Are continuous functions strongly measurable?An example of a space $X$ such that every L-subset of $X^*$ is weakly precompact but not relatively weakly compactNaive example of integral with values in Banach spaceIs my proof that a function is measurable correct?Almost Everywhere pointwise limit measurable functions measurable?What is an example of a function that is measurable but not strongly measurable?Volterra operator and completely continuous operatorsstrongly continuous mapping implies bounded mappingStrongly measurable implies Borel measurable in separable spaceChecking a separability condition for Bochner measurabilityIs weakly continuous function with values in Banach space strongly measurable？
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Are continuous functions strongly measurable?
An example of a space $X$ such that every L-subset of $X^*$ is weakly precompact but not relatively weakly compactNaive example of integral with values in Banach spaceIs my proof that a function is measurable correct?Almost Everywhere pointwise limit measurable functions measurable?What is an example of a function that is measurable but not strongly measurable?Volterra operator and completely continuous operatorsstrongly continuous mapping implies bounded mappingStrongly measurable implies Borel measurable in separable spaceChecking a separability condition for Bochner measurabilityIs weakly continuous function with values in Banach space strongly measurable？
$begingroup$
Measure theory is still quite new to me, and I'm a bit confused about the following.
Suppose we have a continuous function $f: I rightarrow X$, where $I subset mathbbR$ is a closed interval and $X$ is a Banach space. I can show that $f$ is weakly measurable: for each $v in X^*$, we have that the mapping $x mapsto v(f(x))$ is continuous since it is a composition of continuous mappings $v$ and $f$ (is this correct?).
I also know that a continuous function is measurable. In this case, does measurable mean the same as strongly measurable? If it does, how can you show this using Pettis' theorem (i.e. why is there a null set $Nsubset I$ such that the set $ xin Ibackslash N$ is separable)? Or is it easier to prove it without Pettis' theorem?
For continuous $f$, does strong convergence then also imply that it is summable / Bochner integrable, since the mapping $x mapsto ||f(x)||$ is the composition of continuous maps ($f$ and $||.||$)? EDIT: This last sentence is nonsense of course, $x mapsto ||f(x)||$ must be summable, not continuous.
functional-analysis measure-theory lebesgue-integral
asked Apr 27 '13 at 18:17
ScroogeMcDuckScroogeMcDuck
478414
$endgroup$
|
show 1 more comment
$begingroup$
Measure theory is still quite new to me, and I'm a bit confused about the following.
Suppose we have a continuous function $f: I rightarrow X$, where $I subset mathbbR$ is a closed interval and $X$ is a Banach space. I can show that $f$ is weakly measurable: for each $v in X^*$, we have that the mapping $x mapsto v(f(x))$ is continuous since it is a composition of continuous mappings $v$ and $f$ (is this correct?).
I also know that a continuous function is measurable. In this case, does measurable mean the same as strongly measurable? If it does, how can you show this using Pettis' theorem (i.e. why is there a null set $Nsubset I$ such that the set $ xin Ibackslash N$ is separable)? Or is it easier to prove it without Pettis' theorem?
For continuous $f$, does strong convergence then also imply that it is summable / Bochner integrable, since the mapping $x mapsto ||f(x)||$ is the composition of continuous maps ($f$ and $||.||$)? EDIT: This last sentence is nonsense of course, $x mapsto ||f(x)||$ must be summable, not continuous.
functional-analysis measure-theory lebesgue-integral
asked Apr 27 '13 at 18:17
ScroogeMcDuckScroogeMcDuck
478414
$endgroup$
1
$begingroup$
If you are new to measure theory, then you should take a beginning course in measure theory, not a course of measure theory in Banach space.
$endgroup$
– TCL
Apr 27 '13 at 18:31
1
$begingroup$
Do the case $I = [0,1]$ first. Use that $f(I)$ is a compact metric space.
$endgroup$
– Martin
Apr 27 '13 at 19:02
$begingroup$
Is this correct: $I=[0,1]$ is compact, hence $f(I)$ is a compact subset of a metric space and therefore separable. By Pettis' theorem it follows that any continuous function is strongly measurable?
$endgroup$
– ScroogeMcDuck
Apr 28 '13 at 11:18
$begingroup$
Correction: any continuous function with compact domain is strongly measurable?
$endgroup$
– ScroogeMcDuck
Apr 28 '13 at 11:24
2
$begingroup$
Yes, this is correct. The general case follows from this since a closed interval is a countable union of compact intervals. // However, you do not need to use Pettis's theorem. Note that $f$ is uniformly continuous on $[0,1]$. Use this to approximate $f$ by step functions. // Your proof of weak measurability is okay, btw. // Please use@Martinin your future comments so I am notified.
