Mixing
Terence Tao's Construction of the Lebesgue-Stieltjes Measures
$begingroup$
I am having trouble reading the proof of the construction of the Lebesgue-Stieltjes measure in Terence Tao's An Introduction to Measure Theory. See Theorem 1.7.9 on pp. 189-192 on https://terrytao.files.wordpress.com/2011/01/measure-book1.pdf.
Here are my questions:
What is the functional form of the $F-$volume of an interval? If $I$ is, say, $(a,b)$, $- infty < a < b < infty$, then is it $F(b) - F(a)$? If so, is it the case for all intervals, closed, open, half open-half closed, half infinite etc.?
If 1 is true, then the formulas for the Borel measures of the intervals given in the stem of Theorem 1.7.9 don't make much sense.
I think I can pretty much follow the remainder of the proof, but I am very unsure of how an algebra is set up, and how a premeasure is defined on the said algebra, which is then used to exploit the Hahn-Kolmogorov Theorem.
measure-theory
edited Jan 24 at 5:35
J. W. Tanner
3,2701320
asked Jan 24 at 4:41
user82261user82261
27017
$endgroup$
add a comment |
$begingroup$
I am having trouble reading the proof of the construction of the Lebesgue-Stieltjes measure in Terence Tao's An Introduction to Measure Theory. See Theorem 1.7.9 on pp. 189-192 on https://terrytao.files.wordpress.com/2011/01/measure-book1.pdf.
Here are my questions:
What is the functional form of the $F-$volume of an interval? If $I$ is, say, $(a,b)$, $- infty < a < b < infty$, then is it $F(b) - F(a)$? If so, is it the case for all intervals, closed, open, half open-half closed, half infinite etc.?
If 1 is true, then the formulas for the Borel measures of the intervals given in the stem of Theorem 1.7.9 don't make much sense.
I think I can pretty much follow the remainder of the proof, but I am very unsure of how an algebra is set up, and how a premeasure is defined on the said algebra, which is then used to exploit the Hahn-Kolmogorov Theorem.
measure-theory
edited Jan 24 at 5:35
J. W. Tanner
3,2701320
asked Jan 24 at 4:41
user82261user82261
27017
$endgroup$
add a comment |
$begingroup$
I am having trouble reading the proof of the construction of the Lebesgue-Stieltjes measure in Terence Tao's An Introduction to Measure Theory. See Theorem 1.7.9 on pp. 189-192 on https://terrytao.files.wordpress.com/2011/01/measure-book1.pdf.
Here are my questions:
What is the functional form of the $F-$volume of an interval? If $I$ is, say, $(a,b)$, $- infty < a < b < infty$, then is it $F(b) - F(a)$? If so, is it the case for all intervals, closed, open, half open-half closed, half infinite etc.?
If 1 is true, then the formulas for the Borel measures of the intervals given in the stem of Theorem 1.7.9 don't make much sense.
I think I can pretty much follow the remainder of the proof, but I am very unsure of how an algebra is set up, and how a premeasure is defined on the said algebra, which is then used to exploit the Hahn-Kolmogorov Theorem.
measure-theory
edited Jan 24 at 5:35
J. W. Tanner
3,2701320
asked Jan 24 at 4:41
user82261user82261
27017
$endgroup$
I am having trouble reading the proof of the construction of the Lebesgue-Stieltjes measure in Terence Tao's An Introduction to Measure Theory. See Theorem 1.7.9 on pp. 189-192 on https://terrytao.files.wordpress.com/2011/01/measure-book1.pdf.
Here are my questions:
What is the functional form of the $F-$volume of an interval? If $I$ is, say, $(a,b)$, $- infty < a < b < infty$, then is it $F(b) - F(a)$? If so, is it the case for all intervals, closed, open, half open-half closed, half infinite etc.?
If 1 is true, then the formulas for the Borel measures of the intervals given in the stem of Theorem 1.7.9 don't make much sense.
