If $f(x,u(x))$ measuable with conditions
$begingroup$
recently, I've encountered a question about measurable function. after I've searched in MSE, it's been asked in a similar way before, but curiously the answer is the same as I've considered to be wrong. the link is here, so what is the right answer?
copy the question as follows:
Given $f(cdot, y)$ is measurable for each $y$, $f(x, cdot)$ is
continuous for each $x$. If $u(t)$ is continuous, how can I show that
the function $f:[0,1]times mathbb{R} rightarrow mathbb{R}$ defined by $f(t, u(t))$ is
measurable?
The problem of the answer lies in "The continuous of $f$ is unknown", $f(x,u)$ is continuous only for specified $x$, not for all variables. Although the answer is correct in its own way, but no use for the question.
real-analysis functional-analysis measure-theory measurable-functions
asked Dec 13 '18 at 13:11
Larry EppesLarry Eppes
471311
$endgroup$
add a comment |
$begingroup$
recently, I've encountered a question about measurable function. after I've searched in MSE, it's been asked in a similar way before, but curiously the answer is the same as I've considered to be wrong. the link is here, so what is the right answer?
copy the question as follows:
Given $f(cdot, y)$ is measurable for each $y$, $f(x, cdot)$ is
continuous for each $x$. If $u(t)$ is continuous, how can I show that
the function $f:[0,1]times mathbb{R} rightarrow mathbb{R}$ defined by $f(t, u(t))$ is
measurable?
The problem of the answer lies in "The continuous of $f$ is unknown", $f(x,u)$ is continuous only for specified $x$, not for all variables. Although the answer is correct in its own way, but no use for the question.
real-analysis functional-analysis measure-theory measurable-functions
asked Dec 13 '18 at 13:11
Larry EppesLarry Eppes
471311
$endgroup$
add a comment |
$begingroup$
recently, I've encountered a question about measurable function. after I've searched in MSE, it's been asked in a similar way before, but curiously the answer is the same as I've considered to be wrong. the link is here, so what is the right answer?
copy the question as follows:
Given $f(cdot, y)$ is measurable for each $y$, $f(x, cdot)$ is
continuous for each $x$. If $u(t)$ is continuous, how can I show that
the function $f:[0,1]times mathbb{R} rightarrow mathbb{R}$ defined by $f(t, u(t))$ is
measurable?
The problem of the answer lies in "The continuous of $f$ is unknown", $f(x,u)$ is continuous only for specified $x$, not for all variables. Although the answer is correct in its own way, but no use for the question.
real-analysis functional-analysis measure-theory measurable-functions
asked Dec 13 '18 at 13:11
Larry EppesLarry Eppes
471311
$endgroup$
recently, I've encountered a question about measurable function. after I've searched in MSE, it's been asked in a similar way before, but curiously the answer is the same as I've considered to be wrong. the link is here, so what is the right answer?
copy the question as follows:
Given $f(cdot, y)$ is measurable for each $y$, $f(x, cdot)$ is
continuous for each $x$. If $u(t)$ is continuous, how can I show that
the function $f:[0,1]times mathbb{R} rightarrow mathbb{R}$ defined by $f(t, u(t))$ is
measurable?
The problem of the answer lies in "The continuous of $f$ is unknown", $f(x,u)$ is continuous only for specified $x$, not for all variables. Although the answer is correct in its own way, but no use for the question.
real-analysis functional-analysis measure-theory measurable-functions
real-analysis functional-analysis measure-theory measurable-functions
asked Dec 13 '18 at 13:11
Larry EppesLarry Eppes
471311
asked Dec 13 '18 at 13:11
Larry EppesLarry Eppes
471311
asked Dec 13 '18 at 13:11
Larry EppesLarry Eppes
471311
asked Dec 13 '18 at 13:11
Larry EppesLarry Eppes
471311
asked Dec 13 '18 at 13:11
Larry EppesLarry Eppes
471311
471311
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To put this into a little context, a function $f:Omegatimes Bbb R^ntoBbb R$ such that $f(cdot,a)$ is measurable for almost every $xinOmega$ and that $f(x,cdot)$ is continuous is call a Caratheodory function. This kind of function is commonly encountered in Calculus of Variation.
The proof usually goes like this: Let's define
$$
g(x) := f(x,u(x)),
$$
where $u$ is a simple function
$$
u(x)=sum_{i=1}^m a_i mathbf 1_{E_i}(x)
$$
where $E_i$ is a measurable set (I know $u$ is not continuous but please bear with me for a second). For any $tinBbb R$, observe that
$$
{xinOmega : g(x)<t } = bigcup_{i=1}^m { xin E_i : f(x,a_i)<t }.
$$
Since we know that $xmapsto f(x,a_i)$ is measurable, each set on the right hand side is measurable which implies that $g=f(cdot,u(cdot))$ is a measurable function.
