If two functions are equal almost everywhere, the first is continuous a.e., is the second?
$begingroup$
If $f = g$ a.e. in $E in mathfrak{M}$ (the Lebesgue measurable sets)
and $f$ is continuous a.e. in $E$, is $g$ continuous a.e. in $E$?
I think this is true.
My “proof”:
Let us denote $D_1 = { x in E: f(x) text{ discontinuous}}$,
$m(D_1) = 0$ and $D_2 = { x in E: f(x) neq g(x)}$, $m(D_2) = 0$.
Define $D_3 = { x in E: g(x) text{ discontinuous}}$.
If $f$ is identically $g$, then it is clear that the result follows as
$D_3 = D_1$.
Otherwise, we have $D_3 subseteq D_1 cup D_2$, and so $m^*(D_3) leq m^*(D_1 cup D_2) leq m^*(D_1) + m^*(D_2) = 0$.
So, $m(D_3) = 0$ and hence $g$ is continuous almost everywhere.
Does this proof work?
Thanks!
Edit: for clarity, $m$ denotes the Lebesgue measure and $m^*$ the Lebesgue outer measure.
real-analysis analysis measure-theory
asked Dec 12 '18 at 20:43
Jane DoeJane Doe
330113
$endgroup$
add a comment |
$begingroup$
If $f = g$ a.e. in $E in mathfrak{M}$ (the Lebesgue measurable sets)
and $f$ is continuous a.e. in $E$, is $g$ continuous a.e. in $E$?
I think this is true.
My “proof”:
Let us denote $D_1 = { x in E: f(x) text{ discontinuous}}$,
$m(D_1) = 0$ and $D_2 = { x in E: f(x) neq g(x)}$, $m(D_2) = 0$.
Define $D_3 = { x in E: g(x) text{ discontinuous}}$.
If $f$ is identically $g$, then it is clear that the result follows as
$D_3 = D_1$.
Otherwise, we have $D_3 subseteq D_1 cup D_2$, and so $m^*(D_3) leq m^*(D_1 cup D_2) leq m^*(D_1) + m^*(D_2) = 0$.
So, $m(D_3) = 0$ and hence $g$ is continuous almost everywhere.
Does this proof work?
Thanks!
Edit: for clarity, $m$ denotes the Lebesgue measure and $m^*$ the Lebesgue outer measure.
real-analysis analysis measure-theory
asked Dec 12 '18 at 20:43
Jane DoeJane Doe
330113
$endgroup$
2
$begingroup$
How do you justify $D_3 subseteq D_1 cup D_2$ ?
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:46
$begingroup$
@YvesDaoust My reasoning was that if $f(x) = g(x)$ on $Esetminus D_1$ then since $f(x)$ is continuous there, $g(x)$ ought to be as well?
$endgroup$
– Jane Doe
Dec 12 '18 at 20:47
$begingroup$
Continuity of $f$ in $Esetminus D_1$ does not imply continuity of $g$, because of $D_2$.
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:51
4
$begingroup$
Consider the case where $f$ is identically 0, and $g$ is the characteristic function of the rationals.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 20:52
$begingroup$
@Andrés E. Caicedo, you're absolutely right I was just about to fix my mistake with that example. That should be the answer.
$endgroup$
– James Yang
Dec 12 '18 at 20:54
add a comment |
$begingroup$
If $f = g$ a.e. in $E in mathfrak{M}$ (the Lebesgue measurable sets)
and $f$ is continuous a.e. in $E$, is $g$ continuous a.e. in $E$?
I think this is true.
My “proof”:
Let us denote $D_1 = { x in E: f(x) text{ discontinuous}}$,
$m(D_1) = 0$ and $D_2 = { x in E: f(x) neq g(x)}$, $m(D_2) = 0$.
Define $D_3 = { x in E: g(x) text{ discontinuous}}$.
If $f$ is identically $g$, then it is clear that the result follows as
$D_3 = D_1$.
Otherwise, we have $D_3 subseteq D_1 cup D_2$, and so $m^*(D_3) leq m^*(D_1 cup D_2) leq m^*(D_1) + m^*(D_2) = 0$.
So, $m(D_3) = 0$ and hence $g$ is continuous almost everywhere.
Does this proof work?
Thanks!
