Mixing
Does the following set have Lebesgue measure zero?
$begingroup$
Let $F:mathbb R^ntomathbb R$ is a non-constant continuous function. Is it true that $Leb[(x_1,...,x_n)inmathbb R^n:F(x_1,...,x_n)=0]=0?$ Here $Leb$ denotes Lebesgue measure.
I don't know if this is a well known result. I have heard something like graph of a continuous function has Lebesgue measure 0. Is this related to that? I don't even know how to prove this latter fact so if you can include a proof I would be delighted.
analysis measure-theory lebesgue-measure
asked Jan 23 at 17:03
Landon CarterLandon Carter
7,45311544
$endgroup$
add a comment |
$begingroup$
Let $F:mathbb R^ntomathbb R$ is a non-constant continuous function. Is it true that $Leb[(x_1,...,x_n)inmathbb R^n:F(x_1,...,x_n)=0]=0?$ Here $Leb$ denotes Lebesgue measure.
I don't know if this is a well known result. I have heard something like graph of a continuous function has Lebesgue measure 0. Is this related to that? I don't even know how to prove this latter fact so if you can include a proof I would be delighted.
analysis measure-theory lebesgue-measure
asked Jan 23 at 17:03
Landon CarterLandon Carter
7,45311544
$endgroup$
$begingroup$
It's true if $F$ is real analytic. Ekesh's example below shows that $C^infty$ is not enough.
$endgroup$
– Nate Eldredge
Jan 23 at 17:23
$begingroup$
Graph $Gamma(f) = { (x, f(x)) in mathbb{R}^{n+1} : x in mathbb{R}^n }$ of a measurable function $f : mathbb{R}^n to mathbb{R}$ has measure zero in $mathbb{R}^{n+1}$. But this has nothing to do with the measure of level-sets of $f$.
$endgroup$
– Sangchul Lee
Jan 23 at 18:51
add a comment |
$begingroup$
Let $F:mathbb R^ntomathbb R$ is a non-constant continuous function. Is it true that $Leb[(x_1,...,x_n)inmathbb R^n:F(x_1,...,x_n)=0]=0?$ Here $Leb$ denotes Lebesgue measure.
I don't know if this is a well known result. I have heard something like graph of a continuous function has Lebesgue measure 0. Is this related to that? I don't even know how to prove this latter fact so if you can include a proof I would be delighted.
analysis measure-theory lebesgue-measure
asked Jan 23 at 17:03
Landon CarterLandon Carter
7,45311544
$endgroup$
Let $F:mathbb R^ntomathbb R$ is a non-constant continuous function. Is it true that $Leb[(x_1,...,x_n)inmathbb R^n:F(x_1,...,x_n)=0]=0?$ Here $Leb$ denotes Lebesgue measure.
I don't know if this is a well known result. I have heard something like graph of a continuous function has Lebesgue measure 0. Is this related to that? I don't even know how to prove this latter fact so if you can include a proof I would be delighted.
analysis measure-theory lebesgue-measure
analysis measure-theory lebesgue-measure
asked Jan 23 at 17:03
Landon CarterLandon Carter
7,45311544
asked Jan 23 at 17:03
Landon CarterLandon Carter
7,45311544
asked Jan 23 at 17:03
Landon CarterLandon Carter
7,45311544
asked Jan 23 at 17:03
Landon CarterLandon Carter
7,45311544
asked Jan 23 at 17:03
Landon CarterLandon Carter
7,45311544
7,45311544
$begingroup$
It's true if $F$ is real analytic. Ekesh's example below shows that $C^infty$ is not enough.
$endgroup$
– Nate Eldredge
Jan 23 at 17:23
$begingroup$
Graph $Gamma(f) = { (x, f(x)) in mathbb{R}^{n+1} : x in mathbb{R}^n }$ of a measurable function $f : mathbb{R}^n to mathbb{R}$ has measure zero in $mathbb{R}^{n+1}$. But this has nothing to do with the measure of level-sets of $f$.
$endgroup$
– Sangchul Lee
Jan 23 at 18:51
add a comment |
$begingroup$
It's true if $F$ is real analytic. Ekesh's example below shows that $C^infty$ is not enough.
$endgroup$
– Nate Eldredge
Jan 23 at 17:23
$begingroup$
Graph $Gamma(f) = { (x, f(x)) in mathbb{R}^{n+1} : x in mathbb{R}^n }$ of a measurable function $f : mathbb{R}^n to mathbb{R}$ has measure zero in $mathbb{R}^{n+1}$. But this has nothing to do with the measure of level-sets of $f$.
$endgroup$
– Sangchul Lee
Jan 23 at 18:51
$begingroup$
It's true if $F$ is real analytic. Ekesh's example below shows that $C^infty$ is not enough.
$endgroup$
– Nate Eldredge
Jan 23 at 17:23
$begingroup$
It's true if $F$ is real analytic. Ekesh's example below shows that $C^infty$ is not enough.