$endgroup$
– Martin
Apr 29 '13 at 16:40
|
show 1 more comment
$begingroup$
Measure theory is still quite new to me, and I'm a bit confused about the following.
Suppose we have a continuous function $f: I rightarrow X$, where $I subset mathbbR$ is a closed interval and $X$ is a Banach space. I can show that $f$ is weakly measurable: for each $v in X^*$, we have that the mapping $x mapsto v(f(x))$ is continuous since it is a composition of continuous mappings $v$ and $f$ (is this correct?).
I also know that a continuous function is measurable. In this case, does measurable mean the same as strongly measurable? If it does, how can you show this using Pettis' theorem (i.e. why is there a null set $Nsubset I$ such that the set $ xin Ibackslash N$ is separable)? Or is it easier to prove it without Pettis' theorem?
For continuous $f$, does strong convergence then also imply that it is summable / Bochner integrable, since the mapping $x mapsto ||f(x)||$ is the composition of continuous maps ($f$ and $||.||$)? EDIT: This last sentence is nonsense of course, $x mapsto ||f(x)||$ must be summable, not continuous.
functional-analysis measure-theory lebesgue-integral
asked Apr 27 '13 at 18:17
ScroogeMcDuckScroogeMcDuck
478414
$endgroup$
Measure theory is still quite new to me, and I'm a bit confused about the following.
Suppose we have a continuous function $f: I rightarrow X$, where $I subset mathbbR$ is a closed interval and $X$ is a Banach space. I can show that $f$ is weakly measurable: for each $v in X^*$, we have that the mapping $x mapsto v(f(x))$ is continuous since it is a composition of continuous mappings $v$ and $f$ (is this correct?).
I also know that a continuous function is measurable. In this case, does measurable mean the same as strongly measurable? If it does, how can you show this using Pettis' theorem (i.e. why is there a null set $Nsubset I$ such that the set $ xin Ibackslash N$ is separable)? Or is it easier to prove it without Pettis' theorem?
For continuous $f$, does strong convergence then also imply that it is summable / Bochner integrable, since the mapping $x mapsto ||f(x)||$ is the composition of continuous maps ($f$ and $||.||$)? EDIT: This last sentence is nonsense of course, $x mapsto ||f(x)||$ must be summable, not continuous.
functional-analysis measure-theory lebesgue-integral
functional-analysis measure-theory lebesgue-integral
asked Apr 27 '13 at 18:17
ScroogeMcDuckScroogeMcDuck
478414
asked Apr 27 '13 at 18:17
ScroogeMcDuckScroogeMcDuck
478414
edited Apr 28 '13 at 11:37
ScroogeMcDuck
asked Apr 27 '13 at 18:17
ScroogeMcDuckScroogeMcDuck
478414
asked Apr 27 '13 at 18:17
ScroogeMcDuckScroogeMcDuck
478414
asked Apr 27 '13 at 18:17
ScroogeMcDuckScroogeMcDuck
478414
478414
1
$begingroup$
If you are new to measure theory, then you should take a beginning course in measure theory, not a course of measure theory in Banach space.
$endgroup$
– TCL
Apr 27 '13 at 18:31
1
$begingroup$
Do the case $I = [0,1]$ first. Use that $f(I)$ is a compact metric space.
$endgroup$
– Martin
Apr 27 '13 at 19:02
$begingroup$
Is this correct: $I=[0,1]$ is compact, hence $f(I)$ is a compact subset of a metric space and therefore separable. By Pettis' theorem it follows that any continuous function is strongly measurable?
$endgroup$
– ScroogeMcDuck
Apr 28 '13 at 11:18
$begingroup$
Correction: any continuous function with compact domain is strongly measurable?
$endgroup$
– ScroogeMcDuck
Apr 28 '13 at 11:24
2
$begingroup$
Yes, this is correct. The general case follows from this since a closed interval is a countable union of compact intervals. // However, you do not need to use Pettis's theorem. Note that $f$ is uniformly continuous on $[0,1]$. Use this to approximate $f$ by step functions. // Your proof of weak measurability is okay, btw. // Please use@Martinin your future comments so I am notified.
$endgroup$
– Martin
Apr 29 '13 at 16:40
|
show 1 more comment
1
$begingroup$
If you are new to measure theory, then you should take a beginning course in measure theory, not a course of measure theory in Banach space.
$endgroup$
– TCL
Apr 27 '13 at 18:31
1
$begingroup$
Do the case $I = [0,1]$ first. Use that $f(I)$ is a compact metric space.
$endgroup$
– Martin
Apr 27 '13 at 19:02
$begingroup$
Is this correct: $I=[0,1]$ is compact, hence $f(I)$ is a compact subset of a metric space and therefore separable. By Pettis' theorem it follows that any continuous function is strongly measurable?
$endgroup$
– ScroogeMcDuck
Apr 28 '13 at 11:18
$begingroup$
Correction: any continuous function with compact domain is strongly measurable?
