A finite family of integrable functions is uniformly integrable and tight.
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I'm trying to prove a series of questions out of Royden's Real Analysis book. The questions are all extensions of each other.
The first is, Let ${f_j}_{j=1}^n$ be a finite family of functions, which of which is integrable over E. Show that ${f_j}_{k=1}^n$ is uniformly integrable and tight over $E$.
The second problem is an easy extension of the first. It goes let ${h_n}$ be a sequence of nonnegative integrable functions on E. Suppose ${h_n(x)}to 0$ for almost all $x$ in $E$. Then $$lim_{nto infty}int_E h_n = 0$$ iff ${h_n}$ is uniformly integrable and tight over $E$.
Here there is a definition: A family of measurable functions on $E$ is said to be tight over $E$ provided for each $varepsilon>0$, there is a subset $E_0$ of $E$ of finite measure for which $int_{E-E_0} |f|<varepsilon$ for all $f$ in the family.
So there's information on how to solve the second problem (Uniform integrability and tightness.) but I want to solve the first problem.
Anyways, for each of the $j$-many members of the family $f_m$ write $$int_E f_m=sum_{k=-infty}^{infty} int_E u_{m_k},$$ where $f_mchi_{[k,k+1)}=u_{m_k}$ and $kinmathbb{Z}$. Since $f_m$ is integrable on $E$, for any $varepsilon > 0$ there exists an index $N_m$ such that $sum_{k=-infty}^{N_m} int_E u_{m_k} < varepsilon$ (otherwise $sum int u_{m_k}$ diverges.)
I am lucky and have a theorem which gives immediately that the family is uniformly integrable so all I have to prove is tightness.
I have it that $int_E f_n = int_{E-X_0}f_n + int_{X_0}f_n < varepsilon$. I have a list of indices $N_1,N_2,dots,N_n$ which are values in a totally ordered set, i.e., there exists a maximal index say $N^*$. Since $$sum_{k=-infty}^{N^*} int_E u_{m_k} > sum_{k=-infty}^{N_i} int_E u_{m_k} $$ holds for any of the non-maximal indices $N_i$ I can define $X_0$ as $E cap ([-infty, N^*]^c)$. Then $$int_{E-X_0} f_m = sum_{k=-infty}^{N^*}int_E u_{m_k} < varepsilon$$ holds for all $m$ as was desired.
This looks correct?
lebesgue-integral
asked Nov 24 '18 at 1:31
bkbowserbkbowser
11
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add a comment |
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I'm trying to prove a series of questions out of Royden's Real Analysis book. The questions are all extensions of each other.
The first is, Let ${f_j}_{j=1}^n$ be a finite family of functions, which of which is integrable over E. Show that ${f_j}_{k=1}^n$ is uniformly integrable and tight over $E$.
The second problem is an easy extension of the first. It goes let ${h_n}$ be a sequence of nonnegative integrable functions on E. Suppose ${h_n(x)}to 0$ for almost all $x$ in $E$. Then $$lim_{nto infty}int_E h_n = 0$$ iff ${h_n}$ is uniformly integrable and tight over $E$.
Here there is a definition: A family of measurable functions on $E$ is said to be tight over $E$ provided for each $varepsilon>0$, there is a subset $E_0$ of $E$ of finite measure for which $int_{E-E_0} |f|<varepsilon$ for all $f$ in the family.
So there's information on how to solve the second problem (Uniform integrability and tightness.) but I want to solve the first problem.
Anyways, for each of the $j$-many members of the family $f_m$ write $$int_E f_m=sum_{k=-infty}^{infty} int_E u_{m_k},$$ where $f_mchi_{[k,k+1)}=u_{m_k}$ and $kinmathbb{Z}$. Since $f_m$ is integrable on $E$, for any $varepsilon > 0$ there exists an index $N_m$ such that $sum_{k=-infty}^{N_m} int_E u_{m_k} < varepsilon$ (otherwise $sum int u_{m_k}$ diverges.)
I am lucky and have a theorem which gives immediately that the family is uniformly integrable so all I have to prove is tightness.
I have it that $int_E f_n = int_{E-X_0}f_n + int_{X_0}f_n < varepsilon$. I have a list of indices $N_1,N_2,dots,N_n$ which are values in a totally ordered set, i.e., there exists a maximal index say $N^*$. Since $$sum_{k=-infty}^{N^*} int_E u_{m_k} > sum_{k=-infty}^{N_i} int_E u_{m_k} $$ holds for any of the non-maximal indices $N_i$ I can define $X_0$ as $E cap ([-infty, N^*]^c)$. Then $$int_{E-X_0} f_m = sum_{k=-infty}^{N^*}int_E u_{m_k} < varepsilon$$ holds for all $m$ as was desired.
This looks correct?
lebesgue-integral
asked Nov 24 '18 at 1:31
bkbowserbkbowser
11
$endgroup$
add a comment |
$begingroup$
I'm trying to prove a series of questions out of Royden's Real Analysis book. The questions are all extensions of each other.
