Showing a sequence of functions $f_n$ does not converge uniformly to $f$ on an interval.
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Suppose for each $n in mathbb{N}$ we have a function $f_n:[0,1] to [0,1]$ by $f_n(x)=nx$ on the interval $x in [0,frac{1}{n}]$ and $1$ if $x in (frac{1}{n},1]$, and define $f=lim_{n to infty} f_n$.
I want to show that for any Lebesgue measurable $B subseteq [0,1]$ with Lebesgue measure $lambda(B)=0$ that the functions $f_n$ do not converge to $f$ uniformly on $[0,1] backslash B$.
Now clearly we have the the limit $f$ is equal to $0$ at $x=0$ and is equal to $1$ elsewhere, but I am struggling to figure out how to implement the fact that we remove the zero measure interval $B$.
Firstly my thought was to use the uniform convergence theorem, which would work with showing that the functions $f_n$ do not converge uniformly to $f$ as each of the functions $f_n$ is continuous we can use the uniform limit theorem to show that $f_n$ does not converge to $f$.
I am wondering whether there is a simple result that I am missing that ensures this transfers to the same sort of result when we take away a set of measure $0$, for example does continuity of each $f_n$ still hold in this case in which I can still apply the theorem?
Any insight would be much appreciated thanks :)
measure-theory lebesgue-measure
asked Dec 6 '18 at 0:21
MattMatt
16810
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add a comment |
$begingroup$
Suppose for each $n in mathbb{N}$ we have a function $f_n:[0,1] to [0,1]$ by $f_n(x)=nx$ on the interval $x in [0,frac{1}{n}]$ and $1$ if $x in (frac{1}{n},1]$, and define $f=lim_{n to infty} f_n$.
I want to show that for any Lebesgue measurable $B subseteq [0,1]$ with Lebesgue measure $lambda(B)=0$ that the functions $f_n$ do not converge to $f$ uniformly on $[0,1] backslash B$.
Now clearly we have the the limit $f$ is equal to $0$ at $x=0$ and is equal to $1$ elsewhere, but I am struggling to figure out how to implement the fact that we remove the zero measure interval $B$.
Firstly my thought was to use the uniform convergence theorem, which would work with showing that the functions $f_n$ do not converge uniformly to $f$ as each of the functions $f_n$ is continuous we can use the uniform limit theorem to show that $f_n$ does not converge to $f$.
I am wondering whether there is a simple result that I am missing that ensures this transfers to the same sort of result when we take away a set of measure $0$, for example does continuity of each $f_n$ still hold in this case in which I can still apply the theorem?
Any insight would be much appreciated thanks :)
measure-theory lebesgue-measure
asked Dec 6 '18 at 0:21
MattMatt
16810
$endgroup$
add a comment |
$begingroup$
Suppose for each $n in mathbb{N}$ we have a function $f_n:[0,1] to [0,1]$ by $f_n(x)=nx$ on the interval $x in [0,frac{1}{n}]$ and $1$ if $x in (frac{1}{n},1]$, and define $f=lim_{n to infty} f_n$.
I want to show that for any Lebesgue measurable $B subseteq [0,1]$ with Lebesgue measure $lambda(B)=0$ that the functions $f_n$ do not converge to $f$ uniformly on $[0,1] backslash B$.
Now clearly we have the the limit $f$ is equal to $0$ at $x=0$ and is equal to $1$ elsewhere, but I am struggling to figure out how to implement the fact that we remove the zero measure interval $B$.
Firstly my thought was to use the uniform convergence theorem, which would work with showing that the functions $f_n$ do not converge uniformly to $f$ as each of the functions $f_n$ is continuous we can use the uniform limit theorem to show that $f_n$ does not converge to $f$.
I am wondering whether there is a simple result that I am missing that ensures this transfers to the same sort of result when we take away a set of measure $0$, for example does continuity of each $f_n$ still hold in this case in which I can still apply the theorem?
