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The measurability of $f(x)$ given that $lim f_n(x) = f(x)$.
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I have difficulty in interpreting the equality "$x in X : sup_nin N f_n(x) > alpha = cup_n=1^infty x in X : f_n(x) > alpha $". Is this right to say that $sup_nin Nf_n(x)$ is the largest $f_n(x)$, so it contains $x in X$ the most on the condition given, so it is equal to the union of sets?
Furthermore, it is quite difficult for me to understand the proof for $f^*, F^*$. Could you elaborate on either of one?
Thank you in advance.
measure-theory
asked Jul 31 at 10:50
Sihyun Kim
701210
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up vote
1
down vote
favorite
I have difficulty in interpreting the equality "$x in X : sup_nin N f_n(x) > alpha = cup_n=1^infty x in X : f_n(x) > alpha $". Is this right to say that $sup_nin Nf_n(x)$ is the largest $f_n(x)$, so it contains $x in X$ the most on the condition given, so it is equal to the union of sets?
Furthermore, it is quite difficult for me to understand the proof for $f^*, F^*$. Could you elaborate on either of one?
Thank you in advance.
measure-theory
asked Jul 31 at 10:50
Sihyun Kim
701210
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have difficulty in interpreting the equality "$x in X : sup_nin N f_n(x) > alpha = cup_n=1^infty x in X : f_n(x) > alpha $". Is this right to say that $sup_nin Nf_n(x)$ is the largest $f_n(x)$, so it contains $x in X$ the most on the condition given, so it is equal to the union of sets?
Furthermore, it is quite difficult for me to understand the proof for $f^*, F^*$. Could you elaborate on either of one?
Thank you in advance.
measure-theory
asked Jul 31 at 10:50
Sihyun Kim
701210
I have difficulty in interpreting the equality "$x in X : sup_nin N f_n(x) > alpha = cup_n=1^infty x in X : f_n(x) > alpha $". Is this right to say that $sup_nin Nf_n(x)$ is the largest $f_n(x)$, so it contains $x in X$ the most on the condition given, so it is equal to the union of sets?
Furthermore, it is quite difficult for me to understand the proof for $f^*, F^*$. Could you elaborate on either of one?
Thank you in advance.
measure-theory
asked Jul 31 at 10:50
Sihyun Kim
701210
asked Jul 31 at 10:50
Sihyun Kim
701210
asked Jul 31 at 10:50
Sihyun Kim
701210
asked Jul 31 at 10:50
Sihyun Kim
701210
701210
add a comment |Â
add a comment |Â
1 Answer
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Let $z in cup_n=1^infty x in X : f_n(x) > alpha $. This means that for some $m in mathbb N$, $f_m(z) > alpha$, which implies that $sup_n f_n(z) > alpha$. In other words, $$ bigcup_n=1^infty x in X : f_n(x) > alpha subseteq x in X : sup_n f_n(x) > alpha .$$
To see the reverse inclusion, let $z in x in X : sup_n f_n(x) > alpha $. This implies that there is a sequence $ m_i _i=1^infty subseteq mathbb N$ such that
$$ lim_i to infty f_m_i (z) > alpha. $$
In particular, this means that $f_m_i(z) > alpha$ for all large enough $i$. Thus, there exists an $n in mathbb N$ such that $f_n(z) > alpha$. This proves the reverse inclusion.
Now let $g_n(x) = inf_m ge n f_m (x)$. The previous argument shows that $g_n(x)$ is measurable. The same reasoning shows that $f^*(x) = sup_n g_n(x)$ is measurable. The exact same argument applies to $F^*$.
Note that the displayed equalities for $f^*$ and $F^*$ are equivalent definitions for $limsup$ and $liminf$.
answered Jul 31 at 11:07
Theoretical Economist
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let $z in cup_n=1^infty x in X : f_n(x) > alpha $. This means that for some $m in mathbb N$, $f_m(z) > alpha$, which implies that $sup_n f_n(z) > alpha$. In other words, $$ bigcup_n=1^infty x in X : f_n(x) > alpha subseteq x in X : sup_n f_n(x) > alpha .$$
To see the reverse inclusion, let $z in x in X : sup_n f_n(x) > alpha $. This implies that there is a sequence $ m_i _i=1^infty subseteq mathbb N$ such that
$$ lim_i to infty f_m_i (z) > alpha. $$
In particular, this means that $f_m_i(z) > alpha$ for all large enough $i$. Thus, there exists an $n in mathbb N$ such that $f_n(z) > alpha$. This proves the reverse inclusion.