In general, let $u$ be any measurable function then we can find a sequence $u_n$ of simple functions such that $u_nto u$ almost everywhere. By continuity of $amapsto f(x,a)$ (almost every $x$), we have
$$
lim_{ntoinfty} f(x,u_n(x)) = f(x,u(x)) = g(x)
$$
for almost every $x$. This shows that $g$ is a the (almost everywhere) limit of a sequence of measurable functions, hence $g$ is itself measurable.
The case where $u$ is continuous follows immediately.
answered Dec 13 '18 at 16:45
BigbearZzzBigbearZzz
8,67621652
$endgroup$
$begingroup$
+1 for your help, ok I also missed the almost everywhere convergence measurable function converges a measurable function, good tips in proving measurable function.
$endgroup$
– Larry Eppes
Dec 13 '18 at 17:39
$begingroup$
@LarryEppes Glad I could help :)
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:40
$begingroup$
I checked again the condition in the beginning of your answer, I think $f(x,cdot)$ is continuous for almost every $xinOmega$ could be meaningful, and $f(cdot, a)$ is measurable for all $a$. Am I right?
$endgroup$
– Larry Eppes
Dec 13 '18 at 17:43
$begingroup$
@LarryEppes Yes, they're the same conditions as the ones in your question.
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:45
$begingroup$
I am searching on Caratheodory function and thinking about your final statement, when I reached math.stackexchange.com/questions/393111/…, in the answer let $y=x sin(1/x)$, it seem a counterexample of your final line.
$endgroup$
– Larry Eppes
Dec 13 '18 at 19:23
|
show 2 more comments
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$begingroup$
To put this into a little context, a function $f:Omegatimes Bbb R^ntoBbb R$ such that $f(cdot,a)$ is measurable for almost every $xinOmega$ and that $f(x,cdot)$ is continuous is call a Caratheodory function. This kind of function is commonly encountered in Calculus of Variation.
The proof usually goes like this: Let's define
$$
g(x) := f(x,u(x)),
$$
where $u$ is a simple function
$$
u(x)=sum_{i=1}^m a_i mathbf 1_{E_i}(x)
$$
where $E_i$ is a measurable set (I know $u$ is not continuous but please bear with me for a second). For any $tinBbb R$, observe that
$$
{xinOmega : g(x)<t } = bigcup_{i=1}^m { xin E_i : f(x,a_i)<t }.
$$
Since we know that $xmapsto f(x,a_i)$ is measurable, each set on the right hand side is measurable which implies that $g=f(cdot,u(cdot))$ is a measurable function.
In general, let $u$ be any measurable function then we can find a sequence $u_n$ of simple functions such that $u_nto u$ almost everywhere. By continuity of $amapsto f(x,a)$ (almost every $x$), we have
$$
lim_{ntoinfty} f(x,u_n(x)) = f(x,u(x)) = g(x)
$$
for almost every $x$. This shows that $g$ is a the (almost everywhere) limit of a sequence of measurable functions, hence $g$ is itself measurable.
The case where $u$ is continuous follows immediately.
answered Dec 13 '18 at 16:45
BigbearZzzBigbearZzz
8,67621652
$endgroup$
$begingroup$
+1 for your help, ok I also missed the almost everywhere convergence measurable function converges a measurable function, good tips in proving measurable function.
$endgroup$
– Larry Eppes
Dec 13 '18 at 17:39
$begingroup$
@LarryEppes Glad I could help :)
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:40
$begingroup$
I checked again the condition in the beginning of your answer, I think $f(x,cdot)$ is continuous for almost every $xinOmega$ could be meaningful, and $f(cdot, a)$ is measurable for all $a$. Am I right?
$endgroup$
– Larry Eppes
Dec 13 '18 at 17:43
$begingroup$
@LarryEppes Yes, they're the same conditions as the ones in your question.
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:45
$begingroup$
I am searching on Caratheodory function and thinking about your final statement, when I reached math.stackexchange.com/questions/393111/…, in the answer let $y=x sin(1/x)$, it seem a counterexample of your final line.
$endgroup$
– Larry Eppes
Dec 13 '18 at 19:23
|
show 2 more comments
$begingroup$
To put this into a little context, a function $f:Omegatimes Bbb R^ntoBbb R$ such that $f(cdot,a)$ is measurable for almost every $xinOmega$ and that $f(x,cdot)$ is continuous is call a Caratheodory function. This kind of function is commonly encountered in Calculus of Variation.
