Mixing
Measure zero implies volume zero when volume is defined
$begingroup$
I need to prove that if a set has measure zero and it has well-defined volume, then its volume is zero. I have tried to bound the lower sums of the indicator function, but I'm stuck. Given a zero-measure set $A$ which volume is well-defined, if $R$ is a rectangle that contains $A$, we know there exists a partition $P_0$ of $R$ such that:
$$ int_{-} 1_A leq L(1_A, P_0) + frac{epsilon}2 = sum_{Qin P_0| Qsubset A}v(Q)quad+frac{epsilon}2$$
I don't really know how to proceed without assuming things like denumerable subaditivity or monotony of volume function $v$. If you use some of these things to prove this please tell me also how to prove them.
EDIT: when I say "volume of the set $A$" I'm refering to the Riemann integral of the indicator function $1_A$ along $A$: $v(A) = int_A 1_A$
EDIT2: when I say that a set has measure zero I mean that for all $epsilon>0$ there exists a denumerable or finite collection of rectangles $R_1,R_2,...$ such that $Asubsetcup R_i$ and $sum v(A_i)<epsilon$
integration
asked Feb 3 at 10:58
SevenSeven
1249
$endgroup$
|
show 3 more comments
$begingroup$
I need to prove that if a set has measure zero and it has well-defined volume, then its volume is zero. I have tried to bound the lower sums of the indicator function, but I'm stuck. Given a zero-measure set $A$ which volume is well-defined, if $R$ is a rectangle that contains $A$, we know there exists a partition $P_0$ of $R$ such that:
$$ int_{-} 1_A leq L(1_A, P_0) + frac{epsilon}2 = sum_{Qin P_0| Qsubset A}v(Q)quad+frac{epsilon}2$$
I don't really know how to proceed without assuming things like denumerable subaditivity or monotony of volume function $v$. If you use some of these things to prove this please tell me also how to prove them.
EDIT: when I say "volume of the set $A$" I'm refering to the Riemann integral of the indicator function $1_A$ along $A$: $v(A) = int_A 1_A$
EDIT2: when I say that a set has measure zero I mean that for all $epsilon>0$ there exists a denumerable or finite collection of rectangles $R_1,R_2,...$ such that $Asubsetcup R_i$ and $sum v(A_i)<epsilon$
integration
asked Feb 3 at 10:58
SevenSeven
1249
$endgroup$
$begingroup$
What's your definition of volume, if it is not measure?
$endgroup$
– Calvin Khor
Feb 3 at 11:01
$begingroup$
sorry I thought it was standard. I'll edit
$endgroup$
– Seven
Feb 3 at 11:03
$begingroup$
Oh, are you coming at this from the perspective of Riemann integration, rather than Lebesgue integration?
$endgroup$
– Calvin Khor
Feb 3 at 11:05
$begingroup$
yes, my university does it this way. Maybe other approach would be better, but it's not my choice.
$endgroup$
– Seven
Feb 3 at 11:09
1
$begingroup$
I believe this works, math.stackexchange.com/questions/2538567/…
$endgroup$
– Calvin Khor
Feb 3 at 11:22
|
show 3 more comments
$begingroup$
I need to prove that if a set has measure zero and it has well-defined volume, then its volume is zero. I have tried to bound the lower sums of the indicator function, but I'm stuck. Given a zero-measure set $A$ which volume is well-defined, if $R$ is a rectangle that contains $A$, we know there exists a partition $P_0$ of $R$ such that:
$$ int_{-} 1_A leq L(1_A, P_0) + frac{epsilon}2 = sum_{Qin P_0| Qsubset A}v(Q)quad+frac{epsilon}2$$
I don't really know how to proceed without assuming things like denumerable subaditivity or monotony of volume function $v$. If you use some of these things to prove this please tell me also how to prove them.