Now let $g_n(x) = inf_m ge n f_m (x)$. The previous argument shows that $g_n(x)$ is measurable. The same reasoning shows that $f^*(x) = sup_n g_n(x)$ is measurable. The exact same argument applies to $F^*$.
Note that the displayed equalities for $f^*$ and $F^*$ are equivalent definitions for $limsup$ and $liminf$.
answered Jul 31 at 11:07
Theoretical Economist
3,1352626
add a comment |Â
up vote
1
down vote
Let $z in cup_n=1^infty x in X : f_n(x) > alpha $. This means that for some $m in mathbb N$, $f_m(z) > alpha$, which implies that $sup_n f_n(z) > alpha$. In other words, $$ bigcup_n=1^infty x in X : f_n(x) > alpha subseteq x in X : sup_n f_n(x) > alpha .$$
To see the reverse inclusion, let $z in x in X : sup_n f_n(x) > alpha $. This implies that there is a sequence $ m_i _i=1^infty subseteq mathbb N$ such that
$$ lim_i to infty f_m_i (z) > alpha. $$
In particular, this means that $f_m_i(z) > alpha$ for all large enough $i$. Thus, there exists an $n in mathbb N$ such that $f_n(z) > alpha$. This proves the reverse inclusion.
Now let $g_n(x) = inf_m ge n f_m (x)$. The previous argument shows that $g_n(x)$ is measurable. The same reasoning shows that $f^*(x) = sup_n g_n(x)$ is measurable. The exact same argument applies to $F^*$.
Note that the displayed equalities for $f^*$ and $F^*$ are equivalent definitions for $limsup$ and $liminf$.
answered Jul 31 at 11:07
Theoretical Economist
3,1352626
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $z in cup_n=1^infty x in X : f_n(x) > alpha $. This means that for some $m in mathbb N$, $f_m(z) > alpha$, which implies that $sup_n f_n(z) > alpha$. In other words, $$ bigcup_n=1^infty x in X : f_n(x) > alpha subseteq x in X : sup_n f_n(x) > alpha .$$
To see the reverse inclusion, let $z in x in X : sup_n f_n(x) > alpha $. This implies that there is a sequence $ m_i _i=1^infty subseteq mathbb N$ such that
$$ lim_i to infty f_m_i (z) > alpha. $$
In particular, this means that $f_m_i(z) > alpha$ for all large enough $i$. Thus, there exists an $n in mathbb N$ such that $f_n(z) > alpha$. This proves the reverse inclusion.
Now let $g_n(x) = inf_m ge n f_m (x)$. The previous argument shows that $g_n(x)$ is measurable. The same reasoning shows that $f^*(x) = sup_n g_n(x)$ is measurable. The exact same argument applies to $F^*$.
Note that the displayed equalities for $f^*$ and $F^*$ are equivalent definitions for $limsup$ and $liminf$.
answered Jul 31 at 11:07
Theoretical Economist
3,1352626
Let $z in cup_n=1^infty x in X : f_n(x) > alpha $. This means that for some $m in mathbb N$, $f_m(z) > alpha$, which implies that $sup_n f_n(z) > alpha$. In other words, $$ bigcup_n=1^infty x in X : f_n(x) > alpha subseteq x in X : sup_n f_n(x) > alpha .$$
To see the reverse inclusion, let $z in x in X : sup_n f_n(x) > alpha $. This implies that there is a sequence $ m_i _i=1^infty subseteq mathbb N$ such that
$$ lim_i to infty f_m_i (z) > alpha. $$
In particular, this means that $f_m_i(z) > alpha$ for all large enough $i$. Thus, there exists an $n in mathbb N$ such that $f_n(z) > alpha$. This proves the reverse inclusion.
Now let $g_n(x) = inf_m ge n f_m (x)$. The previous argument shows that $g_n(x)$ is measurable. The same reasoning shows that $f^*(x) = sup_n g_n(x)$ is measurable. The exact same argument applies to $F^*$.
Note that the displayed equalities for $f^*$ and $F^*$ are equivalent definitions for $limsup$ and $liminf$.
answered Jul 31 at 11:07
Theoretical Economist
3,1352626
answered Jul 31 at 11:07
Theoretical Economist
3,1352626
answered Jul 31 at 11:07
Theoretical Economist
3,1352626
answered Jul 31 at 11:07
Theoretical Economist
3,1352626
3,1352626
add a comment |Â
add a comment |Â
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