The proof usually goes like this: Let's define
$$
g(x) := f(x,u(x)),
$$
where $u$ is a simple function
$$
u(x)=sum_{i=1}^m a_i mathbf 1_{E_i}(x)
$$
where $E_i$ is a measurable set (I know $u$ is not continuous but please bear with me for a second). For any $tinBbb R$, observe that
$$
{xinOmega : g(x)<t } = bigcup_{i=1}^m { xin E_i : f(x,a_i)<t }.
$$
Since we know that $xmapsto f(x,a_i)$ is measurable, each set on the right hand side is measurable which implies that $g=f(cdot,u(cdot))$ is a measurable function.
In general, let $u$ be any measurable function then we can find a sequence $u_n$ of simple functions such that $u_nto u$ almost everywhere. By continuity of $amapsto f(x,a)$ (almost every $x$), we have
$$
lim_{ntoinfty} f(x,u_n(x)) = f(x,u(x)) = g(x)
$$
for almost every $x$. This shows that $g$ is a the (almost everywhere) limit of a sequence of measurable functions, hence $g$ is itself measurable.
The case where $u$ is continuous follows immediately.
answered Dec 13 '18 at 16:45
BigbearZzzBigbearZzz
8,67621652
$endgroup$
$begingroup$
+1 for your help, ok I also missed the almost everywhere convergence measurable function converges a measurable function, good tips in proving measurable function.
$endgroup$
– Larry Eppes
Dec 13 '18 at 17:39
$begingroup$
@LarryEppes Glad I could help :)
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:40
$begingroup$
I checked again the condition in the beginning of your answer, I think $f(x,cdot)$ is continuous for almost every $xinOmega$ could be meaningful, and $f(cdot, a)$ is measurable for all $a$. Am I right?
$endgroup$
– Larry Eppes
Dec 13 '18 at 17:43
$begingroup$
@LarryEppes Yes, they're the same conditions as the ones in your question.
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:45
$begingroup$
I am searching on Caratheodory function and thinking about your final statement, when I reached math.stackexchange.com/questions/393111/…, in the answer let $y=x sin(1/x)$, it seem a counterexample of your final line.
$endgroup$
– Larry Eppes
Dec 13 '18 at 19:23
|
show 2 more comments
$begingroup$
To put this into a little context, a function $f:Omegatimes Bbb R^ntoBbb R$ such that $f(cdot,a)$ is measurable for almost every $xinOmega$ and that $f(x,cdot)$ is continuous is call a Caratheodory function. This kind of function is commonly encountered in Calculus of Variation.
The proof usually goes like this: Let's define
$$
g(x) := f(x,u(x)),
$$
where $u$ is a simple function
$$
u(x)=sum_{i=1}^m a_i mathbf 1_{E_i}(x)
$$
where $E_i$ is a measurable set (I know $u$ is not continuous but please bear with me for a second). For any $tinBbb R$, observe that
$$
{xinOmega : g(x)<t } = bigcup_{i=1}^m { xin E_i : f(x,a_i)<t }.
$$
Since we know that $xmapsto f(x,a_i)$ is measurable, each set on the right hand side is measurable which implies that $g=f(cdot,u(cdot))$ is a measurable function.
In general, let $u$ be any measurable function then we can find a sequence $u_n$ of simple functions such that $u_nto u$ almost everywhere. By continuity of $amapsto f(x,a)$ (almost every $x$), we have
$$
lim_{ntoinfty} f(x,u_n(x)) = f(x,u(x)) = g(x)
$$
for almost every $x$. This shows that $g$ is a the (almost everywhere) limit of a sequence of measurable functions, hence $g$ is itself measurable.
The case where $u$ is continuous follows immediately.
answered Dec 13 '18 at 16:45
BigbearZzzBigbearZzz
8,67621652
$endgroup$
To put this into a little context, a function $f:Omegatimes Bbb R^ntoBbb R$ such that $f(cdot,a)$ is measurable for almost every $xinOmega$ and that $f(x,cdot)$ is continuous is call a Caratheodory function. This kind of function is commonly encountered in Calculus of Variation.
The proof usually goes like this: Let's define
$$
g(x) := f(x,u(x)),
$$
where $u$ is a simple function
$$
u(x)=sum_{i=1}^m a_i mathbf 1_{E_i}(x)
$$
where $E_i$ is a measurable set (I know $u$ is not continuous but please bear with me for a second). For any $tinBbb R$, observe that
$$
{xinOmega : g(x)<t } = bigcup_{i=1}^m { xin E_i : f(x,a_i)<t }.
$$
Since we know that $xmapsto f(x,a_i)$ is measurable, each set on the right hand side is measurable which implies that $g=f(cdot,u(cdot))$ is a measurable function.