I think I can pretty much follow the remainder of the proof, but I am very unsure of how an algebra is set up, and how a premeasure is defined on the said algebra, which is then used to exploit the Hahn-Kolmogorov Theorem.
measure-theory
measure-theory
edited Jan 24 at 5:35
J. W. Tanner
3,2701320
asked Jan 24 at 4:41
user82261user82261
27017
edited Jan 24 at 5:35
J. W. Tanner
3,2701320
asked Jan 24 at 4:41
user82261user82261
27017
edited Jan 24 at 5:35
J. W. Tanner
3,2701320
edited Jan 24 at 5:35
J. W. Tanner
3,2701320
edited Jan 24 at 5:35
J. W. Tanner
3,2701320
3,2701320
asked Jan 24 at 4:41
user82261user82261
27017
asked Jan 24 at 4:41
user82261user82261
27017
asked Jan 24 at 4:41
user82261user82261
27017
27017
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The $F$-volume of $I$ is defined by the equalities in $(1.33)$. If I have to guess, what might be confusing you is that you imagine that $F$ is continuous. But $F$ may have jumps, and where $F$ has a jump, the measure will have an atom.
For instance, take $F=1_{[1,infty)}$, so there's a jump at $1$. Now
$$
|[0,1]|_F=F_+(1)-F_-(0)=1-0-1,
$$
while
$$
|[0,1)|_F=F_-(1)-F_-(0)=0-0=0.
$$
The measure $mu_F$ will have an atom at $1$, i.e. $mu_F({1})=1$.
answered Jan 24 at 4:55
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
$begingroup$
Is there any particular motivation for defining the $F's$ the way they are? For example, couldn't we have defined the $F-$ volume of $(a,b]$ as $F_{-}(b) - F_{-}(a)$ instead? Why the $+'s$ instead of $-'s$.
$endgroup$
– user82261
Jan 24 at 4:58
$begingroup$
Because you need to take into account whether you are using the jump or not. If you define $|(0,1]|_F=0$ in my example above, you get that $|(1,1+delta)|_F=1$ for all $delta>0$. This contradicts continuity of the measure, since you have $bigcap_n(1,1+1/n)=emptyset$, but all sets in the intersection have measure $1$.
$endgroup$
– Martin Argerami
Jan 24 at 5:02
$begingroup$
This is how I am thinking about it. The Borel sigma algebra is generated by any of the four bounded intervals. Now, we have that $(a,b], [a,b), [a,b]$ can be written as countable intersections/unions of intervals of the form $(a,b)$. So, the choice isn't unique; once the definition of $F-$ volume of $(a,b)$ is pinned down, the choice for the $F-$ volume of the other intervals is also pinned down. But isn't the choice for $(a,b)$ then arbitrary. Why not $F_{+}(b) - F_{-}(a)$ instead of $F_{-}(b) - F_{+}(a)$?
$endgroup$
– user82261
Jan 24 at 5:35
$begingroup$
Because you get something incoherent. That's why I gave you the example. Take my example and define $|(a,b)|=F_+(b)-F_-(a)$. Then $|(0,1)|=1-0=1$. But $|(0,1-1/n)|=0$ for all $n$, contradicting continuity of the measure: $(0,1)=bigcup_n(0,1-1/n)$, but all the sets of the right have measure zero and the one in the left has measure one.
$endgroup$
– Martin Argerami
Jan 24 at 14:11
add a comment |
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$begingroup$
The $F$-volume of $I$ is defined by the equalities in $(1.33)$. If I have to guess, what might be confusing you is that you imagine that $F$ is continuous. But $F$ may have jumps, and where $F$ has a jump, the measure will have an atom.
For instance, take $F=1_{[1,infty)}$, so there's a jump at $1$. Now
$$
|[0,1]|_F=F_+(1)-F_-(0)=1-0-1,
$$
while
$$
|[0,1)|_F=F_-(1)-F_-(0)=0-0=0.
$$
The measure $mu_F$ will have an atom at $1$, i.e. $mu_F({1})=1$.
answered Jan 24 at 4:55
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
$begingroup$
Is there any particular motivation for defining the $F's$ the way they are? For example, couldn't we have defined the $F-$ volume of $(a,b]$ as $F_{-}(b) - F_{-}(a)$ instead? Why the $+'s$ instead of $-'s$.
$endgroup$
– user82261
Jan 24 at 4:58
$begingroup$
Because you need to take into account whether you are using the jump or not. If you define $|(0,1]|_F=0$ in my example above, you get that $|(1,1+delta)|_F=1$ for all $delta>0$. This contradicts continuity of the measure, since you have $bigcap_n(1,1+1/n)=emptyset$, but all sets in the intersection have measure $1$.
