Prove Lebesgue integrability
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Let $f_n in L_0(X), |f_n|leqslantphi in L_1(X), n in N$ and $f_n rightarrow f$ in measure. Prove that $f in L_1(X)$ and
$$lim_{ntoinfty}int_{X}{f_n}dmu = int_{X} fdmu$$
Here $L_0(X)$ stands for Lebesgue measurability, and $L_1(X)$ — for Lebesgue integrability on $X$.
As far as I understand, we should use the fact that $f_n$ is bounded by a Lebesgue integrable function being itself measurable, so $f$ is also Lebesgue integrable. Is that correct? I remember similar reasoning being used in the classroom.
If the first part is correct, how do I prove the equality?
measure-theory lebesgue-integral lebesgue-measure
asked Dec 25 '18 at 16:17
Don DraperDon Draper
539
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add a comment |
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Let $f_n in L_0(X), |f_n|leqslantphi in L_1(X), n in N$ and $f_n rightarrow f$ in measure. Prove that $f in L_1(X)$ and
$$lim_{ntoinfty}int_{X}{f_n}dmu = int_{X} fdmu$$
Here $L_0(X)$ stands for Lebesgue measurability, and $L_1(X)$ — for Lebesgue integrability on $X$.
As far as I understand, we should use the fact that $f_n$ is bounded by a Lebesgue integrable function being itself measurable, so $f$ is also Lebesgue integrable. Is that correct? I remember similar reasoning being used in the classroom.
If the first part is correct, how do I prove the equality?
measure-theory lebesgue-integral lebesgue-measure
asked Dec 25 '18 at 16:17
Don DraperDon Draper
539
$endgroup$
add a comment |
$begingroup$
Let $f_n in L_0(X), |f_n|leqslantphi in L_1(X), n in N$ and $f_n rightarrow f$ in measure. Prove that $f in L_1(X)$ and
$$lim_{ntoinfty}int_{X}{f_n}dmu = int_{X} fdmu$$
Here $L_0(X)$ stands for Lebesgue measurability, and $L_1(X)$ — for Lebesgue integrability on $X$.
As far as I understand, we should use the fact that $f_n$ is bounded by a Lebesgue integrable function being itself measurable, so $f$ is also Lebesgue integrable. Is that correct? I remember similar reasoning being used in the classroom.
If the first part is correct, how do I prove the equality?
measure-theory lebesgue-integral lebesgue-measure
asked Dec 25 '18 at 16:17
Don DraperDon Draper
539
$endgroup$
Let $f_n in L_0(X), |f_n|leqslantphi in L_1(X), n in N$ and $f_n rightarrow f$ in measure. Prove that $f in L_1(X)$ and
$$lim_{ntoinfty}int_{X}{f_n}dmu = int_{X} fdmu$$
Here $L_0(X)$ stands for Lebesgue measurability, and $L_1(X)$ — for Lebesgue integrability on $X$.
As far as I understand, we should use the fact that $f_n$ is bounded by a Lebesgue integrable function being itself measurable, so $f$ is also Lebesgue integrable. Is that correct? I remember similar reasoning being used in the classroom.
If the first part is correct, how do I prove the equality?
measure-theory lebesgue-integral lebesgue-measure
measure-theory lebesgue-integral lebesgue-measure
asked Dec 25 '18 at 16:17
Don DraperDon Draper
539
asked Dec 25 '18 at 16:17
Don DraperDon Draper
539
asked Dec 25 '18 at 16:17
Don DraperDon Draper
539
asked Dec 25 '18 at 16:17
Don DraperDon Draper
539
asked Dec 25 '18 at 16:17
Don DraperDon Draper
539
539
add a comment |
add a comment |
1 Answer
1
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oldest
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Let ${f_{n_k}}_{k=1}^infty$ be an arbitrary subsequence of ${f_n}_{n=1}^infty$. Then $f_{n_k} to f$ in measure, so one may extract a further subsequence ${f_{n_{k_j}}}_{j=1}^infty$ of ${f_{n_k}}_{k=1}^infty$ such that $f_{n_{k_j}}to f$ pointwise a.e. in $X$. By Fatou's lemma $fin L_1(X)$; by the dominated convergence theorem $int_X f, dmu = limlimits_{jto infty} int_X f_{n_{k_j}}, dmu$. Since ${f_{n_k}}_{k=1}^infty$ was arbitrary, the result follows.
answered Dec 25 '18 at 17:21
kobekobe
34.9k22248
$endgroup$
$begingroup$
Why does the result follow from $f_{n_k}$ being arbitrary?
