Bound on $L^1$ norm of $limsup$ gives convergence in measure?
$begingroup$
Suppose ${f_n}$ is a sequence of nonnegative measurable functions on $[0,1]$ such that $f_n(x)leq K$ for all $nin mathbb{N}$ and $xin [0,1]$. Let $f=limsup f_n$ and assume that $||f_n||_1geq ||f||_1$ for each $n$. Then, is it true that $f_nrightarrow f$ in measure?
It is true via Fatou's Lemma and the uniform bound that
$$int f(x) ;dx geq limsup int f_n(x); dx geq liminf int f_n(x);dxgeq int f(x) ; dx,$$
so we get $lim int f_n(x)dx=int f(x)$. This is about where I get stuck. Would it be useful to look at the sets
$$A_n={xin [0,1]: f_n(x)>f(x)+epsilon} qquad text{and} qquad B_n={xin [0,1]:f_n(x)<f(x)-epsilon},$$
since we need to show $lambda(A_ncup B_n)rightarrow 0$ as $nrightarrow infty$? Or is there a better way? I would appreciate any hints or solutions.
real-analysis measure-theory convergence
asked Dec 2 '18 at 18:20
confused_walletconfused_wallet
478311
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add a comment |
$begingroup$
Suppose ${f_n}$ is a sequence of nonnegative measurable functions on $[0,1]$ such that $f_n(x)leq K$ for all $nin mathbb{N}$ and $xin [0,1]$. Let $f=limsup f_n$ and assume that $||f_n||_1geq ||f||_1$ for each $n$. Then, is it true that $f_nrightarrow f$ in measure?
It is true via Fatou's Lemma and the uniform bound that
$$int f(x) ;dx geq limsup int f_n(x); dx geq liminf int f_n(x);dxgeq int f(x) ; dx,$$
so we get $lim int f_n(x)dx=int f(x)$. This is about where I get stuck. Would it be useful to look at the sets
$$A_n={xin [0,1]: f_n(x)>f(x)+epsilon} qquad text{and} qquad B_n={xin [0,1]:f_n(x)<f(x)-epsilon},$$
since we need to show $lambda(A_ncup B_n)rightarrow 0$ as $nrightarrow infty$? Or is there a better way? I would appreciate any hints or solutions.
real-analysis measure-theory convergence
asked Dec 2 '18 at 18:20
confused_walletconfused_wallet
478311
$endgroup$
add a comment |
$begingroup$
Suppose ${f_n}$ is a sequence of nonnegative measurable functions on $[0,1]$ such that $f_n(x)leq K$ for all $nin mathbb{N}$ and $xin [0,1]$. Let $f=limsup f_n$ and assume that $||f_n||_1geq ||f||_1$ for each $n$. Then, is it true that $f_nrightarrow f$ in measure?
It is true via Fatou's Lemma and the uniform bound that
$$int f(x) ;dx geq limsup int f_n(x); dx geq liminf int f_n(x);dxgeq int f(x) ; dx,$$
so we get $lim int f_n(x)dx=int f(x)$. This is about where I get stuck. Would it be useful to look at the sets
$$A_n={xin [0,1]: f_n(x)>f(x)+epsilon} qquad text{and} qquad B_n={xin [0,1]:f_n(x)<f(x)-epsilon},$$
since we need to show $lambda(A_ncup B_n)rightarrow 0$ as $nrightarrow infty$? Or is there a better way? I would appreciate any hints or solutions.
real-analysis measure-theory convergence
asked Dec 2 '18 at 18:20
confused_walletconfused_wallet
478311
$endgroup$
Suppose ${f_n}$ is a sequence of nonnegative measurable functions on $[0,1]$ such that $f_n(x)leq K$ for all $nin mathbb{N}$ and $xin [0,1]$. Let $f=limsup f_n$ and assume that $||f_n||_1geq ||f||_1$ for each $n$. Then, is it true that $f_nrightarrow f$ in measure?
It is true via Fatou's Lemma and the uniform bound that
$$int f(x) ;dx geq limsup int f_n(x); dx geq liminf int f_n(x);dxgeq int f(x) ; dx,$$
so we get $lim int f_n(x)dx=int f(x)$. This is about where I get stuck. Would it be useful to look at the sets
$$A_n={xin [0,1]: f_n(x)>f(x)+epsilon} qquad text{and} qquad B_n={xin [0,1]:f_n(x)<f(x)-epsilon},$$
since we need to show $lambda(A_ncup B_n)rightarrow 0$ as $nrightarrow infty$? Or is there a better way? I would appreciate any hints or solutions.
real-analysis measure-theory convergence
real-analysis measure-theory convergence
asked Dec 2 '18 at 18:20
confused_walletconfused_wallet
478311
asked Dec 2 '18 at 18:20
confused_walletconfused_wallet
478311
asked Dec 2 '18 at 18:20
confused_walletconfused_wallet
478311
asked Dec 2 '18 at 18:20
confused_walletconfused_wallet
478311
asked Dec 2 '18 at 18:20
confused_walletconfused_wallet
478311
478311
add a comment |
add a comment |
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