Support of measurable function is measurable?If $f$ is measurable, then support of $f$ is measurable?Support of measurable function regular?Equivalent conditions for a measurable functionComposition of 2 Lebesgue measurable functions is not lebesgue measurable: Are these two functions Borel Measurable?A function that is Lebesgue integrable but not measurable (not absurd obviously)Equivalency of two forms of Lebesgue measurabilityA simple proof that there exixsts lebesgue measurable sets, not borel measurableLebesgue Measurable via open setIs the composition of two Lebesgue Measurable functions necessarily Lebesgue measurable?Existence/examples of a non measurable function measurable on its supportIf $f_n_n=1^infty$ is a sequence of measurable functions, then $sup_n f_n(x)$ is measurable
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Support of measurable function is measurable?
If $f$ is measurable, then support of $f$ is measurable?Support of measurable function regular?Equivalent conditions for a measurable functionComposition of 2 Lebesgue measurable functions is not lebesgue measurable: Are these two functions Borel Measurable?A function that is Lebesgue integrable but not measurable (not absurd obviously)Equivalency of two forms of Lebesgue measurabilityA simple proof that there exixsts lebesgue measurable sets, not borel measurableLebesgue Measurable via open setIs the composition of two Lebesgue Measurable functions necessarily Lebesgue measurable?Existence/examples of a non measurable function measurable on its supportIf $f_n_n=1^infty$ is a sequence of measurable functions, then $sup_n f_n(x)$ is measurable
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Let $f$ be a Lebesgue measurable function defined on a set $Esubset mathbbR^d$. Let $xin Emid f(x)neq 0 $ be the support of the function on $E$.
Stein and Shakarchi claim without proof that the support is measurable. But why is this true? It's not just due to the fact that the support is a subset of a measurable set.
real-analysis measure-theory lebesgue-measure
asked 8 hours ago
ponchanponchan
4163 silver badges9 bronze badges
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add a comment |
$begingroup$
Let $f$ be a Lebesgue measurable function defined on a set $Esubset mathbbR^d$. Let $xin Emid f(x)neq 0 $ be the support of the function on $E$.
Stein and Shakarchi claim without proof that the support is measurable. But why is this true? It's not just due to the fact that the support is a subset of a measurable set.
real-analysis measure-theory lebesgue-measure
asked 8 hours ago
ponchanponchan
4163 silver badges9 bronze badges
$endgroup$
3
$begingroup$
Possible duplicate of If $f$ is measurable, then support of $f$ is measurable?
$endgroup$
– Meni
8 hours ago
add a comment |
$begingroup$
Let $f$ be a Lebesgue measurable function defined on a set $Esubset mathbbR^d$. Let $xin Emid f(x)neq 0 $ be the support of the function on $E$.
Stein and Shakarchi claim without proof that the support is measurable. But why is this true? It's not just due to the fact that the support is a subset of a measurable set.
real-analysis measure-theory lebesgue-measure
asked 8 hours ago
ponchanponchan
4163 silver badges9 bronze badges
$endgroup$
Let $f$ be a Lebesgue measurable function defined on a set $Esubset mathbbR^d$. Let $xin Emid f(x)neq 0 $ be the support of the function on $E$.
Stein and Shakarchi claim without proof that the support is measurable. But why is this true? It's not just due to the fact that the support is a subset of a measurable set.
real-analysis measure-theory lebesgue-measure
real-analysis measure-theory lebesgue-measure
asked 8 hours ago
ponchanponchan
4163 silver badges9 bronze badges
asked 8 hours ago
ponchanponchan
4163 silver badges9 bronze badges
asked 8 hours ago
ponchanponchan
4163 silver badges9 bronze badges
asked 8 hours ago
ponchanponchan
4163 silver badges9 bronze badges
asked 8 hours ago
ponchanponchan
4163 silver badges9 bronze badges
4163 silver badges9 bronze badges
3
$begingroup$
Possible duplicate of If $f$ is measurable, then support of $f$ is measurable?
$endgroup$
– Meni
8 hours ago
add a comment |
3
$begingroup$
Possible duplicate of If $f$ is measurable, then support of $f$ is measurable?
$endgroup$
– Meni
8 hours ago
3
3
$begingroup$
Possible duplicate of If $f$ is measurable, then support of $f$ is measurable?
$endgroup$
– Meni
8 hours ago
$begingroup$
Possible duplicate of If $f$ is measurable, then support of $f$ is measurable?
