Egorov’s theorem: Small compact set
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Let $K subset mathbb R^n, n in mathbb N$, be compact and suppose the measurable $f_n: K to mathbb R$ converge almost everywhere to a function $f: K to mathbb R$.
By Egorov‘s theorem, for any $varepsilon gt 0$ we may find a measurable $A_varepsilon$ with $|A_varepsilon| lt varepsilon$ and $f_n to f$ uniformly on $K setminus A_varepsilon$.
One can always find an $A_varepsilon$ additionally being open.
Question: Can one also always find a compact set $A_varepsilon$ (satisfying $|A_varepsilon| lt varepsilon$ and $f_n to f$ uniformly on $K setminus A_varepsilon$)?
Sadly, the measure of the closure of an open set may be much larger than the measure of the open set, so I am not really sure where to start here. However, I was also unable to come up with a counter example.
real-analysis measure-theory
asked Jan 22 at 14:33
KebaKeba
1,399618
$endgroup$
|
show 2 more comments
$begingroup$
Let $K subset mathbb R^n, n in mathbb N$, be compact and suppose the measurable $f_n: K to mathbb R$ converge almost everywhere to a function $f: K to mathbb R$.
By Egorov‘s theorem, for any $varepsilon gt 0$ we may find a measurable $A_varepsilon$ with $|A_varepsilon| lt varepsilon$ and $f_n to f$ uniformly on $K setminus A_varepsilon$.
One can always find an $A_varepsilon$ additionally being open.
Question: Can one also always find a compact set $A_varepsilon$ (satisfying $|A_varepsilon| lt varepsilon$ and $f_n to f$ uniformly on $K setminus A_varepsilon$)?
Sadly, the measure of the closure of an open set may be much larger than the measure of the open set, so I am not really sure where to start here. However, I was also unable to come up with a counter example.
real-analysis measure-theory
asked Jan 22 at 14:33
KebaKeba
1,399618
$endgroup$
2
$begingroup$
No you can't hope for that because $A_varepsilon$ may be dense. Think about that you can change countably many points as you wish.
$endgroup$
– Yanko
Jan 22 at 14:36
$begingroup$
@Yanko: Sorry, I fail to imagine an example where all such $A_varepsilon$ are dense – can you help me out?
$endgroup$
– Keba
Jan 22 at 14:40
1
$begingroup$
Choose a dense set of points ${k_n:ninmathbb{N}}$ such that $f_n(k_m)notrightarrow f(k_m)$ ($f_n$ is only defined almost everywhere so changing $f_n$ so that $f_n(k_m)=m$ will do the trick). If you want I can write more details but there's no much more than that.
$endgroup$
– Yanko
Jan 22 at 14:41
$begingroup$
Ah, that is indeed an easy example; I agree. However, that does not really answer the question I had in mind: For instance, is there more hope if we require $f_n to f$ pointwise (instead of almost everywhere)?
$endgroup$
– Keba
Jan 22 at 14:48
$begingroup$
It doesn't matter much. Instead of $f_n(k_m)=m$ let $f_n(k_m)=1-frac{m}{n}$ so it convergence pointwise to $1$ but not uniformly on all $k_m$. An interesting question would be whether you can find $Bsubseteq A_varepsilon$ of measure zero such that $A_varepsilonbackslash B$ is compact (I don't know the answer to that one, but I'd guess it's a no).
$endgroup$
– Yanko
Jan 22 at 14:49
|
show 2 more comments
$begingroup$
Let $K subset mathbb R^n, n in mathbb N$, be compact and suppose the measurable $f_n: K to mathbb R$ converge almost everywhere to a function $f: K to mathbb R$.
By Egorov‘s theorem, for any $varepsilon gt 0$ we may find a measurable $A_varepsilon$ with $|A_varepsilon| lt varepsilon$ and $f_n to f$ uniformly on $K setminus A_varepsilon$.
One can always find an $A_varepsilon$ additionally being open.
Question: Can one also always find a compact set $A_varepsilon$ (satisfying $|A_varepsilon| lt varepsilon$ and $f_n to f$ uniformly on $K setminus A_varepsilon$)?
Sadly, the measure of the closure of an open set may be much larger than the measure of the open set, so I am not really sure where to start here. However, I was also unable to come up with a counter example.
real-analysis measure-theory
asked Jan 22 at 14:33
KebaKeba
1,399618
$endgroup$
Let $K subset mathbb R^n, n in mathbb N$, be compact and suppose the measurable $f_n: K to mathbb R$ converge almost everywhere to a function $f: K to mathbb R$.