EDIT: when I say "volume of the set $A$" I'm refering to the Riemann integral of the indicator function $1_A$ along $A$: $v(A) = int_A 1_A$
EDIT2: when I say that a set has measure zero I mean that for all $epsilon>0$ there exists a denumerable or finite collection of rectangles $R_1,R_2,...$ such that $Asubsetcup R_i$ and $sum v(A_i)<epsilon$
integration
asked Feb 3 at 10:58
SevenSeven
1249
$endgroup$
I need to prove that if a set has measure zero and it has well-defined volume, then its volume is zero. I have tried to bound the lower sums of the indicator function, but I'm stuck. Given a zero-measure set $A$ which volume is well-defined, if $R$ is a rectangle that contains $A$, we know there exists a partition $P_0$ of $R$ such that:
$$ int_{-} 1_A leq L(1_A, P_0) + frac{epsilon}2 = sum_{Qin P_0| Qsubset A}v(Q)quad+frac{epsilon}2$$
I don't really know how to proceed without assuming things like denumerable subaditivity or monotony of volume function $v$. If you use some of these things to prove this please tell me also how to prove them.
EDIT: when I say "volume of the set $A$" I'm refering to the Riemann integral of the indicator function $1_A$ along $A$: $v(A) = int_A 1_A$
EDIT2: when I say that a set has measure zero I mean that for all $epsilon>0$ there exists a denumerable or finite collection of rectangles $R_1,R_2,...$ such that $Asubsetcup R_i$ and $sum v(A_i)<epsilon$
integration
integration
asked Feb 3 at 10:58
SevenSeven
1249
asked Feb 3 at 10:58
SevenSeven
1249
edited Feb 3 at 11:41
Seven
asked Feb 3 at 10:58
SevenSeven
1249
asked Feb 3 at 10:58
SevenSeven
1249
asked Feb 3 at 10:58
SevenSeven
1249
1249
$begingroup$
What's your definition of volume, if it is not measure?
$endgroup$
– Calvin Khor
Feb 3 at 11:01
$begingroup$
sorry I thought it was standard. I'll edit
$endgroup$
– Seven
Feb 3 at 11:03
$begingroup$
Oh, are you coming at this from the perspective of Riemann integration, rather than Lebesgue integration?
$endgroup$
– Calvin Khor
Feb 3 at 11:05
$begingroup$
yes, my university does it this way. Maybe other approach would be better, but it's not my choice.
$endgroup$
– Seven
Feb 3 at 11:09
1
$begingroup$
I believe this works, math.stackexchange.com/questions/2538567/…
$endgroup$
– Calvin Khor
Feb 3 at 11:22
|
show 3 more comments
$begingroup$
What's your definition of volume, if it is not measure?
$endgroup$
– Calvin Khor
Feb 3 at 11:01
$begingroup$
sorry I thought it was standard. I'll edit
$endgroup$
– Seven
Feb 3 at 11:03
$begingroup$
Oh, are you coming at this from the perspective of Riemann integration, rather than Lebesgue integration?
$endgroup$
– Calvin Khor
Feb 3 at 11:05
$begingroup$
yes, my university does it this way. Maybe other approach would be better, but it's not my choice.
$endgroup$
– Seven
Feb 3 at 11:09
1
$begingroup$
I believe this works, math.stackexchange.com/questions/2538567/…
$endgroup$
– Calvin Khor
Feb 3 at 11:22
$begingroup$
What's your definition of volume, if it is not measure?
$endgroup$
– Calvin Khor
Feb 3 at 11:01
$begingroup$
What's your definition of volume, if it is not measure?
$endgroup$
– Calvin Khor
Feb 3 at 11:01
$begingroup$
sorry I thought it was standard. I'll edit
$endgroup$
– Seven
Feb 3 at 11:03
$begingroup$
sorry I thought it was standard. I'll edit
$endgroup$
– Seven
Feb 3 at 11:03
$begingroup$
Oh, are you coming at this from the perspective of Riemann integration, rather than Lebesgue integration?
$endgroup$
– Calvin Khor
Feb 3 at 11:05
$begingroup$
Oh, are you coming at this from the perspective of Riemann integration, rather than Lebesgue integration?
$endgroup$
– Calvin Khor
Feb 3 at 11:05
$begingroup$
yes, my university does it this way. Maybe other approach would be better, but it's not my choice.
$endgroup$
– Seven
Feb 3 at 11:09
$begingroup$
yes, my university does it this way. Maybe other approach would be better, but it's not my choice.
