a.e. point wise convergence on finite measured set implies convergence in measure
$begingroup$
Let $f,f_k$ for $k in mathbb{N}$ be measurable finite a.e. on measurable set $E$. Suppose $f_k rightarrow f$ point wise a.e., then $f_k$ converges to $f$ in measure on $E$.
This is a basic result and the proof follows from basic definitions I believe, I want to know why we need finite valued and finite measure, is it for the obvious reasons that infinite measure does not converge and finite valued so that the limit as $k$ tends to $infty$ makes sense?
because converging pointwise a.e. means that for $x in E$ with $E$ having positive finite measure, one has
lim$_{k rightarrow infty}$ $f_k(x) = f(x)$
so
lim$_{k rightarrow infty}$ $f_k(x) - f(x)$ = $0$,
and we wish to show for any $epsilon > 0$ that
$m({x : vert f_k(x) - f(x) vert > epsilon }) rightarrow 0$ as $k rightarrow infty$
and since $x$ is taken out of some set of finite positive measure, we obtain (and I am not sure this works)
$m({x : vert 0 vert > epsilon }) rightarrow 0$ as $k rightarrow infty$
I get confused here, So sorry for being naive in measure theory, I apologize and thank you in advance! if there is any simple small subtle silly errors I am making, any intuitive assumptions I am incorrectly making please let me know, I am very eager to strengthen my measure theory skills.
real-analysis measure-theory convergence lebesgue-measure
asked Dec 4 '18 at 22:29
eyeheartmatheyeheartmath
747
$endgroup$
add a comment |
$begingroup$
Let $f,f_k$ for $k in mathbb{N}$ be measurable finite a.e. on measurable set $E$. Suppose $f_k rightarrow f$ point wise a.e., then $f_k$ converges to $f$ in measure on $E$.
This is a basic result and the proof follows from basic definitions I believe, I want to know why we need finite valued and finite measure, is it for the obvious reasons that infinite measure does not converge and finite valued so that the limit as $k$ tends to $infty$ makes sense?
because converging pointwise a.e. means that for $x in E$ with $E$ having positive finite measure, one has
lim$_{k rightarrow infty}$ $f_k(x) = f(x)$
so
lim$_{k rightarrow infty}$ $f_k(x) - f(x)$ = $0$,
and we wish to show for any $epsilon > 0$ that
$m({x : vert f_k(x) - f(x) vert > epsilon }) rightarrow 0$ as $k rightarrow infty$
and since $x$ is taken out of some set of finite positive measure, we obtain (and I am not sure this works)
$m({x : vert 0 vert > epsilon }) rightarrow 0$ as $k rightarrow infty$
I get confused here, So sorry for being naive in measure theory, I apologize and thank you in advance! if there is any simple small subtle silly errors I am making, any intuitive assumptions I am incorrectly making please let me know, I am very eager to strengthen my measure theory skills.
real-analysis measure-theory convergence lebesgue-measure
asked Dec 4 '18 at 22:29
eyeheartmatheyeheartmath
747
$endgroup$
$begingroup$
there are a lot of ways to show this, depending on the theoretic background. You can show that $(f_j)to f$ point-wise a.e. implies that $(f_j)to f$ almost uniformly when the space have finite measure. And from here that $(f_j)to f$ almost uniformly implies that $(f_j)to f$ in measure
$endgroup$
– Masacroso
Dec 4 '18 at 22:40
$begingroup$
is almost uniform like saying uniform almost everywhere? or is it a slightly weaker notion of uniform convergence?
$endgroup$
– eyeheartmath
Dec 4 '18 at 23:04
1
$begingroup$
a sequence of measurable functions converges almost uniformly in the measure space $(X,mu)$ if, for any chosen $epsilon>0$, there is some measurable $A_epsilonsubset X$ such that $(f_n)to f$ converges uniformly in $A_epsilon$ and $mu(A_epsilon^complement)<epsilon$. However this doesn't mean that $(f_n)to f$ uniformly a.e.!
$endgroup$
– Masacroso
Dec 4 '18 at 23:11
add a comment |
$begingroup$
Let $f,f_k$ for $k in mathbb{N}$ be measurable finite a.e. on measurable set $E$. Suppose $f_k rightarrow f$ point wise a.e., then $f_k$ converges to $f$ in measure on $E$.