Edit: for clarity, $m$ denotes the Lebesgue measure and $m^*$ the Lebesgue outer measure.
real-analysis analysis measure-theory
asked Dec 12 '18 at 20:43
Jane DoeJane Doe
330113
$endgroup$
If $f = g$ a.e. in $E in mathfrak{M}$ (the Lebesgue measurable sets)
and $f$ is continuous a.e. in $E$, is $g$ continuous a.e. in $E$?
I think this is true.
My “proof”:
Let us denote $D_1 = { x in E: f(x) text{ discontinuous}}$,
$m(D_1) = 0$ and $D_2 = { x in E: f(x) neq g(x)}$, $m(D_2) = 0$.
Define $D_3 = { x in E: g(x) text{ discontinuous}}$.
If $f$ is identically $g$, then it is clear that the result follows as
$D_3 = D_1$.
Otherwise, we have $D_3 subseteq D_1 cup D_2$, and so $m^*(D_3) leq m^*(D_1 cup D_2) leq m^*(D_1) + m^*(D_2) = 0$.
So, $m(D_3) = 0$ and hence $g$ is continuous almost everywhere.
Does this proof work?
Thanks!
Edit: for clarity, $m$ denotes the Lebesgue measure and $m^*$ the Lebesgue outer measure.
real-analysis analysis measure-theory
real-analysis analysis measure-theory
asked Dec 12 '18 at 20:43
Jane DoeJane Doe
330113
asked Dec 12 '18 at 20:43
Jane DoeJane Doe
330113
asked Dec 12 '18 at 20:43
Jane DoeJane Doe
330113
asked Dec 12 '18 at 20:43
Jane DoeJane Doe
330113
asked Dec 12 '18 at 20:43
Jane DoeJane Doe
330113
330113
2
$begingroup$
How do you justify $D_3 subseteq D_1 cup D_2$ ?
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:46
$begingroup$
@YvesDaoust My reasoning was that if $f(x) = g(x)$ on $Esetminus D_1$ then since $f(x)$ is continuous there, $g(x)$ ought to be as well?
$endgroup$
– Jane Doe
Dec 12 '18 at 20:47
$begingroup$
Continuity of $f$ in $Esetminus D_1$ does not imply continuity of $g$, because of $D_2$.
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:51
4
$begingroup$
Consider the case where $f$ is identically 0, and $g$ is the characteristic function of the rationals.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 20:52
$begingroup$
@Andrés E. Caicedo, you're absolutely right I was just about to fix my mistake with that example. That should be the answer.
$endgroup$
– James Yang
Dec 12 '18 at 20:54
add a comment |
2
$begingroup$
How do you justify $D_3 subseteq D_1 cup D_2$ ?
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:46
$begingroup$
@YvesDaoust My reasoning was that if $f(x) = g(x)$ on $Esetminus D_1$ then since $f(x)$ is continuous there, $g(x)$ ought to be as well?
$endgroup$
– Jane Doe
Dec 12 '18 at 20:47
$begingroup$
Continuity of $f$ in $Esetminus D_1$ does not imply continuity of $g$, because of $D_2$.
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:51
4
$begingroup$
Consider the case where $f$ is identically 0, and $g$ is the characteristic function of the rationals.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 20:52
$begingroup$
@Andrés E. Caicedo, you're absolutely right I was just about to fix my mistake with that example. That should be the answer.
$endgroup$
– James Yang
Dec 12 '18 at 20:54
2
2
$begingroup$
How do you justify $D_3 subseteq D_1 cup D_2$ ?
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:46
$begingroup$
How do you justify $D_3 subseteq D_1 cup D_2$ ?
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:46
$begingroup$
@YvesDaoust My reasoning was that if $f(x) = g(x)$ on $Esetminus D_1$ then since $f(x)$ is continuous there, $g(x)$ ought to be as well?
$endgroup$
– Jane Doe
Dec 12 '18 at 20:47
$begingroup$
@YvesDaoust My reasoning was that if $f(x) = g(x)$ on $Esetminus D_1$ then since $f(x)$ is continuous there, $g(x)$ ought to be as well?
$endgroup$
– Jane Doe
Dec 12 '18 at 20:47
$begingroup$
Continuity of $f$ in $Esetminus D_1$ does not imply continuity of $g$, because of $D_2$.
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:51
$begingroup$
Continuity of $f$ in $Esetminus D_1$ does not imply continuity of $g$, because of $D_2$.