$endgroup$
– Seven
Feb 3 at 11:09
1
1
$begingroup$
I believe this works, math.stackexchange.com/questions/2538567/…
$endgroup$
– Calvin Khor
Feb 3 at 11:22
$begingroup$
I believe this works, math.stackexchange.com/questions/2538567/…
$endgroup$
– Calvin Khor
Feb 3 at 11:22
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The indicator function of $A$ is Riemann integrable iff $A$ is Jordan measurable, so the result linked in the comments applies. A sketch-
A (bounded) set $A$ here is defined to have volume $v(A)$ if for any bounded rectangle $R$ containing $A$, $1_A$ is Riemann integrable, and
$ v(A) = int_R 1_A$.
From Riemann integrability and the measure of the boundary , $v(A)$ is well defined iff $partial A$ has (Lebesgue) measure $m(partial A)= 0$. (We only need the direction $1_A$ integrable implies $m(partial A)=0$, whose proof is the first one there.)
From Closure, Interior, and Boundary of Jordan Measurable Sets. , we see that $m(partial A) = 0$ iff $A$ is Jordan measurable. However this page is a mess so I'll give a direct proof (sketch) of the important direction-
First note that $partial A$ is a compact set; thus by passing to finite covers, $m(partial A)=0$ implies its jordan content is $c(partial A) = 0$. Now it should not be hard to find finite collections of rectangles $L_i,U_i$ such that $L=bigcup_i L_i subseteq A subseteq U = bigcup_i U_i$ and $c(Usetminus L)$ is arbitrarily small. Thus $A$ is Jordan measurable.
Since for Jordan measurable sets, $c(A) = m(A)$, we now know that $c(A)=0$. Now we can use the result here, Prove Riemann Integral is Zero.
answered Feb 3 at 13:51
Calvin KhorCalvin Khor
12.5k21439
$endgroup$
$begingroup$
wow, thank you. However, I hope to find a more direct way. I shouldn't need to use Jordan measure.
$endgroup$
– Seven
Feb 3 at 14:05
$begingroup$
@Seven The main difficulty in your question is the nebulous way in which "volume is defined". It turns out by the above characterisation to not be nebulous, and also not a generalisation of the linked question. If one wanted a more general result, you can certainly use Lebesgue's measure theory to define $int 1_A$ /as/ the volume of $A$ for any Lebesgue measurable set, which defines the question away. That said, there's probably a more direct route, I just used the above because my Riemann theory is rusty :) so I relied on quoting results with proofs, which needed to be pieced together.
$endgroup$
– Calvin Khor
Feb 4 at 10:49
add a comment |
$begingroup$
Well, I had to solve it for myself.Maybe Calvin Khor's answer was right, but it wasn't was I was looking for. Continuing from where I stopped at the question, we notice that every rectangle $Qsubset A$ , as it is a subset of a zero-measure set , is also a zero-measure set. As a rectangle, it is compact, and for compact sets zero measure is equivalent to zero volume. Then: $$int_-1_A = sum_{Qin P_0| Qsubset A} v(Q) = 0 $$ As $A$ has well-defined volume, it is neccessarily $v(A)=0$
answered Feb 9 at 14:01
SevenSeven
1249
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The indicator function of $A$ is Riemann integrable iff $A$ is Jordan measurable, so the result linked in the comments applies. A sketch-
A (bounded) set $A$ here is defined to have volume $v(A)$ if for any bounded rectangle $R$ containing $A$, $1_A$ is Riemann integrable, and
$ v(A) = int_R 1_A$.
From Riemann integrability and the measure of the boundary , $v(A)$ is well defined iff $partial A$ has (Lebesgue) measure $m(partial A)= 0$. (We only need the direction $1_A$ integrable implies $m(partial A)=0$, whose proof is the first one there.)
From Closure, Interior, and Boundary of Jordan Measurable Sets. , we see that $m(partial A) = 0$ iff $A$ is Jordan measurable. However this page is a mess so I'll give a direct proof (sketch) of the important direction-
First note that $partial A$ is a compact set; thus by passing to finite covers, $m(partial A)=0$ implies its jordan content is $c(partial A) = 0$. Now it should not be hard to find finite collections of rectangles $L_i,U_i$ such that $L=bigcup_i L_i subseteq A subseteq U = bigcup_i U_i$ and $c(Usetminus L)$ is arbitrarily small. Thus $A$ is Jordan measurable.