$endgroup$
– Guacho Perez
Dec 25 '18 at 19:29
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@Gaucho If every subsequence of a sequence of real numbers converges to some number $A$, then the sequence converges to $A$. In this case, the sequence under consideration is $int_X f_n, dmu$.
$endgroup$
– kobe
Dec 25 '18 at 19:30
$begingroup$
But it is only proved that a subsequence of every subsequence converges to $int f$.
$endgroup$
– Guacho Perez
Dec 25 '18 at 19:45
$begingroup$
@GuachoPerez no, it has been shown that every subsequence of $int_X f_n, dmu$ has a further subsequence which converges to $int_X f, dmu$, so $int_X f_n, dmu to int_X f, dmu$. Indeed, since $int_X f_{n_j}, dmu$ has a subsequence which converges to $int_X f, dmu$, then $int_X f_{n_j}, dmu$ to converges to the same number. But since $int_X f_{n_j}, dmu$ is an arbitrary subsequence of $int_X f_n, dmu$, $lim_n int_X f_n, dmu = int_X f, dmu$.
$endgroup$
– kobe
Dec 25 '18 at 19:53
$begingroup$
@Gaucho take a look here: math.stackexchange.com/questions/397978/….
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– kobe
Dec 25 '18 at 19:55
|
show 1 more comment
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Let ${f_{n_k}}_{k=1}^infty$ be an arbitrary subsequence of ${f_n}_{n=1}^infty$. Then $f_{n_k} to f$ in measure, so one may extract a further subsequence ${f_{n_{k_j}}}_{j=1}^infty$ of ${f_{n_k}}_{k=1}^infty$ such that $f_{n_{k_j}}to f$ pointwise a.e. in $X$. By Fatou's lemma $fin L_1(X)$; by the dominated convergence theorem $int_X f, dmu = limlimits_{jto infty} int_X f_{n_{k_j}}, dmu$. Since ${f_{n_k}}_{k=1}^infty$ was arbitrary, the result follows.
answered Dec 25 '18 at 17:21
kobekobe
34.9k22248
$endgroup$
$begingroup$
Why does the result follow from $f_{n_k}$ being arbitrary?
$endgroup$
– Guacho Perez
Dec 25 '18 at 19:29
$begingroup$
@Gaucho If every subsequence of a sequence of real numbers converges to some number $A$, then the sequence converges to $A$. In this case, the sequence under consideration is $int_X f_n, dmu$.
$endgroup$
– kobe
Dec 25 '18 at 19:30
$begingroup$
But it is only proved that a subsequence of every subsequence converges to $int f$.
$endgroup$
– Guacho Perez
Dec 25 '18 at 19:45
$begingroup$
@GuachoPerez no, it has been shown that every subsequence of $int_X f_n, dmu$ has a further subsequence which converges to $int_X f, dmu$, so $int_X f_n, dmu to int_X f, dmu$. Indeed, since $int_X f_{n_j}, dmu$ has a subsequence which converges to $int_X f, dmu$, then $int_X f_{n_j}, dmu$ to converges to the same number. But since $int_X f_{n_j}, dmu$ is an arbitrary subsequence of $int_X f_n, dmu$, $lim_n int_X f_n, dmu = int_X f, dmu$.
$endgroup$
– kobe
Dec 25 '18 at 19:53
$begingroup$
@Gaucho take a look here: math.stackexchange.com/questions/397978/….
$endgroup$
– kobe
Dec 25 '18 at 19:55
|
show 1 more comment
$begingroup$
Let ${f_{n_k}}_{k=1}^infty$ be an arbitrary subsequence of ${f_n}_{n=1}^infty$. Then $f_{n_k} to f$ in measure, so one may extract a further subsequence ${f_{n_{k_j}}}_{j=1}^infty$ of ${f_{n_k}}_{k=1}^infty$ such that $f_{n_{k_j}}to f$ pointwise a.e. in $X$. By Fatou's lemma $fin L_1(X)$; by the dominated convergence theorem $int_X f, dmu = limlimits_{jto infty} int_X f_{n_{k_j}}, dmu$. Since ${f_{n_k}}_{k=1}^infty$ was arbitrary, the result follows.
answered Dec 25 '18 at 17:21
kobekobe
34.9k22248
$endgroup$
$begingroup$
Why does the result follow from $f_{n_k}$ being arbitrary?