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:51
4
4
$begingroup$
Consider the case where $f$ is identically 0, and $g$ is the characteristic function of the rationals.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 20:52
$begingroup$
Consider the case where $f$ is identically 0, and $g$ is the characteristic function of the rationals.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 20:52
$begingroup$
@Andrés E. Caicedo, you're absolutely right I was just about to fix my mistake with that example. That should be the answer.
$endgroup$
– James Yang
Dec 12 '18 at 20:54
$begingroup$
@Andrés E. Caicedo, you're absolutely right I was just about to fix my mistake with that example. That should be the answer.
$endgroup$
– James Yang
Dec 12 '18 at 20:54
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The claim is not true. An easy counterexample is obtained by letting $f$ be identically 0 and $g$ be the characteristic function of the rationals. We have $f=g$ a.e., $f$ is everywhere continuous and $g$ is nowhere continuous.
(Clearly, the restriction of $g$ to the irrationals is continuous (being constant), but this is not enough to ensure that $g$ is continuous on the irrationals.)
answered Dec 12 '18 at 20:59
Andrés E. CaicedoAndrés E. Caicedo
65.4k8158247
$endgroup$
add a comment |
$begingroup$
$f$ continuous on $Esetminus D_1$ implies $f$ continuous on $Esetminus(D_1cup D_2)$. And of course, $g$ continuous on $Esetminus(D_1cup D_2)$. Then the measure of $D_3$ is at most that of $D_1cup D_2$.
answered Dec 12 '18 at 20:55
Yves DaoustYves Daoust
127k673226
$endgroup$
1
$begingroup$
Not quite. The restriction of $g$ to a set $F$ may be continuous without $g$ itself being continuous on $F$.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 20:57
1
$begingroup$
@AndrésE.Caicedo: what difference do you make between "restricted to $F$" and "on $F$" ??
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:59
$begingroup$
The domain of $g$ restricted to $F$ is $F$. The domain of $g$ may be larger, and points not in $F$ may affect whether $g$ is continuous at some point in $F$. See my answer for an example. There, with $F$ the irrationals, $g|F$ is continuous because it is constant, but $g$, defined everywhere, is not continuous at any point.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 21:02
add a comment |
$begingroup$
No your proof does not work because the fact that $f(x)=g(x)$ on $Esetminus D_1$ implies $g(x)$ is continuous from the continuity of $f(x)$ tells you that:
$D_2cap (E setminus D_1) subseteq (E setminus D_3)$
which means:
$D_3 subseteq (E setminus D_2) cup D_1$
answered Dec 12 '18 at 21:00
John11John11
1,0321821
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The claim is not true. An easy counterexample is obtained by letting $f$ be identically 0 and $g$ be the characteristic function of the rationals. We have $f=g$ a.e., $f$ is everywhere continuous and $g$ is nowhere continuous.
(Clearly, the restriction of $g$ to the irrationals is continuous (being constant), but this is not enough to ensure that $g$ is continuous on the irrationals.)
answered Dec 12 '18 at 20:59
Andrés E. CaicedoAndrés E. Caicedo
65.4k8158247
$endgroup$
add a comment |
$begingroup$
The claim is not true. An easy counterexample is obtained by letting $f$ be identically 0 and $g$ be the characteristic function of the rationals. We have $f=g$ a.e., $f$ is everywhere continuous and $g$ is nowhere continuous.
(Clearly, the restriction of $g$ to the irrationals is continuous (being constant), but this is not enough to ensure that $g$ is continuous on the irrationals.)
answered Dec 12 '18 at 20:59
Andrés E. CaicedoAndrés E. Caicedo
65.4k8158247
$endgroup$
add a comment |
$begingroup$
The claim is not true. An easy counterexample is obtained by letting $f$ be identically 0 and $g$ be the characteristic function of the rationals. We have $f=g$ a.e., $f$ is everywhere continuous and $g$ is nowhere continuous.
(Clearly, the restriction of $g$ to the irrationals is continuous (being constant), but this is not enough to ensure that $g$ is continuous on the irrationals.)
answered Dec 12 '18 at 20:59
Andrés E. CaicedoAndrés E. Caicedo
65.4k8158247
$endgroup$
The claim is not true. An easy counterexample is obtained by letting $f$ be identically 0 and $g$ be the characteristic function of the rationals. We have $f=g$ a.e., $f$ is everywhere continuous and $g$ is nowhere continuous.
