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Let $f$ be defined on a measurable set $E subset mathbb R^n$. If $a<f<+infty$ and $f=-infty$ are measurable, then $f$ is measurable
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Let $f$ be defined on a measurable set $E subset mathbb R^n$. How to show that if $a<f<+infty$ and $f=-infty$ are measurable for every finite $a$, then $f$ is measurable? I think I need to separate the set $a<f<+infty$, then do some set operations. Any hint?
EDIT: This question is part of a corollary.
The definition of measurable function is that $f$ is called a Lebesgue measurable function on $E$, or simply a measurable function, if for every finite $a$, the set $$mathbfxin E: f(mathbfx)>a $$ is a measurable subset of $mathbb R^n$.
measure-theory
asked Jul 14 at 15:45
user398843
394215
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up vote
2
down vote
favorite
Let $f$ be defined on a measurable set $E subset mathbb R^n$. How to show that if $a<f<+infty$ and $f=-infty$ are measurable for every finite $a$, then $f$ is measurable? I think I need to separate the set $a<f<+infty$, then do some set operations. Any hint?
EDIT: This question is part of a corollary.
The definition of measurable function is that $f$ is called a Lebesgue measurable function on $E$, or simply a measurable function, if for every finite $a$, the set $$mathbfxin E: f(mathbfx)>a $$ is a measurable subset of $mathbb R^n$.
measure-theory
asked Jul 14 at 15:45
user398843
394215
Isn't it $f=+infty $ measurable instead of $f=-infty $ ? Moreover, what is your definition of measurable function ? $-infty leq f <a$ measurable for all $a$ ?
– Surb
Jul 14 at 15:53
@Surb I have posted a corollary and a definition of measurable function.
– user398843
Jul 14 at 16:23
@Yanko Does it really duplicate the other post?
– user398843
Jul 14 at 16:51
Oh I see, the main difference is the infinity part. I'll see if I can remove that close vote.
– Yanko
Jul 14 at 17:55
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $f$ be defined on a measurable set $E subset mathbb R^n$. How to show that if $a<f<+infty$ and $f=-infty$ are measurable for every finite $a$, then $f$ is measurable? I think I need to separate the set $a<f<+infty$, then do some set operations. Any hint?
EDIT: This question is part of a corollary.
The definition of measurable function is that $f$ is called a Lebesgue measurable function on $E$, or simply a measurable function, if for every finite $a$, the set $$mathbfxin E: f(mathbfx)>a $$ is a measurable subset of $mathbb R^n$.
measure-theory
asked Jul 14 at 15:45
user398843
394215
Let $f$ be defined on a measurable set $E subset mathbb R^n$. How to show that if $a<f<+infty$ and $f=-infty$ are measurable for every finite $a$, then $f$ is measurable? I think I need to separate the set $a<f<+infty$, then do some set operations. Any hint?
EDIT: This question is part of a corollary.
The definition of measurable function is that $f$ is called a Lebesgue measurable function on $E$, or simply a measurable function, if for every finite $a$, the set $$mathbfxin E: f(mathbfx)>a $$ is a measurable subset of $mathbb R^n$.
measure-theory
asked Jul 14 at 15:45
user398843
394215
edited Jul 14 at 16:22
asked Jul 14 at 15:45
user398843
394215
asked Jul 14 at 15:45
user398843
394215
asked Jul 14 at 15:45
user398843
394215
394215
Isn't it $f=+infty $ measurable instead of $f=-infty $ ? Moreover, what is your definition of measurable function ? $-infty leq f <a$ measurable for all $a$ ?
– Surb
Jul 14 at 15:53
@Surb I have posted a corollary and a definition of measurable function.
– user398843
Jul 14 at 16:23
@Yanko Does it really duplicate the other post?
– user398843
Jul 14 at 16:51
Oh I see, the main difference is the infinity part. I'll see if I can remove that close vote.
– Yanko
Jul 14 at 17:55
add a comment |Â
Isn't it $f=+infty $ measurable instead of $f=-infty $ ? Moreover, what is your definition of measurable function ? $-infty leq f <a$ measurable for all $a$ ?
– Surb
Jul 14 at 15:53
@Surb I have posted a corollary and a definition of measurable function.
– user398843
Jul 14 at 16:23
@Yanko Does it really duplicate the other post?
– user398843
Jul 14 at 16:51
Oh I see, the main difference is the infinity part. I'll see if I can remove that close vote.
– Yanko
Jul 14 at 17:55
Isn't it $f=+infty $ measurable instead of $f=-infty $ ? Moreover, what is your definition of measurable function ? $-infty leq f <a$ measurable for all $a$ ?
– Surb
Jul 14 at 15:53
Isn't it $f=+infty $ measurable instead of $f=-infty $ ? Moreover, what is your definition of measurable function ? $-infty leq f <a$ measurable for all $a$ ?
– Surb
Jul 14 at 15:53
@Surb I have posted a corollary and a definition of measurable function.
– user398843
Jul 14 at 16:23
@Surb I have posted a corollary and a definition of measurable function.
– user398843
Jul 14 at 16:23
@Yanko Does it really duplicate the other post?
– user398843
Jul 14 at 16:51
@Yanko Does it really duplicate the other post?