$endgroup$
– Guacho Perez
Dec 25 '18 at 19:29
$begingroup$
@Gaucho If every subsequence of a sequence of real numbers converges to some number $A$, then the sequence converges to $A$. In this case, the sequence under consideration is $int_X f_n, dmu$.
$endgroup$
– kobe
Dec 25 '18 at 19:30
$begingroup$
But it is only proved that a subsequence of every subsequence converges to $int f$.
$endgroup$
– Guacho Perez
Dec 25 '18 at 19:45
$begingroup$
@GuachoPerez no, it has been shown that every subsequence of $int_X f_n, dmu$ has a further subsequence which converges to $int_X f, dmu$, so $int_X f_n, dmu to int_X f, dmu$. Indeed, since $int_X f_{n_j}, dmu$ has a subsequence which converges to $int_X f, dmu$, then $int_X f_{n_j}, dmu$ to converges to the same number. But since $int_X f_{n_j}, dmu$ is an arbitrary subsequence of $int_X f_n, dmu$, $lim_n int_X f_n, dmu = int_X f, dmu$.
$endgroup$
– kobe
Dec 25 '18 at 19:53
$begingroup$
@Gaucho take a look here: math.stackexchange.com/questions/397978/….
$endgroup$
– kobe
Dec 25 '18 at 19:55
|
show 1 more comment
$begingroup$
Let ${f_{n_k}}_{k=1}^infty$ be an arbitrary subsequence of ${f_n}_{n=1}^infty$. Then $f_{n_k} to f$ in measure, so one may extract a further subsequence ${f_{n_{k_j}}}_{j=1}^infty$ of ${f_{n_k}}_{k=1}^infty$ such that $f_{n_{k_j}}to f$ pointwise a.e. in $X$. By Fatou's lemma $fin L_1(X)$; by the dominated convergence theorem $int_X f, dmu = limlimits_{jto infty} int_X f_{n_{k_j}}, dmu$. Since ${f_{n_k}}_{k=1}^infty$ was arbitrary, the result follows.
answered Dec 25 '18 at 17:21
kobekobe
34.9k22248
$endgroup$
Let ${f_{n_k}}_{k=1}^infty$ be an arbitrary subsequence of ${f_n}_{n=1}^infty$. Then $f_{n_k} to f$ in measure, so one may extract a further subsequence ${f_{n_{k_j}}}_{j=1}^infty$ of ${f_{n_k}}_{k=1}^infty$ such that $f_{n_{k_j}}to f$ pointwise a.e. in $X$. By Fatou's lemma $fin L_1(X)$; by the dominated convergence theorem $int_X f, dmu = limlimits_{jto infty} int_X f_{n_{k_j}}, dmu$. Since ${f_{n_k}}_{k=1}^infty$ was arbitrary, the result follows.
answered Dec 25 '18 at 17:21
kobekobe
34.9k22248
answered Dec 25 '18 at 17:21
kobekobe
34.9k22248
answered Dec 25 '18 at 17:21
kobekobe
34.9k22248
answered Dec 25 '18 at 17:21
kobekobe
34.9k22248
34.9k22248
$begingroup$
Why does the result follow from $f_{n_k}$ being arbitrary?
$endgroup$
– Guacho Perez
Dec 25 '18 at 19:29
$begingroup$
@Gaucho If every subsequence of a sequence of real numbers converges to some number $A$, then the sequence converges to $A$. In this case, the sequence under consideration is $int_X f_n, dmu$.
$endgroup$
– kobe
Dec 25 '18 at 19:30
$begingroup$
But it is only proved that a subsequence of every subsequence converges to $int f$.
$endgroup$
– Guacho Perez
Dec 25 '18 at 19:45
$begingroup$
@GuachoPerez no, it has been shown that every subsequence of $int_X f_n, dmu$ has a further subsequence which converges to $int_X f, dmu$, so $int_X f_n, dmu to int_X f, dmu$. Indeed, since $int_X f_{n_j}, dmu$ has a subsequence which converges to $int_X f, dmu$, then $int_X f_{n_j}, dmu$ to converges to the same number. But since $int_X f_{n_j}, dmu$ is an arbitrary subsequence of $int_X f_n, dmu$, $lim_n int_X f_n, dmu = int_X f, dmu$.