$endgroup$
– ScroogeMcDuck
Apr 28 '13 at 11:24
2
$begingroup$
Yes, this is correct. The general case follows from this since a closed interval is a countable union of compact intervals. // However, you do not need to use Pettis's theorem. Note that $f$ is uniformly continuous on $[0,1]$. Use this to approximate $f$ by step functions. // Your proof of weak measurability is okay, btw. // Please use@Martinin your future comments so I am notified.
$endgroup$
– Martin
Apr 29 '13 at 16:40
1
1
$begingroup$
If you are new to measure theory, then you should take a beginning course in measure theory, not a course of measure theory in Banach space.
$endgroup$
– TCL
Apr 27 '13 at 18:31
$begingroup$
If you are new to measure theory, then you should take a beginning course in measure theory, not a course of measure theory in Banach space.
$endgroup$
– TCL
Apr 27 '13 at 18:31
1
1
$begingroup$
Do the case $I = [0,1]$ first. Use that $f(I)$ is a compact metric space.
$endgroup$
– Martin
Apr 27 '13 at 19:02
$begingroup$
Do the case $I = [0,1]$ first. Use that $f(I)$ is a compact metric space.
$endgroup$
– Martin
Apr 27 '13 at 19:02
$begingroup$
Is this correct: $I=[0,1]$ is compact, hence $f(I)$ is a compact subset of a metric space and therefore separable. By Pettis' theorem it follows that any continuous function is strongly measurable?
$endgroup$
– ScroogeMcDuck
Apr 28 '13 at 11:18
$begingroup$
Is this correct: $I=[0,1]$ is compact, hence $f(I)$ is a compact subset of a metric space and therefore separable. By Pettis' theorem it follows that any continuous function is strongly measurable?
$endgroup$
– ScroogeMcDuck
Apr 28 '13 at 11:18
$begingroup$
Correction: any continuous function with compact domain is strongly measurable?
$endgroup$
– ScroogeMcDuck
Apr 28 '13 at 11:24
$begingroup$
Correction: any continuous function with compact domain is strongly measurable?
$endgroup$
– ScroogeMcDuck
Apr 28 '13 at 11:24
2
2
$begingroup$
Yes, this is correct. The general case follows from this since a closed interval is a countable union of compact intervals. // However, you do not need to use Pettis's theorem. Note that $f$ is uniformly continuous on $[0,1]$. Use this to approximate $f$ by step functions. // Your proof of weak measurability is okay, btw. // Please use
@Martin in your future comments so I am notified.$endgroup$
– Martin
Apr 29 '13 at 16:40
$begingroup$
Yes, this is correct. The general case follows from this since a closed interval is a countable union of compact intervals. // However, you do not need to use Pettis's theorem. Note that $f$ is uniformly continuous on $[0,1]$. Use this to approximate $f$ by step functions. // Your proof of weak measurability is okay, btw. // Please use
@Martin in your future comments so I am notified.$endgroup$
– Martin
Apr 29 '13 at 16:40
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Let $0=t_0le t_1le...le t_n=1$ and suppose that $Delta=max_iin1,...,n(t_i-t_i-1)to 0$ as $nto infty$. Let $E_i=[t_i-1, t_i)$ and for each $i=1,...,n$ fix $xi_iin E_i$. Let us now define
$$f_n(t)=sum_i=1^nf(xi_i)chi_E_i(t)$$
for all $tin [0,1]$ and $nin mathbbN$. Fix $tin [0,1]$, $varepsilon>0$ and observe that
$$|f(t)-f_n(t)| le sum_i=1^n|f(t)-f(xi_i)|chi_E_i(t)=|f(t)-f(xi_i(t))|<varepsilon$$
for sufficiently large $n$ which is a consequance of uniform continuity of $f$ on compact set $[0,1]$. This imples that
$$lim_ntoinfty|f(t)-f_n(t)|=0$$
for all $tin [0,1]$.
answered Mar 22 at 15:15
zorro47zorro47
631515
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
Let $0=t_0le t_1le...le t_n=1$ and suppose that $Delta=max_iin1,...,n(t_i-t_i-1)to 0$ as $nto infty$. Let $E_i=[t_i-1, t_i)$ and for each $i=1,...,n$ fix $xi_iin E_i$. Let us now define
$$f_n(t)=sum_i=1^nf(xi_i)chi_E_i(t)$$
for all $tin [0,1]$ and $nin mathbbN$. Fix $tin [0,1]$, $varepsilon>0$ and observe that
$$|f(t)-f_n(t)| le sum_i=1^n|f(t)-f(xi_i)|chi_E_i(t)=|f(t)-f(xi_i(t))|<varepsilon$$
for sufficiently large $n$ which is a consequance of uniform continuity of $f$ on compact set $[0,1]$. This imples that
$$lim_ntoinfty|f(t)-f_n(t)|=0$$
for all $tin [0,1]$.