Any insight would be much appreciated thanks :)
measure-theory lebesgue-measure
asked Dec 6 '18 at 0:21
MattMatt
16810
$endgroup$
Suppose for each $n in mathbb{N}$ we have a function $f_n:[0,1] to [0,1]$ by $f_n(x)=nx$ on the interval $x in [0,frac{1}{n}]$ and $1$ if $x in (frac{1}{n},1]$, and define $f=lim_{n to infty} f_n$.
I want to show that for any Lebesgue measurable $B subseteq [0,1]$ with Lebesgue measure $lambda(B)=0$ that the functions $f_n$ do not converge to $f$ uniformly on $[0,1] backslash B$.
Now clearly we have the the limit $f$ is equal to $0$ at $x=0$ and is equal to $1$ elsewhere, but I am struggling to figure out how to implement the fact that we remove the zero measure interval $B$.
Firstly my thought was to use the uniform convergence theorem, which would work with showing that the functions $f_n$ do not converge uniformly to $f$ as each of the functions $f_n$ is continuous we can use the uniform limit theorem to show that $f_n$ does not converge to $f$.
I am wondering whether there is a simple result that I am missing that ensures this transfers to the same sort of result when we take away a set of measure $0$, for example does continuity of each $f_n$ still hold in this case in which I can still apply the theorem?
Any insight would be much appreciated thanks :)
measure-theory lebesgue-measure
measure-theory lebesgue-measure
asked Dec 6 '18 at 0:21
MattMatt
16810
asked Dec 6 '18 at 0:21
MattMatt
16810
asked Dec 6 '18 at 0:21
MattMatt
16810
asked Dec 6 '18 at 0:21
MattMatt
16810
asked Dec 6 '18 at 0:21
MattMatt
16810
16810
add a comment |
add a comment |
1 Answer
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Take $B subset [0,1]$ of measure zero and set $l = inf (0,1]setminus B$. If it were $l > 0$, then $B$ would contain some neighbourhood of $0$ (maybe without $0$ itself) which contradicts that $|B| = 0$. Hence $l = 0$ and thus we have a sequence $x_n to 0$ contained in $(0,1] setminus B$. Without loss of generality we can assume that $0 < x_n leq frac{1}{2n}$. Thus, $f_n(x_n) leq frac{1}{2}$ and in particular,
$$
d_{[0,1] setminus B}(f,f_n) geq |f(x_n) - f_n(x_n)| = f(x_n) - f_n(x_n) geq f(x_n) - frac{1}{2} stackrel{(x_n neq 0)}{=} frac{1}{2}.
$$
Taking limits, we get that $lim_n d_{[0,1] setminus B}(f,f_n) geq frac{1}{2} > 0$ as desired.
Edit: note that we only used that $B$ can't contain a neighbourhood of zero, which is weaker than being of measure zero. Thus uniform convergence fails for a (in some sense) larger family of sets.
answered Dec 6 '18 at 0:46
Guido A.Guido A.
7,2861730
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1 Answer
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$begingroup$
Take $B subset [0,1]$ of measure zero and set $l = inf (0,1]setminus B$. If it were $l > 0$, then $B$ would contain some neighbourhood of $0$ (maybe without $0$ itself) which contradicts that $|B| = 0$. Hence $l = 0$ and thus we have a sequence $x_n to 0$ contained in $(0,1] setminus B$. Without loss of generality we can assume that $0 < x_n leq frac{1}{2n}$. Thus, $f_n(x_n) leq frac{1}{2}$ and in particular,
$$
d_{[0,1] setminus B}(f,f_n) geq |f(x_n) - f_n(x_n)| = f(x_n) - f_n(x_n) geq f(x_n) - frac{1}{2} stackrel{(x_n neq 0)}{=} frac{1}{2}.
$$
Taking limits, we get that $lim_n d_{[0,1] setminus B}(f,f_n) geq frac{1}{2} > 0$ as desired.
Edit: note that we only used that $B$ can't contain a neighbourhood of zero, which is weaker than being of measure zero. Thus uniform convergence fails for a (in some sense) larger family of sets.
answered Dec 6 '18 at 0:46
Guido A.Guido A.