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– Meni
8 hours ago
add a comment |
3 Answers
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The support is usually defined as the closure of that set, but I'll answer your specific question.
Note that your set is precisely $$f^-1 ((-infty,0)cup (0,infty))=f^-1((-infty,0))cup f^-1((0,infty)).$$ Now, just recall the definition of a Lebesgue measurable function.
answered 8 hours ago
cmkcmk
4,1971 gold badge7 silver badges25 bronze badges
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As a general rule, anything that you can define explicitly in terms of a measurable function or sequence of functions is going to turn out to be measurable; it takes a little trick to show that not every set is measurable.
That's too broad to be provable, but this particular example is trivial. Say $f=u+iv$ where $u$ and $v$ are real-valued. Then $f(x)ne0=u(x)ne0cupv(x)ne0$. So you can assume wlog that $f$ is real-valued. So $f(x)ne0=f(x)>0cupf(x)<0$, the union of two measurable sets.
answered 8 hours ago
David C. UllrichDavid C. Ullrich
64.1k4 gold badges44 silver badges99 bronze badges
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A function $f:mathbbR^dmapsto mathbbR$ is Lebesque measurable by definition if
$$xin mathbbR^d $$
is measureable for all $ain mathbbR$. Equivilently you may use $leq, <$ or $geq$ instead of $>$. Therefor
$$ xin mathbbR^d = f(x) > 0 cupxin mathbbR^d $$
answered 8 hours ago
DaytonDayton
3361 silver badge13 bronze badges
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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$begingroup$
The support is usually defined as the closure of that set, but I'll answer your specific question.
Note that your set is precisely $$f^-1 ((-infty,0)cup (0,infty))=f^-1((-infty,0))cup f^-1((0,infty)).$$ Now, just recall the definition of a Lebesgue measurable function.
answered 8 hours ago
cmkcmk
4,1971 gold badge7 silver badges25 bronze badges
$endgroup$
add a comment |
$begingroup$
The support is usually defined as the closure of that set, but I'll answer your specific question.
Note that your set is precisely $$f^-1 ((-infty,0)cup (0,infty))=f^-1((-infty,0))cup f^-1((0,infty)).$$ Now, just recall the definition of a Lebesgue measurable function.
answered 8 hours ago
cmkcmk
4,1971 gold badge7 silver badges25 bronze badges
$endgroup$
add a comment |
$begingroup$
The support is usually defined as the closure of that set, but I'll answer your specific question.
Note that your set is precisely $$f^-1 ((-infty,0)cup (0,infty))=f^-1((-infty,0))cup f^-1((0,infty)).$$ Now, just recall the definition of a Lebesgue measurable function.
answered 8 hours ago
cmkcmk
4,1971 gold badge7 silver badges25 bronze badges
$endgroup$
The support is usually defined as the closure of that set, but I'll answer your specific question.
Note that your set is precisely $$f^-1 ((-infty,0)cup (0,infty))=f^-1((-infty,0))cup f^-1((0,infty)).$$ Now, just recall the definition of a Lebesgue measurable function.
answered 8 hours ago
cmkcmk
4,1971 gold badge7 silver badges25 bronze badges
edited 8 hours ago
answered 8 hours ago
cmkcmk
4,1971 gold badge7 silver badges25 bronze badges
answered 8 hours ago
cmkcmk
4,1971 gold badge7 silver badges25 bronze badges
answered 8 hours ago
cmkcmk
4,1971 gold badge7 silver badges25 bronze badges
4,1971 gold badge7 silver badges25 bronze badges
add a comment |
add a comment |
$begingroup$
As a general rule, anything that you can define explicitly in terms of a measurable function or sequence of functions is going to turn out to be measurable; it takes a little trick to show that not every set is measurable.
That's too broad to be provable, but this particular example is trivial. Say $f=u+iv$ where $u$ and $v$ are real-valued. Then $f(x)ne0=u(x)ne0cupv(x)ne0$. So you can assume wlog that $f$ is real-valued. So $f(x)ne0=f(x)>0cupf(x)<0$, the union of two measurable sets.
answered 8 hours ago
David C. UllrichDavid C. Ullrich
64.1k4 gold badges44 silver badges99 bronze badges
$endgroup$
add a comment |
$begingroup$
As a general rule, anything that you can define explicitly in terms of a measurable function or sequence of functions is going to turn out to be measurable; it takes a little trick to show that not every set is measurable.