– user398843
Jul 14 at 16:51
Oh I see, the main difference is the infinity part. I'll see if I can remove that close vote.
– Yanko
Jul 14 at 17:55
Oh I see, the main difference is the infinity part. I'll see if I can remove that close vote.
– Yanko
Jul 14 at 17:55
add a comment |Â
1 Answer
1
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oldest
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up vote
0
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accepted
It is enough to show that the set $xin E: f = +infty$ is measurable, since then for each finite $a$ we get that the set
$$
x in E: f(x) > a = xin E: f = +infty cup x in E: +infty>f(x) > a
$$
is a union of two measurable sets, hence measurable itself.
To see that $xin E: f = +infty$ is measurable, observe that in view of the measurability of $xin E: f = -infty$ its complement is also measurable, hence
$$
tag1 xin E: f = -infty^c = xin E: f > - infty = \xin E: f = +infty cup left( bigcuplimits_n=-infty^+infty x in E: n<f(x) < +infty right).
$$
Now the union in big brackets is measurable as a countable union of measurable sets. Since the left-hand side of $(1)$ is also measurable, the claim follows.
answered Jul 14 at 18:38
Hayk
1,39129
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
It is enough to show that the set $xin E: f = +infty$ is measurable, since then for each finite $a$ we get that the set
$$
x in E: f(x) > a = xin E: f = +infty cup x in E: +infty>f(x) > a
$$
is a union of two measurable sets, hence measurable itself.
To see that $xin E: f = +infty$ is measurable, observe that in view of the measurability of $xin E: f = -infty$ its complement is also measurable, hence
$$
tag1 xin E: f = -infty^c = xin E: f > - infty = \xin E: f = +infty cup left( bigcuplimits_n=-infty^+infty x in E: n<f(x) < +infty right).
$$
Now the union in big brackets is measurable as a countable union of measurable sets. Since the left-hand side of $(1)$ is also measurable, the claim follows.
answered Jul 14 at 18:38
Hayk
1,39129
add a comment |Â
up vote
0
down vote
accepted
It is enough to show that the set $xin E: f = +infty$ is measurable, since then for each finite $a$ we get that the set
$$
x in E: f(x) > a = xin E: f = +infty cup x in E: +infty>f(x) > a
$$
is a union of two measurable sets, hence measurable itself.
To see that $xin E: f = +infty$ is measurable, observe that in view of the measurability of $xin E: f = -infty$ its complement is also measurable, hence
$$
tag1 xin E: f = -infty^c = xin E: f > - infty = \xin E: f = +infty cup left( bigcuplimits_n=-infty^+infty x in E: n<f(x) < +infty right).
$$
Now the union in big brackets is measurable as a countable union of measurable sets. Since the left-hand side of $(1)$ is also measurable, the claim follows.
answered Jul 14 at 18:38
Hayk
1,39129
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
It is enough to show that the set $xin E: f = +infty$ is measurable, since then for each finite $a$ we get that the set
$$
x in E: f(x) > a = xin E: f = +infty cup x in E: +infty>f(x) > a
$$
is a union of two measurable sets, hence measurable itself.
To see that $xin E: f = +infty$ is measurable, observe that in view of the measurability of $xin E: f = -infty$ its complement is also measurable, hence
$$
tag1 xin E: f = -infty^c = xin E: f > - infty = \xin E: f = +infty cup left( bigcuplimits_n=-infty^+infty x in E: n<f(x) < +infty right).
$$
Now the union in big brackets is measurable as a countable union of measurable sets. Since the left-hand side of $(1)$ is also measurable, the claim follows.
answered Jul 14 at 18:38
Hayk
1,39129
It is enough to show that the set $xin E: f = +infty$ is measurable, since then for each finite $a$ we get that the set
$$
x in E: f(x) > a = xin E: f = +infty cup x in E: +infty>f(x) > a
$$
is a union of two measurable sets, hence measurable itself.
To see that $xin E: f = +infty$ is measurable, observe that in view of the measurability of $xin E: f = -infty$ its complement is also measurable, hence
$$
tag1 xin E: f = -infty^c = xin E: f > - infty = \xin E: f = +infty cup left( bigcuplimits_n=-infty^+infty x in E: n<f(x) < +infty right).
$$
Now the union in big brackets is measurable as a countable union of measurable sets. Since the left-hand side of $(1)$ is also measurable, the claim follows.
answered Jul 14 at 18:38
Hayk
1,39129
answered Jul 14 at 18:38
Hayk
1,39129
answered Jul 14 at 18:38
Hayk
1,39129
answered Jul 14 at 18:38
Hayk
1,39129
1,39129
add a comment |Â
add a comment |Â
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Isn't it $f=+infty $ measurable instead of $f=-infty $ ? Moreover, what is your definition of measurable function ? $-infty leq f <a$ measurable for all $a$ ?
– Surb
Jul 14 at 15:53
@Surb I have posted a corollary and a definition of measurable function.
– user398843
Jul 14 at 16:23
@Yanko Does it really duplicate the other post?
– user398843
Jul 14 at 16:51
Oh I see, the main difference is the infinity part. I'll see if I can remove that close vote.
– Yanko
Jul 14 at 17:55