Why does $mu(f_n to f)=mu(bigcap_epsilon in mathbb Q^+bigcup_ninmathbb Nbigcap_mge nmid f_m-fmid<epsilon)$ holds?
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If $f_n to f$ a.e, i.e. $forall epsilon >0 exists n_0inmathbb N ,forall xin N^c forall nge n_0:mid f_n(x)-f(x)mid <epsilon$.
Why does $mu(f_n to f)=mu(bigcap_epsilon in mathbb Q^+bigcup_ninmathbb Nbigcap_mge nmid f_m-fmid<epsilon)$ holds?
real-analysis convergence
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If $f_n to f$ a.e, i.e. $forall epsilon >0 exists n_0inmathbb N ,forall xin N^c forall nge n_0:mid f_n(x)-f(x)mid <epsilon$.
Why does $mu(f_n to f)=mu(bigcap_epsilon in mathbb Q^+bigcup_ninmathbb Nbigcap_mge nmid f_m-fmid<epsilon)$ holds?
real-analysis convergence
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up vote
0
down vote
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up vote
0
down vote
favorite
If $f_n to f$ a.e, i.e. $forall epsilon >0 exists n_0inmathbb N ,forall xin N^c forall nge n_0:mid f_n(x)-f(x)mid <epsilon$.
Why does $mu(f_n to f)=mu(bigcap_epsilon in mathbb Q^+bigcup_ninmathbb Nbigcap_mge nmid f_m-fmid<epsilon)$ holds?
real-analysis convergence
If $f_n to f$ a.e, i.e. $forall epsilon >0 exists n_0inmathbb N ,forall xin N^c forall nge n_0:mid f_n(x)-f(x)mid <epsilon$.
Why does $mu(f_n to f)=mu(bigcap_epsilon in mathbb Q^+bigcup_ninmathbb Nbigcap_mge nmid f_m-fmid<epsilon)$ holds?
real-analysis convergence
real-analysis convergence
asked Sep 5 at 11:21
Joey Doey
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1 Answer
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$xinbigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f$ is exactly the statement that for every positive $epsiloninmathbb Q$ an integer $n$ exists such that for every $mgeq n$ we have $|f_m(x)-f(x)|<epsilon$.
This is evidently equivalent with the same statement with the only difference that $mathbb Q$ is replaced by $mathbb R$, and that statement on its own is exactly the statement that $lim_ntoinftyf_n(x)=f(x)$ or equivalently $xinf_nto f$.
So we have:$$f_nto f=bigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f$$and consequently:$$mu(f_nto f)=mu(bigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f)$$
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$xinbigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f$ is exactly the statement that for every positive $epsiloninmathbb Q$ an integer $n$ exists such that for every $mgeq n$ we have $|f_m(x)-f(x)|<epsilon$.
This is evidently equivalent with the same statement with the only difference that $mathbb Q$ is replaced by $mathbb R$, and that statement on its own is exactly the statement that $lim_ntoinftyf_n(x)=f(x)$ or equivalently $xinf_nto f$.
So we have:$$f_nto f=bigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f$$and consequently:$$mu(f_nto f)=mu(bigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f)$$
add a comment |Â
up vote
2
down vote
accepted
$xinbigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f$ is exactly the statement that for every positive $epsiloninmathbb Q$ an integer $n$ exists such that for every $mgeq n$ we have $|f_m(x)-f(x)|<epsilon$.
This is evidently equivalent with the same statement with the only difference that $mathbb Q$ is replaced by $mathbb R$, and that statement on its own is exactly the statement that $lim_ntoinftyf_n(x)=f(x)$ or equivalently $xinf_nto f$.
So we have:$$f_nto f=bigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f$$and consequently:$$mu(f_nto f)=mu(bigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f)$$
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$xinbigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f$ is exactly the statement that for every positive $epsiloninmathbb Q$ an integer $n$ exists such that for every $mgeq n$ we have $|f_m(x)-f(x)|<epsilon$.
This is evidently equivalent with the same statement with the only difference that $mathbb Q$ is replaced by $mathbb R$, and that statement on its own is exactly the statement that $lim_ntoinftyf_n(x)=f(x)$ or equivalently $xinf_nto f$.
So we have:$$f_nto f=bigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f$$and consequently:$$mu(f_nto f)=mu(bigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f)$$
$xinbigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f$ is exactly the statement that for every positive $epsiloninmathbb Q$ an integer $n$ exists such that for every $mgeq n$ we have $|f_m(x)-f(x)|<epsilon$.
This is evidently equivalent with the same statement with the only difference that $mathbb Q$ is replaced by $mathbb R$, and that statement on its own is exactly the statement that $lim_ntoinftyf_n(x)=f(x)$ or equivalently $xinf_nto f$.
So we have:$$f_nto f=bigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f$$and consequently:$$mu(f_nto f)=mu(bigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f)$$
answered Sep 5 at 11:47
drhab
89.1k541122
89.1k541122
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