Mixing
Have I understood Compact Set correctly
$begingroup$
In our current Measure Theory Class, we bought up the notion for a function $f:mathbb R to mathbb R$ that is continuous to have a compact support, is equivalent to the fact that $overline{{x in mathbb R: f(x)neq 0}}$ needs to be compact.
Note $C_{c}(mathbb R)={ f: mathbb R to mathbb R:f$ is continuous and $supp f$ compact $}$
I assume the most important aspect to take out of the notion of compact support (at least for my class in Measure Theory) is the fact that: $forall p in [1,infty]:$
$C_{c}(mathbb R)subseteq L^{p}$
So if I need to prove a function is $p-$integrable then I can simply show that it has compact support $supp f$.
Is this the most applicable case to use the notion of a compact support in Measure Theory?
Other Question:
Considering the factor $partial {x in mathbb R: f(x)neq 0}$, does this mean that the set $overline{{x in mathbb R: f(x)neq 0}}$ can contain points $y in mathbb R$ such $f(y)=0$ but these sets of points need to be countable? Or can they be uncountably infinite?
real-analysis measure-theory compactness lp-spaces
asked Jan 22 at 13:17
SABOYSABOY
656311
$endgroup$
add a comment |
$begingroup$
In our current Measure Theory Class, we bought up the notion for a function $f:mathbb R to mathbb R$ that is continuous to have a compact support, is equivalent to the fact that $overline{{x in mathbb R: f(x)neq 0}}$ needs to be compact.
Note $C_{c}(mathbb R)={ f: mathbb R to mathbb R:f$ is continuous and $supp f$ compact $}$
I assume the most important aspect to take out of the notion of compact support (at least for my class in Measure Theory) is the fact that: $forall p in [1,infty]:$
$C_{c}(mathbb R)subseteq L^{p}$
So if I need to prove a function is $p-$integrable then I can simply show that it has compact support $supp f$.
Is this the most applicable case to use the notion of a compact support in Measure Theory?
Other Question:
Considering the factor $partial {x in mathbb R: f(x)neq 0}$, does this mean that the set $overline{{x in mathbb R: f(x)neq 0}}$ can contain points $y in mathbb R$ such $f(y)=0$ but these sets of points need to be countable? Or can they be uncountably infinite?
real-analysis measure-theory compactness lp-spaces
asked Jan 22 at 13:17
SABOYSABOY
656311
$endgroup$
$begingroup$
What is $;C_C(Bbb R);$ ?
$endgroup$
– DonAntonio
Jan 22 at 13:21
add a comment |
$begingroup$
In our current Measure Theory Class, we bought up the notion for a function $f:mathbb R to mathbb R$ that is continuous to have a compact support, is equivalent to the fact that $overline{{x in mathbb R: f(x)neq 0}}$ needs to be compact.
Note $C_{c}(mathbb R)={ f: mathbb R to mathbb R:f$ is continuous and $supp f$ compact $}$
I assume the most important aspect to take out of the notion of compact support (at least for my class in Measure Theory) is the fact that: $forall p in [1,infty]:$
$C_{c}(mathbb R)subseteq L^{p}$
So if I need to prove a function is $p-$integrable then I can simply show that it has compact support $supp f$.
Is this the most applicable case to use the notion of a compact support in Measure Theory?
Other Question:
Considering the factor $partial {x in mathbb R: f(x)neq 0}$, does this mean that the set $overline{{x in mathbb R: f(x)neq 0}}$ can contain points $y in mathbb R$ such $f(y)=0$ but these sets of points need to be countable? Or can they be uncountably infinite?
real-analysis measure-theory compactness lp-spaces
asked Jan 22 at 13:17
SABOYSABOY
656311
$endgroup$
In our current Measure Theory Class, we bought up the notion for a function $f:mathbb R to mathbb R$ that is continuous to have a compact support, is equivalent to the fact that $overline{{x in mathbb R: f(x)neq 0}}$ needs to be compact.
Note $C_{c}(mathbb R)={ f: mathbb R to mathbb R:f$ is continuous and $supp f$ compact $}$
I assume the most important aspect to take out of the notion of compact support (at least for my class in Measure Theory) is the fact that: $forall p in [1,infty]:$
$C_{c}(mathbb R)subseteq L^{p}$
So if I need to prove a function is $p-$integrable then I can simply show that it has compact support $supp f$.
Is this the most applicable case to use the notion of a compact support in Measure Theory?