The first is, Let ${f_j}_{j=1}^n$ be a finite family of functions, which of which is integrable over E. Show that ${f_j}_{k=1}^n$ is uniformly integrable and tight over $E$.
The second problem is an easy extension of the first. It goes let ${h_n}$ be a sequence of nonnegative integrable functions on E. Suppose ${h_n(x)}to 0$ for almost all $x$ in $E$. Then $$lim_{nto infty}int_E h_n = 0$$ iff ${h_n}$ is uniformly integrable and tight over $E$.
Here there is a definition: A family of measurable functions on $E$ is said to be tight over $E$ provided for each $varepsilon>0$, there is a subset $E_0$ of $E$ of finite measure for which $int_{E-E_0} |f|<varepsilon$ for all $f$ in the family.
So there's information on how to solve the second problem (Uniform integrability and tightness.) but I want to solve the first problem.
Anyways, for each of the $j$-many members of the family $f_m$ write $$int_E f_m=sum_{k=-infty}^{infty} int_E u_{m_k},$$ where $f_mchi_{[k,k+1)}=u_{m_k}$ and $kinmathbb{Z}$. Since $f_m$ is integrable on $E$, for any $varepsilon > 0$ there exists an index $N_m$ such that $sum_{k=-infty}^{N_m} int_E u_{m_k} < varepsilon$ (otherwise $sum int u_{m_k}$ diverges.)
I am lucky and have a theorem which gives immediately that the family is uniformly integrable so all I have to prove is tightness.
I have it that $int_E f_n = int_{E-X_0}f_n + int_{X_0}f_n < varepsilon$. I have a list of indices $N_1,N_2,dots,N_n$ which are values in a totally ordered set, i.e., there exists a maximal index say $N^*$. Since $$sum_{k=-infty}^{N^*} int_E u_{m_k} > sum_{k=-infty}^{N_i} int_E u_{m_k} $$ holds for any of the non-maximal indices $N_i$ I can define $X_0$ as $E cap ([-infty, N^*]^c)$. Then $$int_{E-X_0} f_m = sum_{k=-infty}^{N^*}int_E u_{m_k} < varepsilon$$ holds for all $m$ as was desired.
This looks correct?
lebesgue-integral
asked Nov 24 '18 at 1:31
bkbowserbkbowser
11
$endgroup$
I'm trying to prove a series of questions out of Royden's Real Analysis book. The questions are all extensions of each other.
The first is, Let ${f_j}_{j=1}^n$ be a finite family of functions, which of which is integrable over E. Show that ${f_j}_{k=1}^n$ is uniformly integrable and tight over $E$.
The second problem is an easy extension of the first. It goes let ${h_n}$ be a sequence of nonnegative integrable functions on E. Suppose ${h_n(x)}to 0$ for almost all $x$ in $E$. Then $$lim_{nto infty}int_E h_n = 0$$ iff ${h_n}$ is uniformly integrable and tight over $E$.
Here there is a definition: A family of measurable functions on $E$ is said to be tight over $E$ provided for each $varepsilon>0$, there is a subset $E_0$ of $E$ of finite measure for which $int_{E-E_0} |f|<varepsilon$ for all $f$ in the family.
So there's information on how to solve the second problem (Uniform integrability and tightness.) but I want to solve the first problem.
Anyways, for each of the $j$-many members of the family $f_m$ write $$int_E f_m=sum_{k=-infty}^{infty} int_E u_{m_k},$$ where $f_mchi_{[k,k+1)}=u_{m_k}$ and $kinmathbb{Z}$. Since $f_m$ is integrable on $E$, for any $varepsilon > 0$ there exists an index $N_m$ such that $sum_{k=-infty}^{N_m} int_E u_{m_k} < varepsilon$ (otherwise $sum int u_{m_k}$ diverges.)
I am lucky and have a theorem which gives immediately that the family is uniformly integrable so all I have to prove is tightness.
I have it that $int_E f_n = int_{E-X_0}f_n + int_{X_0}f_n < varepsilon$. I have a list of indices $N_1,N_2,dots,N_n$ which are values in a totally ordered set, i.e., there exists a maximal index say $N^*$. Since $$sum_{k=-infty}^{N^*} int_E u_{m_k} > sum_{k=-infty}^{N_i} int_E u_{m_k} $$ holds for any of the non-maximal indices $N_i$ I can define $X_0$ as $E cap ([-infty, N^*]^c)$. Then $$int_{E-X_0} f_m = sum_{k=-infty}^{N^*}int_E u_{m_k} < varepsilon$$ holds for all $m$ as was desired.
This looks correct?
lebesgue-integral
lebesgue-integral
asked Nov 24 '18 at 1:31
bkbowserbkbowser
11
asked Nov 24 '18 at 1:31
bkbowserbkbowser
11
asked Nov 24 '18 at 1:31
bkbowserbkbowser
11
asked Nov 24 '18 at 1:31
bkbowserbkbowser
11
asked Nov 24 '18 at 1:31
bkbowserbkbowser
11
11
add a comment |
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