$endgroup$
– Nate Eldredge
Jan 23 at 17:23
$begingroup$
Graph $Gamma(f) = { (x, f(x)) in mathbb{R}^{n+1} : x in mathbb{R}^n }$ of a measurable function $f : mathbb{R}^n to mathbb{R}$ has measure zero in $mathbb{R}^{n+1}$. But this has nothing to do with the measure of level-sets of $f$.
$endgroup$
– Sangchul Lee
Jan 23 at 18:51
$begingroup$
Graph $Gamma(f) = { (x, f(x)) in mathbb{R}^{n+1} : x in mathbb{R}^n }$ of a measurable function $f : mathbb{R}^n to mathbb{R}$ has measure zero in $mathbb{R}^{n+1}$. But this has nothing to do with the measure of level-sets of $f$.
$endgroup$
– Sangchul Lee
Jan 23 at 18:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The statement is false.
A counterexample is $$f(x) = begin{cases}
0 & text{ if } x leq 0 \
e^{-1/x^{2}} & text { if } x > 0.
end{cases} $$
answered Jan 23 at 17:14
Ekesh KumarEkesh Kumar
1,04928
$endgroup$
$begingroup$
This example is even $C^infty$. A simpler example could just be $f(x) = x$ when $x > 0$ (and still $0$ when $x le 0$).
$endgroup$
– Nate Eldredge
Jan 23 at 17:21
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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votes
$begingroup$
The statement is false.
A counterexample is $$f(x) = begin{cases}
0 & text{ if } x leq 0 \
e^{-1/x^{2}} & text { if } x > 0.
end{cases} $$
answered Jan 23 at 17:14
Ekesh KumarEkesh Kumar
1,04928
$endgroup$
$begingroup$
This example is even $C^infty$. A simpler example could just be $f(x) = x$ when $x > 0$ (and still $0$ when $x le 0$).
$endgroup$
– Nate Eldredge
Jan 23 at 17:21
add a comment |
$begingroup$
The statement is false.
A counterexample is $$f(x) = begin{cases}
0 & text{ if } x leq 0 \
e^{-1/x^{2}} & text { if } x > 0.
end{cases} $$
answered Jan 23 at 17:14
Ekesh KumarEkesh Kumar
1,04928
$endgroup$
$begingroup$
This example is even $C^infty$. A simpler example could just be $f(x) = x$ when $x > 0$ (and still $0$ when $x le 0$).
$endgroup$
– Nate Eldredge
Jan 23 at 17:21
add a comment |
$begingroup$
The statement is false.
A counterexample is $$f(x) = begin{cases}
0 & text{ if } x leq 0 \
e^{-1/x^{2}} & text { if } x > 0.
end{cases} $$
answered Jan 23 at 17:14
Ekesh KumarEkesh Kumar
1,04928
$endgroup$
The statement is false.
A counterexample is $$f(x) = begin{cases}
0 & text{ if } x leq 0 \
e^{-1/x^{2}} & text { if } x > 0.
end{cases} $$
answered Jan 23 at 17:14
Ekesh KumarEkesh Kumar
1,04928
answered Jan 23 at 17:14
Ekesh KumarEkesh Kumar
1,04928
answered Jan 23 at 17:14
Ekesh KumarEkesh Kumar
1,04928
answered Jan 23 at 17:14
Ekesh KumarEkesh Kumar
1,04928
1,04928
$begingroup$
This example is even $C^infty$. A simpler example could just be $f(x) = x$ when $x > 0$ (and still $0$ when $x le 0$).
$endgroup$
– Nate Eldredge
Jan 23 at 17:21
add a comment |
$begingroup$
This example is even $C^infty$. A simpler example could just be $f(x) = x$ when $x > 0$ (and still $0$ when $x le 0$).
$endgroup$
– Nate Eldredge
Jan 23 at 17:21
$begingroup$
This example is even $C^infty$. A simpler example could just be $f(x) = x$ when $x > 0$ (and still $0$ when $x le 0$).
$endgroup$
– Nate Eldredge
Jan 23 at 17:21
$begingroup$
This example is even $C^infty$. A simpler example could just be $f(x) = x$ when $x > 0$ (and still $0$ when $x le 0$).
$endgroup$
– Nate Eldredge
Jan 23 at 17:21
add a comment |
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$begingroup$
It's true if $F$ is real analytic. Ekesh's example below shows that $C^infty$ is not enough.
$endgroup$
– Nate Eldredge
Jan 23 at 17:23
$begingroup$
Graph $Gamma(f) = { (x, f(x)) in mathbb{R}^{n+1} : x in mathbb{R}^n }$ of a measurable function $f : mathbb{R}^n to mathbb{R}$ has measure zero in $mathbb{R}^{n+1}$. But this has nothing to do with the measure of level-sets of $f$.
$endgroup$
– Sangchul Lee
Jan 23 at 18:51