$endgroup$
– Martin Argerami
Jan 24 at 5:02
$begingroup$
This is how I am thinking about it. The Borel sigma algebra is generated by any of the four bounded intervals. Now, we have that $(a,b], [a,b), [a,b]$ can be written as countable intersections/unions of intervals of the form $(a,b)$. So, the choice isn't unique; once the definition of $F-$ volume of $(a,b)$ is pinned down, the choice for the $F-$ volume of the other intervals is also pinned down. But isn't the choice for $(a,b)$ then arbitrary. Why not $F_{+}(b) - F_{-}(a)$ instead of $F_{-}(b) - F_{+}(a)$?
$endgroup$
– user82261
Jan 24 at 5:35
$begingroup$
Because you get something incoherent. That's why I gave you the example. Take my example and define $|(a,b)|=F_+(b)-F_-(a)$. Then $|(0,1)|=1-0=1$. But $|(0,1-1/n)|=0$ for all $n$, contradicting continuity of the measure: $(0,1)=bigcup_n(0,1-1/n)$, but all the sets of the right have measure zero and the one in the left has measure one.
$endgroup$
– Martin Argerami
Jan 24 at 14:11
add a comment |
$begingroup$
The $F$-volume of $I$ is defined by the equalities in $(1.33)$. If I have to guess, what might be confusing you is that you imagine that $F$ is continuous. But $F$ may have jumps, and where $F$ has a jump, the measure will have an atom.
For instance, take $F=1_{[1,infty)}$, so there's a jump at $1$. Now
$$
|[0,1]|_F=F_+(1)-F_-(0)=1-0-1,
$$
while
$$
|[0,1)|_F=F_-(1)-F_-(0)=0-0=0.
$$
The measure $mu_F$ will have an atom at $1$, i.e. $mu_F({1})=1$.
answered Jan 24 at 4:55
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
$begingroup$
Is there any particular motivation for defining the $F's$ the way they are? For example, couldn't we have defined the $F-$ volume of $(a,b]$ as $F_{-}(b) - F_{-}(a)$ instead? Why the $+'s$ instead of $-'s$.
$endgroup$
– user82261
Jan 24 at 4:58
$begingroup$
Because you need to take into account whether you are using the jump or not. If you define $|(0,1]|_F=0$ in my example above, you get that $|(1,1+delta)|_F=1$ for all $delta>0$. This contradicts continuity of the measure, since you have $bigcap_n(1,1+1/n)=emptyset$, but all sets in the intersection have measure $1$.
$endgroup$
– Martin Argerami
Jan 24 at 5:02
$begingroup$
This is how I am thinking about it. The Borel sigma algebra is generated by any of the four bounded intervals. Now, we have that $(a,b], [a,b), [a,b]$ can be written as countable intersections/unions of intervals of the form $(a,b)$. So, the choice isn't unique; once the definition of $F-$ volume of $(a,b)$ is pinned down, the choice for the $F-$ volume of the other intervals is also pinned down. But isn't the choice for $(a,b)$ then arbitrary. Why not $F_{+}(b) - F_{-}(a)$ instead of $F_{-}(b) - F_{+}(a)$?
$endgroup$
– user82261
Jan 24 at 5:35
$begingroup$
Because you get something incoherent. That's why I gave you the example. Take my example and define $|(a,b)|=F_+(b)-F_-(a)$. Then $|(0,1)|=1-0=1$. But $|(0,1-1/n)|=0$ for all $n$, contradicting continuity of the measure: $(0,1)=bigcup_n(0,1-1/n)$, but all the sets of the right have measure zero and the one in the left has measure one.
$endgroup$
– Martin Argerami
Jan 24 at 14:11
add a comment |
$begingroup$
The $F$-volume of $I$ is defined by the equalities in $(1.33)$. If I have to guess, what might be confusing you is that you imagine that $F$ is continuous. But $F$ may have jumps, and where $F$ has a jump, the measure will have an atom.