(Clearly, the restriction of $g$ to the irrationals is continuous (being constant), but this is not enough to ensure that $g$ is continuous on the irrationals.)
answered Dec 12 '18 at 20:59
Andrés E. CaicedoAndrés E. Caicedo
65.4k8158247
answered Dec 12 '18 at 20:59
Andrés E. CaicedoAndrés E. Caicedo
65.4k8158247
answered Dec 12 '18 at 20:59
Andrés E. CaicedoAndrés E. Caicedo
65.4k8158247
answered Dec 12 '18 at 20:59
Andrés E. CaicedoAndrés E. Caicedo
65.4k8158247
65.4k8158247
add a comment |
add a comment |
$begingroup$
$f$ continuous on $Esetminus D_1$ implies $f$ continuous on $Esetminus(D_1cup D_2)$. And of course, $g$ continuous on $Esetminus(D_1cup D_2)$. Then the measure of $D_3$ is at most that of $D_1cup D_2$.
answered Dec 12 '18 at 20:55
Yves DaoustYves Daoust
127k673226
$endgroup$
1
$begingroup$
Not quite. The restriction of $g$ to a set $F$ may be continuous without $g$ itself being continuous on $F$.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 20:57
1
$begingroup$
@AndrésE.Caicedo: what difference do you make between "restricted to $F$" and "on $F$" ??
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:59
$begingroup$
The domain of $g$ restricted to $F$ is $F$. The domain of $g$ may be larger, and points not in $F$ may affect whether $g$ is continuous at some point in $F$. See my answer for an example. There, with $F$ the irrationals, $g|F$ is continuous because it is constant, but $g$, defined everywhere, is not continuous at any point.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 21:02
add a comment |
$begingroup$
$f$ continuous on $Esetminus D_1$ implies $f$ continuous on $Esetminus(D_1cup D_2)$. And of course, $g$ continuous on $Esetminus(D_1cup D_2)$. Then the measure of $D_3$ is at most that of $D_1cup D_2$.
answered Dec 12 '18 at 20:55
Yves DaoustYves Daoust
127k673226
$endgroup$
1
$begingroup$
Not quite. The restriction of $g$ to a set $F$ may be continuous without $g$ itself being continuous on $F$.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 20:57
1
$begingroup$
@AndrésE.Caicedo: what difference do you make between "restricted to $F$" and "on $F$" ??
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:59
$begingroup$
The domain of $g$ restricted to $F$ is $F$. The domain of $g$ may be larger, and points not in $F$ may affect whether $g$ is continuous at some point in $F$. See my answer for an example. There, with $F$ the irrationals, $g|F$ is continuous because it is constant, but $g$, defined everywhere, is not continuous at any point.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 21:02
add a comment |
$begingroup$
$f$ continuous on $Esetminus D_1$ implies $f$ continuous on $Esetminus(D_1cup D_2)$. And of course, $g$ continuous on $Esetminus(D_1cup D_2)$. Then the measure of $D_3$ is at most that of $D_1cup D_2$.
answered Dec 12 '18 at 20:55
Yves DaoustYves Daoust
127k673226
$endgroup$
$f$ continuous on $Esetminus D_1$ implies $f$ continuous on $Esetminus(D_1cup D_2)$. And of course, $g$ continuous on $Esetminus(D_1cup D_2)$. Then the measure of $D_3$ is at most that of $D_1cup D_2$.
answered Dec 12 '18 at 20:55
Yves DaoustYves Daoust
127k673226
answered Dec 12 '18 at 20:55
Yves DaoustYves Daoust
127k673226
answered Dec 12 '18 at 20:55
Yves DaoustYves Daoust
127k673226
answered Dec 12 '18 at 20:55
Yves DaoustYves Daoust
127k673226
127k673226
1
$begingroup$
Not quite. The restriction of $g$ to a set $F$ may be continuous without $g$ itself being continuous on $F$.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 20:57
1
$begingroup$
@AndrésE.Caicedo: what difference do you make between "restricted to $F$" and "on $F$" ??
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:59
$begingroup$
The domain of $g$ restricted to $F$ is $F$. The domain of $g$ may be larger, and points not in $F$ may affect whether $g$ is continuous at some point in $F$. See my answer for an example. There, with $F$ the irrationals, $g|F$ is continuous because it is constant, but $g$, defined everywhere, is not continuous at any point.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 21:02
add a comment |
1
$begingroup$
Not quite. The restriction of $g$ to a set $F$ may be continuous without $g$ itself being continuous on $F$.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 20:57
1
$begingroup$
@AndrésE.Caicedo: what difference do you make between "restricted to $F$" and "on $F$" ??