That's too broad to be provable, but this particular example is trivial. Say $f=u+iv$ where $u$ and $v$ are real-valued. Then $f(x)ne0=u(x)ne0cupv(x)ne0$. So you can assume wlog that $f$ is real-valued. So $f(x)ne0=f(x)>0cupf(x)<0$, the union of two measurable sets.
answered 8 hours ago
David C. UllrichDavid C. Ullrich
64.1k4 gold badges44 silver badges99 bronze badges
$endgroup$
add a comment |
$begingroup$
As a general rule, anything that you can define explicitly in terms of a measurable function or sequence of functions is going to turn out to be measurable; it takes a little trick to show that not every set is measurable.
That's too broad to be provable, but this particular example is trivial. Say $f=u+iv$ where $u$ and $v$ are real-valued. Then $f(x)ne0=u(x)ne0cupv(x)ne0$. So you can assume wlog that $f$ is real-valued. So $f(x)ne0=f(x)>0cupf(x)<0$, the union of two measurable sets.
answered 8 hours ago
David C. UllrichDavid C. Ullrich
64.1k4 gold badges44 silver badges99 bronze badges
$endgroup$
As a general rule, anything that you can define explicitly in terms of a measurable function or sequence of functions is going to turn out to be measurable; it takes a little trick to show that not every set is measurable.
That's too broad to be provable, but this particular example is trivial. Say $f=u+iv$ where $u$ and $v$ are real-valued. Then $f(x)ne0=u(x)ne0cupv(x)ne0$. So you can assume wlog that $f$ is real-valued. So $f(x)ne0=f(x)>0cupf(x)<0$, the union of two measurable sets.
answered 8 hours ago
David C. UllrichDavid C. Ullrich
64.1k4 gold badges44 silver badges99 bronze badges
answered 8 hours ago
David C. UllrichDavid C. Ullrich
64.1k4 gold badges44 silver badges99 bronze badges
answered 8 hours ago
David C. UllrichDavid C. Ullrich
64.1k4 gold badges44 silver badges99 bronze badges
answered 8 hours ago
David C. UllrichDavid C. Ullrich
64.1k4 gold badges44 silver badges99 bronze badges
64.1k4 gold badges44 silver badges99 bronze badges
add a comment |
add a comment |
$begingroup$
A function $f:mathbbR^dmapsto mathbbR$ is Lebesque measurable by definition if
$$xin mathbbR^d $$
is measureable for all $ain mathbbR$. Equivilently you may use $leq, <$ or $geq$ instead of $>$. Therefor
$$ xin mathbbR^d = f(x) > 0 cupxin mathbbR^d $$
answered 8 hours ago
DaytonDayton
3361 silver badge13 bronze badges
$endgroup$
add a comment |
$begingroup$
A function $f:mathbbR^dmapsto mathbbR$ is Lebesque measurable by definition if
$$xin mathbbR^d $$
is measureable for all $ain mathbbR$. Equivilently you may use $leq, <$ or $geq$ instead of $>$. Therefor
$$ xin mathbbR^d = f(x) > 0 cupxin mathbbR^d $$
answered 8 hours ago
DaytonDayton
3361 silver badge13 bronze badges
$endgroup$
add a comment |
$begingroup$
A function $f:mathbbR^dmapsto mathbbR$ is Lebesque measurable by definition if
$$xin mathbbR^d $$
is measureable for all $ain mathbbR$. Equivilently you may use $leq, <$ or $geq$ instead of $>$. Therefor
$$ xin mathbbR^d = f(x) > 0 cupxin mathbbR^d $$
answered 8 hours ago
DaytonDayton
3361 silver badge13 bronze badges
$endgroup$
A function $f:mathbbR^dmapsto mathbbR$ is Lebesque measurable by definition if
$$xin mathbbR^d $$
is measureable for all $ain mathbbR$. Equivilently you may use $leq, <$ or $geq$ instead of $>$. Therefor
$$ xin mathbbR^d = f(x) > 0 cupxin mathbbR^d $$
answered 8 hours ago
DaytonDayton
3361 silver badge13 bronze badges
answered 8 hours ago
DaytonDayton
3361 silver badge13 bronze badges
answered 8 hours ago
DaytonDayton
3361 silver badge13 bronze badges
answered 8 hours ago
DaytonDayton
3361 silver badge13 bronze badges
3361 silver badge13 bronze badges
add a comment |
add a comment |
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Possible duplicate of If $f$ is measurable, then support of $f$ is measurable?
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– Meni
8 hours ago