This is a basic result and the proof follows from basic definitions I believe, I want to know why we need finite valued and finite measure, is it for the obvious reasons that infinite measure does not converge and finite valued so that the limit as $k$ tends to $infty$ makes sense?
because converging pointwise a.e. means that for $x in E$ with $E$ having positive finite measure, one has
lim$_{k rightarrow infty}$ $f_k(x) = f(x)$
so
lim$_{k rightarrow infty}$ $f_k(x) - f(x)$ = $0$,
and we wish to show for any $epsilon > 0$ that
$m({x : vert f_k(x) - f(x) vert > epsilon }) rightarrow 0$ as $k rightarrow infty$
and since $x$ is taken out of some set of finite positive measure, we obtain (and I am not sure this works)
$m({x : vert 0 vert > epsilon }) rightarrow 0$ as $k rightarrow infty$
I get confused here, So sorry for being naive in measure theory, I apologize and thank you in advance! if there is any simple small subtle silly errors I am making, any intuitive assumptions I am incorrectly making please let me know, I am very eager to strengthen my measure theory skills.
real-analysis measure-theory convergence lebesgue-measure
asked Dec 4 '18 at 22:29
eyeheartmatheyeheartmath
747
$endgroup$
Let $f,f_k$ for $k in mathbb{N}$ be measurable finite a.e. on measurable set $E$. Suppose $f_k rightarrow f$ point wise a.e., then $f_k$ converges to $f$ in measure on $E$.
This is a basic result and the proof follows from basic definitions I believe, I want to know why we need finite valued and finite measure, is it for the obvious reasons that infinite measure does not converge and finite valued so that the limit as $k$ tends to $infty$ makes sense?
because converging pointwise a.e. means that for $x in E$ with $E$ having positive finite measure, one has
lim$_{k rightarrow infty}$ $f_k(x) = f(x)$
so
lim$_{k rightarrow infty}$ $f_k(x) - f(x)$ = $0$,
and we wish to show for any $epsilon > 0$ that
$m({x : vert f_k(x) - f(x) vert > epsilon }) rightarrow 0$ as $k rightarrow infty$
and since $x$ is taken out of some set of finite positive measure, we obtain (and I am not sure this works)
$m({x : vert 0 vert > epsilon }) rightarrow 0$ as $k rightarrow infty$
I get confused here, So sorry for being naive in measure theory, I apologize and thank you in advance! if there is any simple small subtle silly errors I am making, any intuitive assumptions I am incorrectly making please let me know, I am very eager to strengthen my measure theory skills.
real-analysis measure-theory convergence lebesgue-measure
real-analysis measure-theory convergence lebesgue-measure
asked Dec 4 '18 at 22:29
eyeheartmatheyeheartmath
747
asked Dec 4 '18 at 22:29
eyeheartmatheyeheartmath
747
asked Dec 4 '18 at 22:29
eyeheartmatheyeheartmath
747
asked Dec 4 '18 at 22:29
eyeheartmatheyeheartmath
747
asked Dec 4 '18 at 22:29
eyeheartmatheyeheartmath
747
747
$begingroup$
there are a lot of ways to show this, depending on the theoretic background. You can show that $(f_j)to f$ point-wise a.e. implies that $(f_j)to f$ almost uniformly when the space have finite measure. And from here that $(f_j)to f$ almost uniformly implies that $(f_j)to f$ in measure
$endgroup$
– Masacroso
Dec 4 '18 at 22:40
$begingroup$
is almost uniform like saying uniform almost everywhere? or is it a slightly weaker notion of uniform convergence?
$endgroup$
– eyeheartmath
Dec 4 '18 at 23:04
1
$begingroup$
a sequence of measurable functions converges almost uniformly in the measure space $(X,mu)$ if, for any chosen $epsilon>0$, there is some measurable $A_epsilonsubset X$ such that $(f_n)to f$ converges uniformly in $A_epsilon$ and $mu(A_epsilon^complement)<epsilon$. However this doesn't mean that $(f_n)to f$ uniformly a.e.!