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:59
$begingroup$
The domain of $g$ restricted to $F$ is $F$. The domain of $g$ may be larger, and points not in $F$ may affect whether $g$ is continuous at some point in $F$. See my answer for an example. There, with $F$ the irrationals, $g|F$ is continuous because it is constant, but $g$, defined everywhere, is not continuous at any point.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 21:02
1
1
$begingroup$
Not quite. The restriction of $g$ to a set $F$ may be continuous without $g$ itself being continuous on $F$.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 20:57
$begingroup$
Not quite. The restriction of $g$ to a set $F$ may be continuous without $g$ itself being continuous on $F$.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 20:57
1
1
$begingroup$
@AndrésE.Caicedo: what difference do you make between "restricted to $F$" and "on $F$" ??
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:59
$begingroup$
@AndrésE.Caicedo: what difference do you make between "restricted to $F$" and "on $F$" ??
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:59
$begingroup$
The domain of $g$ restricted to $F$ is $F$. The domain of $g$ may be larger, and points not in $F$ may affect whether $g$ is continuous at some point in $F$. See my answer for an example. There, with $F$ the irrationals, $g|F$ is continuous because it is constant, but $g$, defined everywhere, is not continuous at any point.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 21:02
$begingroup$
The domain of $g$ restricted to $F$ is $F$. The domain of $g$ may be larger, and points not in $F$ may affect whether $g$ is continuous at some point in $F$. See my answer for an example. There, with $F$ the irrationals, $g|F$ is continuous because it is constant, but $g$, defined everywhere, is not continuous at any point.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 21:02
add a comment |
$begingroup$
No your proof does not work because the fact that $f(x)=g(x)$ on $Esetminus D_1$ implies $g(x)$ is continuous from the continuity of $f(x)$ tells you that:
$D_2cap (E setminus D_1) subseteq (E setminus D_3)$
which means:
$D_3 subseteq (E setminus D_2) cup D_1$
answered Dec 12 '18 at 21:00
John11John11
1,0321821
$endgroup$
add a comment |
$begingroup$
No your proof does not work because the fact that $f(x)=g(x)$ on $Esetminus D_1$ implies $g(x)$ is continuous from the continuity of $f(x)$ tells you that:
$D_2cap (E setminus D_1) subseteq (E setminus D_3)$
which means:
$D_3 subseteq (E setminus D_2) cup D_1$
answered Dec 12 '18 at 21:00
John11John11
1,0321821
$endgroup$
add a comment |
$begingroup$
No your proof does not work because the fact that $f(x)=g(x)$ on $Esetminus D_1$ implies $g(x)$ is continuous from the continuity of $f(x)$ tells you that:
$D_2cap (E setminus D_1) subseteq (E setminus D_3)$
which means:
$D_3 subseteq (E setminus D_2) cup D_1$
answered Dec 12 '18 at 21:00
John11John11
1,0321821
$endgroup$
No your proof does not work because the fact that $f(x)=g(x)$ on $Esetminus D_1$ implies $g(x)$ is continuous from the continuity of $f(x)$ tells you that:
$D_2cap (E setminus D_1) subseteq (E setminus D_3)$
which means:
$D_3 subseteq (E setminus D_2) cup D_1$
answered Dec 12 '18 at 21:00
John11John11
1,0321821
answered Dec 12 '18 at 21:00
John11John11
1,0321821
answered Dec 12 '18 at 21:00
John11John11
1,0321821
answered Dec 12 '18 at 21:00
John11John11
1,0321821
1,0321821
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2
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How do you justify $D_3 subseteq D_1 cup D_2$ ?
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– Yves Daoust
Dec 12 '18 at 20:46
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@YvesDaoust My reasoning was that if $f(x) = g(x)$ on $Esetminus D_1$ then since $f(x)$ is continuous there, $g(x)$ ought to be as well?
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– Jane Doe
Dec 12 '18 at 20:47
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Continuity of $f$ in $Esetminus D_1$ does not imply continuity of $g$, because of $D_2$.
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– Yves Daoust
Dec 12 '18 at 20:51
4
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Consider the case where $f$ is identically 0, and $g$ is the characteristic function of the rationals.
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– Andrés E. Caicedo
Dec 12 '18 at 20:52
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@Andrés E. Caicedo, you're absolutely right I was just about to fix my mistake with that example. That should be the answer.
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– James Yang
Dec 12 '18 at 20:54