);$), we have $sum_i=1^nP(c_i)geq$ $ sum_j=1^KP(c_f(j))>$ $sum_j=1^K U(j)-U(j-1)=$ $=U(K)-U(0)=U(K)-a>b-a=P(I).$
answered Jul 20 at 12:03
DanielWainfleet
31.7k31644
To the proposer: You seem to be having difficulties with Lemma 1. But your main result (Lemma 3) uses Lemma 2 only, so I decided to post a direct proof of Lemma 2 .
– DanielWainfleet
Jul 20 at 12:06
Thanks for the alternative proof! I am still trying to understand all of it. A few questions for clarification: (1) Is your definition of $C$ not that of a finite collection of open intervals? (2) By “let $f(j+1)$ be the least $m$ such that $sup c_f(j) in c_m$†are you essentially specifying an overlap of intervals similar to that in my proof of Lemma 1? Thanks again.
– Moed Pol Bollo
Jul 21 at 7:52
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Prove Lemma 2 directly, using a finite recursion, and you will not need Lemma 1. As follows: Let $I=[a,b]$ with $a<b$ and let $C=c_j:1leq jleq n$ be a family of open intervals that covers $[a,b].$
Let $f(1)$ be the least $m$ such that $0in c_m.$
For $jgeq 1,$ if $sup c_f(j)leq b,$ let $f(j+1)$ be the least $m$ such that $sup c_f(j)in c_m.$ If $sup c_f(j)>b$ then $f(j')$ is not defined for $j'>j$.
By induction on $j:$ If $f(j)$ exists then
$(alpha).;sup (cup_ileq jc_f(i))=sup c_f(j),$ and
$(beta).; cup_ileq jc_f(i)supset [0,sup c_f(j)), $ and
$(gamma).;$ If $1leq i<j$ then $f(j)ne f(i) $ and $c_f(i)ne c_f(j)$.
Now dom($f)=1,..,K$ for some $Kleq n.$ And $sup c_f(K)>b.$ For convenience define $U(0)=a$ and $U(j)=sup c_f(j)$ for $1leq jleq K.$
For $1leq jleq K $ we have $U(j-1)in c_f(j)$ so we have $P(c_f(j))>(sup c_f(j))-U(j-1)=U(j)-U(j-1).$
Therefore, (especially by $(gamma);$), we have $sum_i=1^nP(c_i)geq$ $ sum_j=1^KP(c_f(j))>$ $sum_j=1^K U(j)-U(j-1)=$ $=U(K)-U(0)=U(K)-a>b-a=P(I).$
answered Jul 20 at 12:03
DanielWainfleet
31.7k31644
Prove Lemma 2 directly, using a finite recursion, and you will not need Lemma 1. As follows: Let $I=[a,b]$ with $a<b$ and let $C=c_j:1leq jleq n$ be a family of open intervals that covers $[a,b].$
Let $f(1)$ be the least $m$ such that $0in c_m.$
For $jgeq 1,$ if $sup c_f(j)leq b,$ let $f(j+1)$ be the least $m$ such that $sup c_f(j)in c_m.$ If $sup c_f(j)>b$ then $f(j')$ is not defined for $j'>j$.
By induction on $j:$ If $f(j)$ exists then
$(alpha).;sup (cup_ileq jc_f(i))=sup c_f(j),$ and
$(beta).; cup_ileq jc_f(i)supset [0,sup c_f(j)), $ and
$(gamma).;$ If $1leq i<j$ then $f(j)ne f(i) $ and $c_f(i)ne c_f(j)$.