In general, let $u$ be any measurable function then we can find a sequence $u_n$ of simple functions such that $u_nto u$ almost everywhere. By continuity of $amapsto f(x,a)$ (almost every $x$), we have
$$
lim_{ntoinfty} f(x,u_n(x)) = f(x,u(x)) = g(x)
$$
for almost every $x$. This shows that $g$ is a the (almost everywhere) limit of a sequence of measurable functions, hence $g$ is itself measurable.
The case where $u$ is continuous follows immediately.
answered Dec 13 '18 at 16:45
BigbearZzzBigbearZzz
8,67621652
edited Dec 14 '18 at 2:32
answered Dec 13 '18 at 16:45
BigbearZzzBigbearZzz
8,67621652
answered Dec 13 '18 at 16:45
BigbearZzzBigbearZzz
8,67621652
answered Dec 13 '18 at 16:45
BigbearZzzBigbearZzz
8,67621652
8,67621652
$begingroup$
+1 for your help, ok I also missed the almost everywhere convergence measurable function converges a measurable function, good tips in proving measurable function.
$endgroup$
– Larry Eppes
Dec 13 '18 at 17:39
$begingroup$
@LarryEppes Glad I could help :)
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:40
$begingroup$
I checked again the condition in the beginning of your answer, I think $f(x,cdot)$ is continuous for almost every $xinOmega$ could be meaningful, and $f(cdot, a)$ is measurable for all $a$. Am I right?
$endgroup$
– Larry Eppes
Dec 13 '18 at 17:43
$begingroup$
@LarryEppes Yes, they're the same conditions as the ones in your question.
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:45
$begingroup$
I am searching on Caratheodory function and thinking about your final statement, when I reached math.stackexchange.com/questions/393111/…, in the answer let $y=x sin(1/x)$, it seem a counterexample of your final line.
$endgroup$
– Larry Eppes
Dec 13 '18 at 19:23
|
show 2 more comments
$begingroup$
+1 for your help, ok I also missed the almost everywhere convergence measurable function converges a measurable function, good tips in proving measurable function.
$endgroup$
– Larry Eppes
Dec 13 '18 at 17:39
$begingroup$
@LarryEppes Glad I could help :)
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:40
$begingroup$
I checked again the condition in the beginning of your answer, I think $f(x,cdot)$ is continuous for almost every $xinOmega$ could be meaningful, and $f(cdot, a)$ is measurable for all $a$. Am I right?
$endgroup$
– Larry Eppes
Dec 13 '18 at 17:43
$begingroup$
@LarryEppes Yes, they're the same conditions as the ones in your question.
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:45
$begingroup$
I am searching on Caratheodory function and thinking about your final statement, when I reached math.stackexchange.com/questions/393111/…, in the answer let $y=x sin(1/x)$, it seem a counterexample of your final line.
$endgroup$
– Larry Eppes
Dec 13 '18 at 19:23
$begingroup$
+1 for your help, ok I also missed the almost everywhere convergence measurable function converges a measurable function, good tips in proving measurable function.
$endgroup$
– Larry Eppes
Dec 13 '18 at 17:39
$begingroup$
+1 for your help, ok I also missed the almost everywhere convergence measurable function converges a measurable function, good tips in proving measurable function.
$endgroup$
– Larry Eppes
Dec 13 '18 at 17:39
$begingroup$
@LarryEppes Glad I could help :)
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:40
$begingroup$
@LarryEppes Glad I could help :)
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:40
$begingroup$
I checked again the condition in the beginning of your answer, I think $f(x,cdot)$ is continuous for almost every $xinOmega$ could be meaningful, and $f(cdot, a)$ is measurable for all $a$. Am I right?
$endgroup$
– Larry Eppes
Dec 13 '18 at 17:43
$begingroup$
I checked again the condition in the beginning of your answer, I think $f(x,cdot)$ is continuous for almost every $xinOmega$ could be meaningful, and $f(cdot, a)$ is measurable for all $a$. Am I right?
$endgroup$
– Larry Eppes
Dec 13 '18 at 17:43
$begingroup$
@LarryEppes Yes, they're the same conditions as the ones in your question.
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:45
$begingroup$
@LarryEppes Yes, they're the same conditions as the ones in your question.
$endgroup$
– BigbearZzz
Dec 13 '18 at 17:45
$begingroup$
I am searching on Caratheodory function and thinking about your final statement, when I reached math.stackexchange.com/questions/393111/…, in the answer let $y=x sin(1/x)$, it seem a counterexample of your final line.
$endgroup$
– Larry Eppes
Dec 13 '18 at 19:23
$begingroup$
I am searching on Caratheodory function and thinking about your final statement, when I reached math.stackexchange.com/questions/393111/…, in the answer let $y=x sin(1/x)$, it seem a counterexample of your final line.
$endgroup$
– Larry Eppes
Dec 13 '18 at 19:23
|
show 2 more comments
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