Since for Jordan measurable sets, $c(A) = m(A)$, we now know that $c(A)=0$. Now we can use the result here, Prove Riemann Integral is Zero.
answered Feb 3 at 13:51
Calvin KhorCalvin Khor
12.5k21439
$endgroup$
$begingroup$
wow, thank you. However, I hope to find a more direct way. I shouldn't need to use Jordan measure.
$endgroup$
– Seven
Feb 3 at 14:05
$begingroup$
@Seven The main difficulty in your question is the nebulous way in which "volume is defined". It turns out by the above characterisation to not be nebulous, and also not a generalisation of the linked question. If one wanted a more general result, you can certainly use Lebesgue's measure theory to define $int 1_A$ /as/ the volume of $A$ for any Lebesgue measurable set, which defines the question away. That said, there's probably a more direct route, I just used the above because my Riemann theory is rusty :) so I relied on quoting results with proofs, which needed to be pieced together.
$endgroup$
– Calvin Khor
Feb 4 at 10:49
add a comment |
$begingroup$
The indicator function of $A$ is Riemann integrable iff $A$ is Jordan measurable, so the result linked in the comments applies. A sketch-
A (bounded) set $A$ here is defined to have volume $v(A)$ if for any bounded rectangle $R$ containing $A$, $1_A$ is Riemann integrable, and
$ v(A) = int_R 1_A$.
From Riemann integrability and the measure of the boundary , $v(A)$ is well defined iff $partial A$ has (Lebesgue) measure $m(partial A)= 0$. (We only need the direction $1_A$ integrable implies $m(partial A)=0$, whose proof is the first one there.)
From Closure, Interior, and Boundary of Jordan Measurable Sets. , we see that $m(partial A) = 0$ iff $A$ is Jordan measurable. However this page is a mess so I'll give a direct proof (sketch) of the important direction-
First note that $partial A$ is a compact set; thus by passing to finite covers, $m(partial A)=0$ implies its jordan content is $c(partial A) = 0$. Now it should not be hard to find finite collections of rectangles $L_i,U_i$ such that $L=bigcup_i L_i subseteq A subseteq U = bigcup_i U_i$ and $c(Usetminus L)$ is arbitrarily small. Thus $A$ is Jordan measurable.
Since for Jordan measurable sets, $c(A) = m(A)$, we now know that $c(A)=0$. Now we can use the result here, Prove Riemann Integral is Zero.
answered Feb 3 at 13:51
Calvin KhorCalvin Khor
12.5k21439
$endgroup$
$begingroup$
wow, thank you. However, I hope to find a more direct way. I shouldn't need to use Jordan measure.
$endgroup$
– Seven
Feb 3 at 14:05
$begingroup$
@Seven The main difficulty in your question is the nebulous way in which "volume is defined". It turns out by the above characterisation to not be nebulous, and also not a generalisation of the linked question. If one wanted a more general result, you can certainly use Lebesgue's measure theory to define $int 1_A$ /as/ the volume of $A$ for any Lebesgue measurable set, which defines the question away. That said, there's probably a more direct route, I just used the above because my Riemann theory is rusty :) so I relied on quoting results with proofs, which needed to be pieced together.
$endgroup$
– Calvin Khor
Feb 4 at 10:49
add a comment |
$begingroup$
The indicator function of $A$ is Riemann integrable iff $A$ is Jordan measurable, so the result linked in the comments applies. A sketch-
A (bounded) set $A$ here is defined to have volume $v(A)$ if for any bounded rectangle $R$ containing $A$, $1_A$ is Riemann integrable, and
$ v(A) = int_R 1_A$.
From Riemann integrability and the measure of the boundary , $v(A)$ is well defined iff $partial A$ has (Lebesgue) measure $m(partial A)= 0$. (We only need the direction $1_A$ integrable implies $m(partial A)=0$, whose proof is the first one there.)
From Closure, Interior, and Boundary of Jordan Measurable Sets. , we see that $m(partial A) = 0$ iff $A$ is Jordan measurable. However this page is a mess so I'll give a direct proof (sketch) of the important direction-
First note that $partial A$ is a compact set; thus by passing to finite covers, $m(partial A)=0$ implies its jordan content is $c(partial A) = 0$. Now it should not be hard to find finite collections of rectangles $L_i,U_i$ such that $L=bigcup_i L_i subseteq A subseteq U = bigcup_i U_i$ and $c(Usetminus L)$ is arbitrarily small. Thus $A$ is Jordan measurable.