For instance, take $F=1_{[1,infty)}$, so there's a jump at $1$. Now
$$
|[0,1]|_F=F_+(1)-F_-(0)=1-0-1,
$$
while
$$
|[0,1)|_F=F_-(1)-F_-(0)=0-0=0.
$$
The measure $mu_F$ will have an atom at $1$, i.e. $mu_F({1})=1$.
answered Jan 24 at 4:55
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
The $F$-volume of $I$ is defined by the equalities in $(1.33)$. If I have to guess, what might be confusing you is that you imagine that $F$ is continuous. But $F$ may have jumps, and where $F$ has a jump, the measure will have an atom.
For instance, take $F=1_{[1,infty)}$, so there's a jump at $1$. Now
$$
|[0,1]|_F=F_+(1)-F_-(0)=1-0-1,
$$
while
$$
|[0,1)|_F=F_-(1)-F_-(0)=0-0=0.
$$
The measure $mu_F$ will have an atom at $1$, i.e. $mu_F({1})=1$.
answered Jan 24 at 4:55
Martin ArgeramiMartin Argerami
128k1184184
answered Jan 24 at 4:55
Martin ArgeramiMartin Argerami
128k1184184
answered Jan 24 at 4:55
Martin ArgeramiMartin Argerami
128k1184184
answered Jan 24 at 4:55
Martin ArgeramiMartin Argerami
128k1184184
128k1184184
$begingroup$
Is there any particular motivation for defining the $F's$ the way they are? For example, couldn't we have defined the $F-$ volume of $(a,b]$ as $F_{-}(b) - F_{-}(a)$ instead? Why the $+'s$ instead of $-'s$.
$endgroup$
– user82261
Jan 24 at 4:58
$begingroup$
Because you need to take into account whether you are using the jump or not. If you define $|(0,1]|_F=0$ in my example above, you get that $|(1,1+delta)|_F=1$ for all $delta>0$. This contradicts continuity of the measure, since you have $bigcap_n(1,1+1/n)=emptyset$, but all sets in the intersection have measure $1$.
$endgroup$
– Martin Argerami
Jan 24 at 5:02
$begingroup$
This is how I am thinking about it. The Borel sigma algebra is generated by any of the four bounded intervals. Now, we have that $(a,b], [a,b), [a,b]$ can be written as countable intersections/unions of intervals of the form $(a,b)$. So, the choice isn't unique; once the definition of $F-$ volume of $(a,b)$ is pinned down, the choice for the $F-$ volume of the other intervals is also pinned down. But isn't the choice for $(a,b)$ then arbitrary. Why not $F_{+}(b) - F_{-}(a)$ instead of $F_{-}(b) - F_{+}(a)$?
$endgroup$
– user82261
Jan 24 at 5:35
$begingroup$
Because you get something incoherent. That's why I gave you the example. Take my example and define $|(a,b)|=F_+(b)-F_-(a)$. Then $|(0,1)|=1-0=1$. But $|(0,1-1/n)|=0$ for all $n$, contradicting continuity of the measure: $(0,1)=bigcup_n(0,1-1/n)$, but all the sets of the right have measure zero and the one in the left has measure one.
$endgroup$
– Martin Argerami
Jan 24 at 14:11
add a comment |
$begingroup$
Is there any particular motivation for defining the $F's$ the way they are? For example, couldn't we have defined the $F-$ volume of $(a,b]$ as $F_{-}(b) - F_{-}(a)$ instead? Why the $+'s$ instead of $-'s$.
$endgroup$
– user82261
Jan 24 at 4:58
$begingroup$
Because you need to take into account whether you are using the jump or not. If you define $|(0,1]|_F=0$ in my example above, you get that $|(1,1+delta)|_F=1$ for all $delta>0$. This contradicts continuity of the measure, since you have $bigcap_n(1,1+1/n)=emptyset$, but all sets in the intersection have measure $1$.
$endgroup$
– Martin Argerami
Jan 24 at 5:02
$begingroup$
This is how I am thinking about it. The Borel sigma algebra is generated by any of the four bounded intervals. Now, we have that $(a,b], [a,b), [a,b]$ can be written as countable intersections/unions of intervals of the form $(a,b)$. So, the choice isn't unique; once the definition of $F-$ volume of $(a,b)$ is pinned down, the choice for the $F-$ volume of the other intervals is also pinned down. But isn't the choice for $(a,b)$ then arbitrary. Why not $F_{+}(b) - F_{-}(a)$ instead of $F_{-}(b) - F_{+}(a)$?