$endgroup$
– Masacroso
Dec 4 '18 at 23:11
add a comment |
$begingroup$
there are a lot of ways to show this, depending on the theoretic background. You can show that $(f_j)to f$ point-wise a.e. implies that $(f_j)to f$ almost uniformly when the space have finite measure. And from here that $(f_j)to f$ almost uniformly implies that $(f_j)to f$ in measure
$endgroup$
– Masacroso
Dec 4 '18 at 22:40
$begingroup$
is almost uniform like saying uniform almost everywhere? or is it a slightly weaker notion of uniform convergence?
$endgroup$
– eyeheartmath
Dec 4 '18 at 23:04
1
$begingroup$
a sequence of measurable functions converges almost uniformly in the measure space $(X,mu)$ if, for any chosen $epsilon>0$, there is some measurable $A_epsilonsubset X$ such that $(f_n)to f$ converges uniformly in $A_epsilon$ and $mu(A_epsilon^complement)<epsilon$. However this doesn't mean that $(f_n)to f$ uniformly a.e.!
$endgroup$
– Masacroso
Dec 4 '18 at 23:11
$begingroup$
there are a lot of ways to show this, depending on the theoretic background. You can show that $(f_j)to f$ point-wise a.e. implies that $(f_j)to f$ almost uniformly when the space have finite measure. And from here that $(f_j)to f$ almost uniformly implies that $(f_j)to f$ in measure
$endgroup$
– Masacroso
Dec 4 '18 at 22:40
$begingroup$
there are a lot of ways to show this, depending on the theoretic background. You can show that $(f_j)to f$ point-wise a.e. implies that $(f_j)to f$ almost uniformly when the space have finite measure. And from here that $(f_j)to f$ almost uniformly implies that $(f_j)to f$ in measure
$endgroup$
– Masacroso
Dec 4 '18 at 22:40
$begingroup$
is almost uniform like saying uniform almost everywhere? or is it a slightly weaker notion of uniform convergence?
$endgroup$
– eyeheartmath
Dec 4 '18 at 23:04
$begingroup$
is almost uniform like saying uniform almost everywhere? or is it a slightly weaker notion of uniform convergence?
$endgroup$
– eyeheartmath
Dec 4 '18 at 23:04
1
1
$begingroup$
a sequence of measurable functions converges almost uniformly in the measure space $(X,mu)$ if, for any chosen $epsilon>0$, there is some measurable $A_epsilonsubset X$ such that $(f_n)to f$ converges uniformly in $A_epsilon$ and $mu(A_epsilon^complement)<epsilon$. However this doesn't mean that $(f_n)to f$ uniformly a.e.!
$endgroup$
– Masacroso
Dec 4 '18 at 23:11
$begingroup$
a sequence of measurable functions converges almost uniformly in the measure space $(X,mu)$ if, for any chosen $epsilon>0$, there is some measurable $A_epsilonsubset X$ such that $(f_n)to f$ converges uniformly in $A_epsilon$ and $mu(A_epsilon^complement)<epsilon$. However this doesn't mean that $(f_n)to f$ uniformly a.e.!
$endgroup$
– Masacroso
Dec 4 '18 at 23:11
add a comment |
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$begingroup$
there are a lot of ways to show this, depending on the theoretic background. You can show that $(f_j)to f$ point-wise a.e. implies that $(f_j)to f$ almost uniformly when the space have finite measure. And from here that $(f_j)to f$ almost uniformly implies that $(f_j)to f$ in measure
$endgroup$
– Masacroso
Dec 4 '18 at 22:40
$begingroup$
is almost uniform like saying uniform almost everywhere? or is it a slightly weaker notion of uniform convergence?
$endgroup$
– eyeheartmath
Dec 4 '18 at 23:04
1
$begingroup$
a sequence of measurable functions converges almost uniformly in the measure space $(X,mu)$ if, for any chosen $epsilon>0$, there is some measurable $A_epsilonsubset X$ such that $(f_n)to f$ converges uniformly in $A_epsilon$ and $mu(A_epsilon^complement)<epsilon$. However this doesn't mean that $(f_n)to f$ uniformly a.e.!
$endgroup$
– Masacroso
Dec 4 '18 at 23:11