$endgroup$
– kobe
Dec 25 '18 at 19:53
$begingroup$
@Gaucho take a look here: math.stackexchange.com/questions/397978/….
$endgroup$
– kobe
Dec 25 '18 at 19:55
|
show 1 more comment
$begingroup$
Why does the result follow from $f_{n_k}$ being arbitrary?
$endgroup$
– Guacho Perez
Dec 25 '18 at 19:29
$begingroup$
@Gaucho If every subsequence of a sequence of real numbers converges to some number $A$, then the sequence converges to $A$. In this case, the sequence under consideration is $int_X f_n, dmu$.
$endgroup$
– kobe
Dec 25 '18 at 19:30
$begingroup$
But it is only proved that a subsequence of every subsequence converges to $int f$.
$endgroup$
– Guacho Perez
Dec 25 '18 at 19:45
$begingroup$
@GuachoPerez no, it has been shown that every subsequence of $int_X f_n, dmu$ has a further subsequence which converges to $int_X f, dmu$, so $int_X f_n, dmu to int_X f, dmu$. Indeed, since $int_X f_{n_j}, dmu$ has a subsequence which converges to $int_X f, dmu$, then $int_X f_{n_j}, dmu$ to converges to the same number. But since $int_X f_{n_j}, dmu$ is an arbitrary subsequence of $int_X f_n, dmu$, $lim_n int_X f_n, dmu = int_X f, dmu$.
$endgroup$
– kobe
Dec 25 '18 at 19:53
$begingroup$
@Gaucho take a look here: math.stackexchange.com/questions/397978/….
$endgroup$
– kobe
Dec 25 '18 at 19:55
$begingroup$
Why does the result follow from $f_{n_k}$ being arbitrary?
$endgroup$
– Guacho Perez
Dec 25 '18 at 19:29
$begingroup$
Why does the result follow from $f_{n_k}$ being arbitrary?
$endgroup$
– Guacho Perez
Dec 25 '18 at 19:29
$begingroup$
@Gaucho If every subsequence of a sequence of real numbers converges to some number $A$, then the sequence converges to $A$. In this case, the sequence under consideration is $int_X f_n, dmu$.
$endgroup$
– kobe
Dec 25 '18 at 19:30
$begingroup$
@Gaucho If every subsequence of a sequence of real numbers converges to some number $A$, then the sequence converges to $A$. In this case, the sequence under consideration is $int_X f_n, dmu$.
$endgroup$
– kobe
Dec 25 '18 at 19:30
$begingroup$
But it is only proved that a subsequence of every subsequence converges to $int f$.
$endgroup$
– Guacho Perez
Dec 25 '18 at 19:45
$begingroup$
But it is only proved that a subsequence of every subsequence converges to $int f$.
$endgroup$
– Guacho Perez
Dec 25 '18 at 19:45
$begingroup$
@GuachoPerez no, it has been shown that every subsequence of $int_X f_n, dmu$ has a further subsequence which converges to $int_X f, dmu$, so $int_X f_n, dmu to int_X f, dmu$. Indeed, since $int_X f_{n_j}, dmu$ has a subsequence which converges to $int_X f, dmu$, then $int_X f_{n_j}, dmu$ to converges to the same number. But since $int_X f_{n_j}, dmu$ is an arbitrary subsequence of $int_X f_n, dmu$, $lim_n int_X f_n, dmu = int_X f, dmu$.
$endgroup$
– kobe
Dec 25 '18 at 19:53
$begingroup$
@GuachoPerez no, it has been shown that every subsequence of $int_X f_n, dmu$ has a further subsequence which converges to $int_X f, dmu$, so $int_X f_n, dmu to int_X f, dmu$. Indeed, since $int_X f_{n_j}, dmu$ has a subsequence which converges to $int_X f, dmu$, then $int_X f_{n_j}, dmu$ to converges to the same number. But since $int_X f_{n_j}, dmu$ is an arbitrary subsequence of $int_X f_n, dmu$, $lim_n int_X f_n, dmu = int_X f, dmu$.
$endgroup$
– kobe
Dec 25 '18 at 19:53
$begingroup$
@Gaucho take a look here: math.stackexchange.com/questions/397978/….
$endgroup$
– kobe
Dec 25 '18 at 19:55
$begingroup$
@Gaucho take a look here: math.stackexchange.com/questions/397978/….
$endgroup$
– kobe
Dec 25 '18 at 19:55
|
show 1 more comment
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