7,2861730
$endgroup$
add a comment |
$begingroup$
Take $B subset [0,1]$ of measure zero and set $l = inf (0,1]setminus B$. If it were $l > 0$, then $B$ would contain some neighbourhood of $0$ (maybe without $0$ itself) which contradicts that $|B| = 0$. Hence $l = 0$ and thus we have a sequence $x_n to 0$ contained in $(0,1] setminus B$. Without loss of generality we can assume that $0 < x_n leq frac{1}{2n}$. Thus, $f_n(x_n) leq frac{1}{2}$ and in particular,
$$
d_{[0,1] setminus B}(f,f_n) geq |f(x_n) - f_n(x_n)| = f(x_n) - f_n(x_n) geq f(x_n) - frac{1}{2} stackrel{(x_n neq 0)}{=} frac{1}{2}.
$$
Taking limits, we get that $lim_n d_{[0,1] setminus B}(f,f_n) geq frac{1}{2} > 0$ as desired.
Edit: note that we only used that $B$ can't contain a neighbourhood of zero, which is weaker than being of measure zero. Thus uniform convergence fails for a (in some sense) larger family of sets.
answered Dec 6 '18 at 0:46
Guido A.Guido A.
7,2861730
$endgroup$
add a comment |
$begingroup$
Take $B subset [0,1]$ of measure zero and set $l = inf (0,1]setminus B$. If it were $l > 0$, then $B$ would contain some neighbourhood of $0$ (maybe without $0$ itself) which contradicts that $|B| = 0$. Hence $l = 0$ and thus we have a sequence $x_n to 0$ contained in $(0,1] setminus B$. Without loss of generality we can assume that $0 < x_n leq frac{1}{2n}$. Thus, $f_n(x_n) leq frac{1}{2}$ and in particular,
$$
d_{[0,1] setminus B}(f,f_n) geq |f(x_n) - f_n(x_n)| = f(x_n) - f_n(x_n) geq f(x_n) - frac{1}{2} stackrel{(x_n neq 0)}{=} frac{1}{2}.
$$
Taking limits, we get that $lim_n d_{[0,1] setminus B}(f,f_n) geq frac{1}{2} > 0$ as desired.
Edit: note that we only used that $B$ can't contain a neighbourhood of zero, which is weaker than being of measure zero. Thus uniform convergence fails for a (in some sense) larger family of sets.
answered Dec 6 '18 at 0:46
Guido A.Guido A.
7,2861730
$endgroup$
Take $B subset [0,1]$ of measure zero and set $l = inf (0,1]setminus B$. If it were $l > 0$, then $B$ would contain some neighbourhood of $0$ (maybe without $0$ itself) which contradicts that $|B| = 0$. Hence $l = 0$ and thus we have a sequence $x_n to 0$ contained in $(0,1] setminus B$. Without loss of generality we can assume that $0 < x_n leq frac{1}{2n}$. Thus, $f_n(x_n) leq frac{1}{2}$ and in particular,
$$
d_{[0,1] setminus B}(f,f_n) geq |f(x_n) - f_n(x_n)| = f(x_n) - f_n(x_n) geq f(x_n) - frac{1}{2} stackrel{(x_n neq 0)}{=} frac{1}{2}.
$$
Taking limits, we get that $lim_n d_{[0,1] setminus B}(f,f_n) geq frac{1}{2} > 0$ as desired.
Edit: note that we only used that $B$ can't contain a neighbourhood of zero, which is weaker than being of measure zero. Thus uniform convergence fails for a (in some sense) larger family of sets.
answered Dec 6 '18 at 0:46
Guido A.Guido A.
7,2861730
edited Dec 6 '18 at 1:03
answered Dec 6 '18 at 0:46
Guido A.Guido A.
7,2861730
answered Dec 6 '18 at 0:46
Guido A.Guido A.
7,2861730
answered Dec 6 '18 at 0:46
Guido A.Guido A.
7,2861730
7,2861730
add a comment |
add a comment |
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