Now dom($f)=1,..,K$ for some $Kleq n.$ And $sup c_f(K)>b.$ For convenience define $U(0)=a$ and $U(j)=sup c_f(j)$ for $1leq jleq K.$
For $1leq jleq K $ we have $U(j-1)in c_f(j)$ so we have $P(c_f(j))>(sup c_f(j))-U(j-1)=U(j)-U(j-1).$
Therefore, (especially by $(gamma);$), we have $sum_i=1^nP(c_i)geq$ $ sum_j=1^KP(c_f(j))>$ $sum_j=1^K U(j)-U(j-1)=$ $=U(K)-U(0)=U(K)-a>b-a=P(I).$
answered Jul 20 at 12:03
DanielWainfleet
31.7k31644
edited Jul 20 at 12:34
answered Jul 20 at 12:03
DanielWainfleet
31.7k31644
answered Jul 20 at 12:03
DanielWainfleet
31.7k31644
answered Jul 20 at 12:03
DanielWainfleet
31.7k31644
31.7k31644
To the proposer: You seem to be having difficulties with Lemma 1. But your main result (Lemma 3) uses Lemma 2 only, so I decided to post a direct proof of Lemma 2 .
– DanielWainfleet
Jul 20 at 12:06
Thanks for the alternative proof! I am still trying to understand all of it. A few questions for clarification: (1) Is your definition of $C$ not that of a finite collection of open intervals? (2) By “let $f(j+1)$ be the least $m$ such that $sup c_f(j) in c_m$†are you essentially specifying an overlap of intervals similar to that in my proof of Lemma 1? Thanks again.
– Moed Pol Bollo
Jul 21 at 7:52
add a comment |Â
To the proposer: You seem to be having difficulties with Lemma 1. But your main result (Lemma 3) uses Lemma 2 only, so I decided to post a direct proof of Lemma 2 .
– DanielWainfleet
Jul 20 at 12:06
Thanks for the alternative proof! I am still trying to understand all of it. A few questions for clarification: (1) Is your definition of $C$ not that of a finite collection of open intervals? (2) By “let $f(j+1)$ be the least $m$ such that $sup c_f(j) in c_m$†are you essentially specifying an overlap of intervals similar to that in my proof of Lemma 1? Thanks again.
– Moed Pol Bollo
Jul 21 at 7:52
To the proposer: You seem to be having difficulties with Lemma 1. But your main result (Lemma 3) uses Lemma 2 only, so I decided to post a direct proof of Lemma 2 .
– DanielWainfleet
Jul 20 at 12:06
To the proposer: You seem to be having difficulties with Lemma 1. But your main result (Lemma 3) uses Lemma 2 only, so I decided to post a direct proof of Lemma 2 .
– DanielWainfleet
Jul 20 at 12:06
Thanks for the alternative proof! I am still trying to understand all of it. A few questions for clarification: (1) Is your definition of $C$ not that of a finite collection of open intervals? (2) By “let $f(j+1)$ be the least $m$ such that $sup c_f(j) in c_m$†are you essentially specifying an overlap of intervals similar to that in my proof of Lemma 1? Thanks again.
– Moed Pol Bollo
Jul 21 at 7:52
Thanks for the alternative proof! I am still trying to understand all of it. A few questions for clarification: (1) Is your definition of $C$ not that of a finite collection of open intervals? (2) By “let $f(j+1)$ be the least $m$ such that $sup c_f(j) in c_m$†are you essentially specifying an overlap of intervals similar to that in my proof of Lemma 1? Thanks again.
– Moed Pol Bollo
Jul 21 at 7:52
add a comment |Â
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What's the question?
– Lord Shark the Unknown
Jul 17 at 5:37
@LordSharktheUnknown I just want to make sure the proof is correct, as it is necessary for the construction of Lebesgue measure on $[0,1]$.
– Moed Pol Bollo
Jul 17 at 5:39
What does "Choose $a_k$ and $b_k$ such that: $a_kle a_j$, $b_jle b_k$" mean?
– Lord Shark the Unknown
Jul 17 at 5:42
@LordSharktheUnknown That’s a poorly worded way of saying $a_k$ is the smallest element of the set of left end points and $b_k$ is the largest element of the set of right end points.
– Moed Pol Bollo
Jul 17 at 5:45
Then that's an error straight away; the least left endpoint may not belong to the same interval $I_k$ as the largest right endpoint.
– Lord Shark the Unknown
Jul 17 at 5:46