By Egorov‘s theorem, for any $varepsilon gt 0$ we may find a measurable $A_varepsilon$ with $|A_varepsilon| lt varepsilon$ and $f_n to f$ uniformly on $K setminus A_varepsilon$.
One can always find an $A_varepsilon$ additionally being open.
Question: Can one also always find a compact set $A_varepsilon$ (satisfying $|A_varepsilon| lt varepsilon$ and $f_n to f$ uniformly on $K setminus A_varepsilon$)?
Sadly, the measure of the closure of an open set may be much larger than the measure of the open set, so I am not really sure where to start here. However, I was also unable to come up with a counter example.
real-analysis measure-theory
real-analysis measure-theory
asked Jan 22 at 14:33
KebaKeba
1,399618
asked Jan 22 at 14:33
KebaKeba
1,399618
asked Jan 22 at 14:33
KebaKeba
1,399618
asked Jan 22 at 14:33
KebaKeba
1,399618
asked Jan 22 at 14:33
KebaKeba
1,399618
1,399618
2
$begingroup$
No you can't hope for that because $A_varepsilon$ may be dense. Think about that you can change countably many points as you wish.
$endgroup$
– Yanko
Jan 22 at 14:36
$begingroup$
@Yanko: Sorry, I fail to imagine an example where all such $A_varepsilon$ are dense – can you help me out?
$endgroup$
– Keba
Jan 22 at 14:40
1
$begingroup$
Choose a dense set of points ${k_n:ninmathbb{N}}$ such that $f_n(k_m)notrightarrow f(k_m)$ ($f_n$ is only defined almost everywhere so changing $f_n$ so that $f_n(k_m)=m$ will do the trick). If you want I can write more details but there's no much more than that.
$endgroup$
– Yanko
Jan 22 at 14:41
$begingroup$
Ah, that is indeed an easy example; I agree. However, that does not really answer the question I had in mind: For instance, is there more hope if we require $f_n to f$ pointwise (instead of almost everywhere)?
$endgroup$
– Keba
Jan 22 at 14:48
$begingroup$
It doesn't matter much. Instead of $f_n(k_m)=m$ let $f_n(k_m)=1-frac{m}{n}$ so it convergence pointwise to $1$ but not uniformly on all $k_m$. An interesting question would be whether you can find $Bsubseteq A_varepsilon$ of measure zero such that $A_varepsilonbackslash B$ is compact (I don't know the answer to that one, but I'd guess it's a no).
$endgroup$
– Yanko
Jan 22 at 14:49
|
show 2 more comments
2
$begingroup$
No you can't hope for that because $A_varepsilon$ may be dense. Think about that you can change countably many points as you wish.
$endgroup$
– Yanko
Jan 22 at 14:36
$begingroup$
@Yanko: Sorry, I fail to imagine an example where all such $A_varepsilon$ are dense – can you help me out?
$endgroup$
– Keba
Jan 22 at 14:40
1
$begingroup$
Choose a dense set of points ${k_n:ninmathbb{N}}$ such that $f_n(k_m)notrightarrow f(k_m)$ ($f_n$ is only defined almost everywhere so changing $f_n$ so that $f_n(k_m)=m$ will do the trick). If you want I can write more details but there's no much more than that.
$endgroup$
– Yanko
Jan 22 at 14:41
$begingroup$
Ah, that is indeed an easy example; I agree. However, that does not really answer the question I had in mind: For instance, is there more hope if we require $f_n to f$ pointwise (instead of almost everywhere)?
$endgroup$
– Keba
Jan 22 at 14:48
$begingroup$
It doesn't matter much. Instead of $f_n(k_m)=m$ let $f_n(k_m)=1-frac{m}{n}$ so it convergence pointwise to $1$ but not uniformly on all $k_m$. An interesting question would be whether you can find $Bsubseteq A_varepsilon$ of measure zero such that $A_varepsilonbackslash B$ is compact (I don't know the answer to that one, but I'd guess it's a no).
$endgroup$
– Yanko
Jan 22 at 14:49
2
2
$begingroup$
No you can't hope for that because $A_varepsilon$ may be dense. Think about that you can change countably many points as you wish.
