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?










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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?










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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?










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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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      asked Sep 5 at 11:21









      Joey Doey

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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:$$mu(f_nto f)=mu(bigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f)$$






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            1 Answer
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            active

            oldest

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            1 Answer
            1






            active

            oldest

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            active

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            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:$$mu(f_nto f)=mu(bigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f)$$






            share|cite|improve this answer
























              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:$$mu(f_nto f)=mu(bigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f)$$






              share|cite|improve this answer






















                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:$$mu(f_nto f)=mu(bigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f)$$






                share|cite|improve this answer












                $xinbigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f$ 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:$$mu(f_nto f)=mu(bigcap_epsiloninmathbb Q^+bigcup_ninmathbb Nbigcap_mgeq nf_m-f)$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 5 at 11:47









                drhab

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                89.1k541122



























                     

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