Other Question:
Considering the factor $partial {x in mathbb R: f(x)neq 0}$, does this mean that the set $overline{{x in mathbb R: f(x)neq 0}}$ can contain points $y in mathbb R$ such $f(y)=0$ but these sets of points need to be countable? Or can they be uncountably infinite?
real-analysis measure-theory compactness lp-spaces
real-analysis measure-theory compactness lp-spaces
asked Jan 22 at 13:17
SABOYSABOY
656311
asked Jan 22 at 13:17
SABOYSABOY
656311
edited Jan 22 at 13:23
SABOY
asked Jan 22 at 13:17
SABOYSABOY
656311
asked Jan 22 at 13:17
SABOYSABOY
656311
asked Jan 22 at 13:17
SABOYSABOY
656311
656311
$begingroup$
What is $;C_C(Bbb R);$ ?
$endgroup$
– DonAntonio
Jan 22 at 13:21
add a comment |
$begingroup$
What is $;C_C(Bbb R);$ ?
$endgroup$
– DonAntonio
Jan 22 at 13:21
$begingroup$
What is $;C_C(Bbb R);$ ?
$endgroup$
– DonAntonio
Jan 22 at 13:21
$begingroup$
What is $;C_C(Bbb R);$ ?
$endgroup$
– DonAntonio
Jan 22 at 13:21
add a comment |
1 Answer
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votes
$begingroup$
In measure theory in isolation the concern is really more about finite measure support (where support has no closure operation in that setting, in part because there is no guarantee that a topology has been defined on the measure space). The point there is that a bounded measurable function with finite measure support is in all of the $L^p$ spaces. Compact support is important for other reasons but it starts to involve topics that are not strictly measure-theoretic in nature (e.g. distribution theory).
Your other question (how "big" can the boundary of the support of a continuous function be?) should probably be asked separately.
answered Jan 22 at 13:25
IanIan
68.7k25389
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
In measure theory in isolation the concern is really more about finite measure support (where support has no closure operation in that setting, in part because there is no guarantee that a topology has been defined on the measure space). The point there is that a bounded measurable function with finite measure support is in all of the $L^p$ spaces. Compact support is important for other reasons but it starts to involve topics that are not strictly measure-theoretic in nature (e.g. distribution theory).
Your other question (how "big" can the boundary of the support of a continuous function be?) should probably be asked separately.
answered Jan 22 at 13:25
IanIan
68.7k25389
$endgroup$
add a comment |
$begingroup$
In measure theory in isolation the concern is really more about finite measure support (where support has no closure operation in that setting, in part because there is no guarantee that a topology has been defined on the measure space). The point there is that a bounded measurable function with finite measure support is in all of the $L^p$ spaces. Compact support is important for other reasons but it starts to involve topics that are not strictly measure-theoretic in nature (e.g. distribution theory).
Your other question (how "big" can the boundary of the support of a continuous function be?) should probably be asked separately.
answered Jan 22 at 13:25
IanIan
68.7k25389
$endgroup$
add a comment |
$begingroup$
In measure theory in isolation the concern is really more about finite measure support (where support has no closure operation in that setting, in part because there is no guarantee that a topology has been defined on the measure space). The point there is that a bounded measurable function with finite measure support is in all of the $L^p$ spaces. Compact support is important for other reasons but it starts to involve topics that are not strictly measure-theoretic in nature (e.g. distribution theory).
Your other question (how "big" can the boundary of the support of a continuous function be?) should probably be asked separately.
answered Jan 22 at 13:25
IanIan
68.7k25389
$endgroup$
In measure theory in isolation the concern is really more about finite measure support (where support has no closure operation in that setting, in part because there is no guarantee that a topology has been defined on the measure space). The point there is that a bounded measurable function with finite measure support is in all of the $L^p$ spaces. Compact support is important for other reasons but it starts to involve topics that are not strictly measure-theoretic in nature (e.g. distribution theory).
Your other question (how "big" can the boundary of the support of a continuous function be?) should probably be asked separately.
answered Jan 22 at 13:25
IanIan
68.7k25389
edited Jan 22 at 14:27
answered Jan 22 at 13:25
IanIan
68.7k25389
answered Jan 22 at 13:25
IanIan
68.7k25389
answered Jan 22 at 13:25
IanIan
68.7k25389
68.7k25389
add a comment |
add a comment |
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$begingroup$
What is $;C_C(Bbb R);$ ?
$endgroup$
– DonAntonio
Jan 22 at 13:21