$endgroup$
– Yanko
Jan 22 at 14:36
$begingroup$
No you can't hope for that because $A_varepsilon$ may be dense. Think about that you can change countably many points as you wish.
$endgroup$
– Yanko
Jan 22 at 14:36
$begingroup$
@Yanko: Sorry, I fail to imagine an example where all such $A_varepsilon$ are dense – can you help me out?
$endgroup$
– Keba
Jan 22 at 14:40
$begingroup$
@Yanko: Sorry, I fail to imagine an example where all such $A_varepsilon$ are dense – can you help me out?
$endgroup$
– Keba
Jan 22 at 14:40
1
1
$begingroup$
Choose a dense set of points ${k_n:ninmathbb{N}}$ such that $f_n(k_m)notrightarrow f(k_m)$ ($f_n$ is only defined almost everywhere so changing $f_n$ so that $f_n(k_m)=m$ will do the trick). If you want I can write more details but there's no much more than that.
$endgroup$
– Yanko
Jan 22 at 14:41
$begingroup$
Choose a dense set of points ${k_n:ninmathbb{N}}$ such that $f_n(k_m)notrightarrow f(k_m)$ ($f_n$ is only defined almost everywhere so changing $f_n$ so that $f_n(k_m)=m$ will do the trick). If you want I can write more details but there's no much more than that.
$endgroup$
– Yanko
Jan 22 at 14:41
$begingroup$
Ah, that is indeed an easy example; I agree. However, that does not really answer the question I had in mind: For instance, is there more hope if we require $f_n to f$ pointwise (instead of almost everywhere)?
$endgroup$
– Keba
Jan 22 at 14:48
$begingroup$
Ah, that is indeed an easy example; I agree. However, that does not really answer the question I had in mind: For instance, is there more hope if we require $f_n to f$ pointwise (instead of almost everywhere)?
$endgroup$
– Keba
Jan 22 at 14:48
$begingroup$
It doesn't matter much. Instead of $f_n(k_m)=m$ let $f_n(k_m)=1-frac{m}{n}$ so it convergence pointwise to $1$ but not uniformly on all $k_m$. An interesting question would be whether you can find $Bsubseteq A_varepsilon$ of measure zero such that $A_varepsilonbackslash B$ is compact (I don't know the answer to that one, but I'd guess it's a no).
$endgroup$
– Yanko
Jan 22 at 14:49
$begingroup$
It doesn't matter much. Instead of $f_n(k_m)=m$ let $f_n(k_m)=1-frac{m}{n}$ so it convergence pointwise to $1$ but not uniformly on all $k_m$. An interesting question would be whether you can find $Bsubseteq A_varepsilon$ of measure zero such that $A_varepsilonbackslash B$ is compact (I don't know the answer to that one, but I'd guess it's a no).
$endgroup$
– Yanko
Jan 22 at 14:49
|
show 2 more comments
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$begingroup$
No you can't hope for that because $A_varepsilon$ may be dense. Think about that you can change countably many points as you wish.
$endgroup$
– Yanko
Jan 22 at 14:36
$begingroup$
@Yanko: Sorry, I fail to imagine an example where all such $A_varepsilon$ are dense – can you help me out?
$endgroup$
– Keba
Jan 22 at 14:40
1
$begingroup$
Choose a dense set of points ${k_n:ninmathbb{N}}$ such that $f_n(k_m)notrightarrow f(k_m)$ ($f_n$ is only defined almost everywhere so changing $f_n$ so that $f_n(k_m)=m$ will do the trick). If you want I can write more details but there's no much more than that.
$endgroup$
– Yanko
Jan 22 at 14:41
$begingroup$
Ah, that is indeed an easy example; I agree. However, that does not really answer the question I had in mind: For instance, is there more hope if we require $f_n to f$ pointwise (instead of almost everywhere)?
$endgroup$
– Keba
Jan 22 at 14:48
$begingroup$
It doesn't matter much. Instead of $f_n(k_m)=m$ let $f_n(k_m)=1-frac{m}{n}$ so it convergence pointwise to $1$ but not uniformly on all $k_m$. An interesting question would be whether you can find $Bsubseteq A_varepsilon$ of measure zero such that $A_varepsilonbackslash B$ is compact (I don't know the answer to that one, but I'd guess it's a no).
$endgroup$
– Yanko
Jan 22 at 14:49