answered Mar 22 at 15:15
zorro47zorro47
631515
$endgroup$
add a comment |
$begingroup$
Let $0=t_0le t_1le...le t_n=1$ and suppose that $Delta=max_iin1,...,n(t_i-t_i-1)to 0$ as $nto infty$. Let $E_i=[t_i-1, t_i)$ and for each $i=1,...,n$ fix $xi_iin E_i$. Let us now define
$$f_n(t)=sum_i=1^nf(xi_i)chi_E_i(t)$$
for all $tin [0,1]$ and $nin mathbbN$. Fix $tin [0,1]$, $varepsilon>0$ and observe that
$$|f(t)-f_n(t)| le sum_i=1^n|f(t)-f(xi_i)|chi_E_i(t)=|f(t)-f(xi_i(t))|<varepsilon$$
for sufficiently large $n$ which is a consequance of uniform continuity of $f$ on compact set $[0,1]$. This imples that
$$lim_ntoinfty|f(t)-f_n(t)|=0$$
for all $tin [0,1]$.
answered Mar 22 at 15:15
zorro47zorro47
631515
$endgroup$
add a comment |
$begingroup$
Let $0=t_0le t_1le...le t_n=1$ and suppose that $Delta=max_iin1,...,n(t_i-t_i-1)to 0$ as $nto infty$. Let $E_i=[t_i-1, t_i)$ and for each $i=1,...,n$ fix $xi_iin E_i$. Let us now define
$$f_n(t)=sum_i=1^nf(xi_i)chi_E_i(t)$$
for all $tin [0,1]$ and $nin mathbbN$. Fix $tin [0,1]$, $varepsilon>0$ and observe that
$$|f(t)-f_n(t)| le sum_i=1^n|f(t)-f(xi_i)|chi_E_i(t)=|f(t)-f(xi_i(t))|<varepsilon$$
for sufficiently large $n$ which is a consequance of uniform continuity of $f$ on compact set $[0,1]$. This imples that
$$lim_ntoinfty|f(t)-f_n(t)|=0$$
for all $tin [0,1]$.
answered Mar 22 at 15:15
zorro47zorro47
631515
$endgroup$
Let $0=t_0le t_1le...le t_n=1$ and suppose that $Delta=max_iin1,...,n(t_i-t_i-1)to 0$ as $nto infty$. Let $E_i=[t_i-1, t_i)$ and for each $i=1,...,n$ fix $xi_iin E_i$. Let us now define
$$f_n(t)=sum_i=1^nf(xi_i)chi_E_i(t)$$
for all $tin [0,1]$ and $nin mathbbN$. Fix $tin [0,1]$, $varepsilon>0$ and observe that
$$|f(t)-f_n(t)| le sum_i=1^n|f(t)-f(xi_i)|chi_E_i(t)=|f(t)-f(xi_i(t))|<varepsilon$$
for sufficiently large $n$ which is a consequance of uniform continuity of $f$ on compact set $[0,1]$. This imples that
$$lim_ntoinfty|f(t)-f_n(t)|=0$$
for all $tin [0,1]$.
answered Mar 22 at 15:15
zorro47zorro47
631515
answered Mar 22 at 15:15
zorro47zorro47
631515
answered Mar 22 at 15:15
zorro47zorro47
631515
answered Mar 22 at 15:15
zorro47zorro47
631515
631515
add a comment |
add a comment |
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1
$begingroup$
If you are new to measure theory, then you should take a beginning course in measure theory, not a course of measure theory in Banach space.
$endgroup$
– TCL
Apr 27 '13 at 18:31
1
$begingroup$
Do the case $I = [0,1]$ first. Use that $f(I)$ is a compact metric space.
$endgroup$
– Martin
Apr 27 '13 at 19:02
$begingroup$
Is this correct: $I=[0,1]$ is compact, hence $f(I)$ is a compact subset of a metric space and therefore separable. By Pettis' theorem it follows that any continuous function is strongly measurable?
$endgroup$
– ScroogeMcDuck
Apr 28 '13 at 11:18
$begingroup$
Correction: any continuous function with compact domain is strongly measurable?
$endgroup$
– ScroogeMcDuck
Apr 28 '13 at 11:24
2
$begingroup$
Yes, this is correct. The general case follows from this since a closed interval is a countable union of compact intervals. // However, you do not need to use Pettis's theorem. Note that $f$ is uniformly continuous on $[0,1]$. Use this to approximate $f$ by step functions. // Your proof of weak measurability is okay, btw. // Please use
@Martinin your future comments so I am notified.$endgroup$
– Martin
Apr 29 '13 at 16:40