$endgroup$
– user82261
Jan 24 at 5:35
$begingroup$
Because you get something incoherent. That's why I gave you the example. Take my example and define $|(a,b)|=F_+(b)-F_-(a)$. Then $|(0,1)|=1-0=1$. But $|(0,1-1/n)|=0$ for all $n$, contradicting continuity of the measure: $(0,1)=bigcup_n(0,1-1/n)$, but all the sets of the right have measure zero and the one in the left has measure one.
$endgroup$
– Martin Argerami
Jan 24 at 14:11
$begingroup$
Is there any particular motivation for defining the $F's$ the way they are? For example, couldn't we have defined the $F-$ volume of $(a,b]$ as $F_{-}(b) - F_{-}(a)$ instead? Why the $+'s$ instead of $-'s$.
$endgroup$
– user82261
Jan 24 at 4:58
$begingroup$
Is there any particular motivation for defining the $F's$ the way they are? For example, couldn't we have defined the $F-$ volume of $(a,b]$ as $F_{-}(b) - F_{-}(a)$ instead? Why the $+'s$ instead of $-'s$.
$endgroup$
– user82261
Jan 24 at 4:58
$begingroup$
Because you need to take into account whether you are using the jump or not. If you define $|(0,1]|_F=0$ in my example above, you get that $|(1,1+delta)|_F=1$ for all $delta>0$. This contradicts continuity of the measure, since you have $bigcap_n(1,1+1/n)=emptyset$, but all sets in the intersection have measure $1$.
$endgroup$
– Martin Argerami
Jan 24 at 5:02
$begingroup$
Because you need to take into account whether you are using the jump or not. If you define $|(0,1]|_F=0$ in my example above, you get that $|(1,1+delta)|_F=1$ for all $delta>0$. This contradicts continuity of the measure, since you have $bigcap_n(1,1+1/n)=emptyset$, but all sets in the intersection have measure $1$.
$endgroup$
– Martin Argerami
Jan 24 at 5:02
$begingroup$
This is how I am thinking about it. The Borel sigma algebra is generated by any of the four bounded intervals. Now, we have that $(a,b], [a,b), [a,b]$ can be written as countable intersections/unions of intervals of the form $(a,b)$. So, the choice isn't unique; once the definition of $F-$ volume of $(a,b)$ is pinned down, the choice for the $F-$ volume of the other intervals is also pinned down. But isn't the choice for $(a,b)$ then arbitrary. Why not $F_{+}(b) - F_{-}(a)$ instead of $F_{-}(b) - F_{+}(a)$?
$endgroup$
– user82261
Jan 24 at 5:35
$begingroup$
This is how I am thinking about it. The Borel sigma algebra is generated by any of the four bounded intervals. Now, we have that $(a,b], [a,b), [a,b]$ can be written as countable intersections/unions of intervals of the form $(a,b)$. So, the choice isn't unique; once the definition of $F-$ volume of $(a,b)$ is pinned down, the choice for the $F-$ volume of the other intervals is also pinned down. But isn't the choice for $(a,b)$ then arbitrary. Why not $F_{+}(b) - F_{-}(a)$ instead of $F_{-}(b) - F_{+}(a)$?
$endgroup$
– user82261
Jan 24 at 5:35
$begingroup$
Because you get something incoherent. That's why I gave you the example. Take my example and define $|(a,b)|=F_+(b)-F_-(a)$. Then $|(0,1)|=1-0=1$. But $|(0,1-1/n)|=0$ for all $n$, contradicting continuity of the measure: $(0,1)=bigcup_n(0,1-1/n)$, but all the sets of the right have measure zero and the one in the left has measure one.
$endgroup$
– Martin Argerami
Jan 24 at 14:11
$begingroup$
Because you get something incoherent. That's why I gave you the example. Take my example and define $|(a,b)|=F_+(b)-F_-(a)$. Then $|(0,1)|=1-0=1$. But $|(0,1-1/n)|=0$ for all $n$, contradicting continuity of the measure: $(0,1)=bigcup_n(0,1-1/n)$, but all the sets of the right have measure zero and the one in the left has measure one.
$endgroup$
– Martin Argerami
Jan 24 at 14:11
add a comment |
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