Since for Jordan measurable sets, $c(A) = m(A)$, we now know that $c(A)=0$. Now we can use the result here, Prove Riemann Integral is Zero.
answered Feb 3 at 13:51
Calvin KhorCalvin Khor
12.5k21439
$endgroup$
The indicator function of $A$ is Riemann integrable iff $A$ is Jordan measurable, so the result linked in the comments applies. A sketch-
A (bounded) set $A$ here is defined to have volume $v(A)$ if for any bounded rectangle $R$ containing $A$, $1_A$ is Riemann integrable, and
$ v(A) = int_R 1_A$.
From Riemann integrability and the measure of the boundary , $v(A)$ is well defined iff $partial A$ has (Lebesgue) measure $m(partial A)= 0$. (We only need the direction $1_A$ integrable implies $m(partial A)=0$, whose proof is the first one there.)
From Closure, Interior, and Boundary of Jordan Measurable Sets. , we see that $m(partial A) = 0$ iff $A$ is Jordan measurable. However this page is a mess so I'll give a direct proof (sketch) of the important direction-
First note that $partial A$ is a compact set; thus by passing to finite covers, $m(partial A)=0$ implies its jordan content is $c(partial A) = 0$. Now it should not be hard to find finite collections of rectangles $L_i,U_i$ such that $L=bigcup_i L_i subseteq A subseteq U = bigcup_i U_i$ and $c(Usetminus L)$ is arbitrarily small. Thus $A$ is Jordan measurable.
Since for Jordan measurable sets, $c(A) = m(A)$, we now know that $c(A)=0$. Now we can use the result here, Prove Riemann Integral is Zero.
answered Feb 3 at 13:51
Calvin KhorCalvin Khor
12.5k21439
answered Feb 3 at 13:51
Calvin KhorCalvin Khor
12.5k21439
answered Feb 3 at 13:51
Calvin KhorCalvin Khor
12.5k21439
answered Feb 3 at 13:51
Calvin KhorCalvin Khor
12.5k21439
12.5k21439
$begingroup$
wow, thank you. However, I hope to find a more direct way. I shouldn't need to use Jordan measure.
$endgroup$
– Seven
Feb 3 at 14:05
$begingroup$
@Seven The main difficulty in your question is the nebulous way in which "volume is defined". It turns out by the above characterisation to not be nebulous, and also not a generalisation of the linked question. If one wanted a more general result, you can certainly use Lebesgue's measure theory to define $int 1_A$ /as/ the volume of $A$ for any Lebesgue measurable set, which defines the question away. That said, there's probably a more direct route, I just used the above because my Riemann theory is rusty :) so I relied on quoting results with proofs, which needed to be pieced together.
$endgroup$
– Calvin Khor
Feb 4 at 10:49
add a comment |
$begingroup$
wow, thank you. However, I hope to find a more direct way. I shouldn't need to use Jordan measure.
$endgroup$
– Seven
Feb 3 at 14:05
$begingroup$
@Seven The main difficulty in your question is the nebulous way in which "volume is defined". It turns out by the above characterisation to not be nebulous, and also not a generalisation of the linked question. If one wanted a more general result, you can certainly use Lebesgue's measure theory to define $int 1_A$ /as/ the volume of $A$ for any Lebesgue measurable set, which defines the question away. That said, there's probably a more direct route, I just used the above because my Riemann theory is rusty :) so I relied on quoting results with proofs, which needed to be pieced together.
$endgroup$
– Calvin Khor
Feb 4 at 10:49
$begingroup$
wow, thank you. However, I hope to find a more direct way. I shouldn't need to use Jordan measure.
$endgroup$
– Seven
Feb 3 at 14:05
$begingroup$
wow, thank you. However, I hope to find a more direct way. I shouldn't need to use Jordan measure.
$endgroup$
– Seven
Feb 3 at 14:05
$begingroup$
@Seven The main difficulty in your question is the nebulous way in which "volume is defined". It turns out by the above characterisation to not be nebulous, and also not a generalisation of the linked question. If one wanted a more general result, you can certainly use Lebesgue's measure theory to define $int 1_A$ /as/ the volume of $A$ for any Lebesgue measurable set, which defines the question away. That said, there's probably a more direct route, I just used the above because my Riemann theory is rusty :) so I relied on quoting results with proofs, which needed to be pieced together.
