$f$ bounded almost everywhere $Longrightarrow$ $f$ bounded or $f$ discontinuous
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0
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Let $f:mathbb{R}tomathbb{R}$ be a function. I am just asking if the following result is true:
$f$ is bounded almost everywhere $Longrightarrow$ $f$ is bounded or $f$ is discontinuous
real-analysis measure-theory continuity
asked Nov 24 at 14:13
David Lingard
534
add a comment |
up vote
0
down vote
favorite
Let $f:mathbb{R}tomathbb{R}$ be a function. I am just asking if the following result is true:
$f$ is bounded almost everywhere $Longrightarrow$ $f$ is bounded or $f$ is discontinuous
real-analysis measure-theory continuity
asked Nov 24 at 14:13
David Lingard
534
$f$ has to be unbounded at finite number of points.
– Anik Bhowmick
Nov 24 at 14:27
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f:mathbb{R}tomathbb{R}$ be a function. I am just asking if the following result is true:
$f$ is bounded almost everywhere $Longrightarrow$ $f$ is bounded or $f$ is discontinuous
real-analysis measure-theory continuity
asked Nov 24 at 14:13
David Lingard
534
Let $f:mathbb{R}tomathbb{R}$ be a function. I am just asking if the following result is true:
$f$ is bounded almost everywhere $Longrightarrow$ $f$ is bounded or $f$ is discontinuous
real-analysis measure-theory continuity
real-analysis measure-theory continuity
asked Nov 24 at 14:13
David Lingard
534
asked Nov 24 at 14:13
David Lingard
534
asked Nov 24 at 14:13
David Lingard
534
asked Nov 24 at 14:13
David Lingard
534
asked Nov 24 at 14:13
David Lingard
534
534
$f$ has to be unbounded at finite number of points.
– Anik Bhowmick
Nov 24 at 14:27
add a comment |
$f$ has to be unbounded at finite number of points.
– Anik Bhowmick
Nov 24 at 14:27
$f$ has to be unbounded at finite number of points.
– Anik Bhowmick
Nov 24 at 14:27
$f$ has to be unbounded at finite number of points.
– Anik Bhowmick
Nov 24 at 14:27
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
If $f$ is continuous and unbounded, then for each $M$ there exists $x_0$ such that $|f(x_0)|>M$ and by continuity, $|f(x)|>M$ in an open neighbourhood $(x_0-epsilon, x_0+epsilon)$ of $x_0$. As that interval is not a zero-set, $M$ fals to be an almost-bound of $f$.
answered Nov 24 at 14:25
Hagen von Eitzen
274k21266494
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If $f$ is continuous and unbounded, then for each $M$ there exists $x_0$ such that $|f(x_0)|>M$ and by continuity, $|f(x)|>M$ in an open neighbourhood $(x_0-epsilon, x_0+epsilon)$ of $x_0$. As that interval is not a zero-set, $M$ fals to be an almost-bound of $f$.
answered Nov 24 at 14:25
Hagen von Eitzen
274k21266494
add a comment |
up vote
2
down vote
accepted
If $f$ is continuous and unbounded, then for each $M$ there exists $x_0$ such that $|f(x_0)|>M$ and by continuity, $|f(x)|>M$ in an open neighbourhood $(x_0-epsilon, x_0+epsilon)$ of $x_0$. As that interval is not a zero-set, $M$ fals to be an almost-bound of $f$.
answered Nov 24 at 14:25
Hagen von Eitzen
274k21266494
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If $f$ is continuous and unbounded, then for each $M$ there exists $x_0$ such that $|f(x_0)|>M$ and by continuity, $|f(x)|>M$ in an open neighbourhood $(x_0-epsilon, x_0+epsilon)$ of $x_0$. As that interval is not a zero-set, $M$ fals to be an almost-bound of $f$.
answered Nov 24 at 14:25
Hagen von Eitzen
274k21266494
If $f$ is continuous and unbounded, then for each $M$ there exists $x_0$ such that $|f(x_0)|>M$ and by continuity, $|f(x)|>M$ in an open neighbourhood $(x_0-epsilon, x_0+epsilon)$ of $x_0$. As that interval is not a zero-set, $M$ fals to be an almost-bound of $f$.
answered Nov 24 at 14:25
Hagen von Eitzen
274k21266494
answered Nov 24 at 14:25
Hagen von Eitzen
274k21266494
answered Nov 24 at 14:25
Hagen von Eitzen
274k21266494
answered Nov 24 at 14:25
Hagen von Eitzen
274k21266494
274k21266494
add a comment |
add a comment |
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$f$ has to be unbounded at finite number of points.
– Anik Bhowmick
Nov 24 at 14:27