$endgroup$
– Calvin Khor
Feb 4 at 10:49
$begingroup$
@Seven The main difficulty in your question is the nebulous way in which "volume is defined". It turns out by the above characterisation to not be nebulous, and also not a generalisation of the linked question. If one wanted a more general result, you can certainly use Lebesgue's measure theory to define $int 1_A$ /as/ the volume of $A$ for any Lebesgue measurable set, which defines the question away. That said, there's probably a more direct route, I just used the above because my Riemann theory is rusty :) so I relied on quoting results with proofs, which needed to be pieced together.
$endgroup$
– Calvin Khor
Feb 4 at 10:49
add a comment |
$begingroup$
Well, I had to solve it for myself.Maybe Calvin Khor's answer was right, but it wasn't was I was looking for. Continuing from where I stopped at the question, we notice that every rectangle $Qsubset A$ , as it is a subset of a zero-measure set , is also a zero-measure set. As a rectangle, it is compact, and for compact sets zero measure is equivalent to zero volume. Then: $$int_-1_A = sum_{Qin P_0| Qsubset A} v(Q) = 0 $$ As $A$ has well-defined volume, it is neccessarily $v(A)=0$
answered Feb 9 at 14:01
SevenSeven
1249
$endgroup$
add a comment |
$begingroup$
Well, I had to solve it for myself.Maybe Calvin Khor's answer was right, but it wasn't was I was looking for. Continuing from where I stopped at the question, we notice that every rectangle $Qsubset A$ , as it is a subset of a zero-measure set , is also a zero-measure set. As a rectangle, it is compact, and for compact sets zero measure is equivalent to zero volume. Then: $$int_-1_A = sum_{Qin P_0| Qsubset A} v(Q) = 0 $$ As $A$ has well-defined volume, it is neccessarily $v(A)=0$
answered Feb 9 at 14:01
SevenSeven
1249
$endgroup$
add a comment |
$begingroup$
Well, I had to solve it for myself.Maybe Calvin Khor's answer was right, but it wasn't was I was looking for. Continuing from where I stopped at the question, we notice that every rectangle $Qsubset A$ , as it is a subset of a zero-measure set , is also a zero-measure set. As a rectangle, it is compact, and for compact sets zero measure is equivalent to zero volume. Then: $$int_-1_A = sum_{Qin P_0| Qsubset A} v(Q) = 0 $$ As $A$ has well-defined volume, it is neccessarily $v(A)=0$
answered Feb 9 at 14:01
SevenSeven
1249
$endgroup$
Well, I had to solve it for myself.Maybe Calvin Khor's answer was right, but it wasn't was I was looking for. Continuing from where I stopped at the question, we notice that every rectangle $Qsubset A$ , as it is a subset of a zero-measure set , is also a zero-measure set. As a rectangle, it is compact, and for compact sets zero measure is equivalent to zero volume. Then: $$int_-1_A = sum_{Qin P_0| Qsubset A} v(Q) = 0 $$ As $A$ has well-defined volume, it is neccessarily $v(A)=0$
answered Feb 9 at 14:01
SevenSeven
1249
answered Feb 9 at 14:01
SevenSeven
1249
answered Feb 9 at 14:01
SevenSeven
1249
answered Feb 9 at 14:01
SevenSeven
1249
1249
add a comment |
add a comment |
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$begingroup$
What's your definition of volume, if it is not measure?
$endgroup$
– Calvin Khor
Feb 3 at 11:01
$begingroup$
sorry I thought it was standard. I'll edit
$endgroup$
– Seven
Feb 3 at 11:03
$begingroup$
Oh, are you coming at this from the perspective of Riemann integration, rather than Lebesgue integration?
$endgroup$
– Calvin Khor
Feb 3 at 11:05
$begingroup$
yes, my university does it this way. Maybe other approach would be better, but it's not my choice.
$endgroup$
– Seven
Feb 3 at 11:09
1
$begingroup$
I believe this works, math.stackexchange.com/questions/2538567/…
$endgroup$
– Calvin Khor
Feb 3 at 11:22