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Equivalent condition of a non-negative real-valued function on $[0,1]$ being measurable
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I could use some help for the problem:
Let $f$ be a non-negative real-valued function defined on $[0,1]$.
True or False? $f$ is measurable if and only if there is a sequence ${E_n}$ (finite or infinite) of measurable subsets of $[0,1]$ and a sequence of non-negative constants ${c_n}$ such that $f(x) = sum_n c_n chi_{E_n}(x)$ for every $xin[0,1]$, where $chi_{E_n}$ is the the characteristic function of $E_n$ for each $n$.
I think the statement is true. For the direction $(Longleftarrow$), since $f(x) = sum_n c_n chi_{E_n}(x)$ for every $xin[0,1]$, we see that ${sum_n^k c_n chi_{E_n}}_k$ is a sequence of simple functions on $[0,1]$ converging pointwise to $f$ and $|sum_n^k c_n chi_{E_n}|le |f|$ on $[0,1]$ for all $k$. By simple approximation theorem, $f$ is measurable.
But I got stuck on the forward ($Longrightarrow$) direction. Also, if the above proof is not right, please point it out. Thank you!
real-analysis measure-theory lebesgue-measure
asked Jan 7 at 3:45
AlexAlex
567
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add a comment |
$begingroup$
I could use some help for the problem:
Let $f$ be a non-negative real-valued function defined on $[0,1]$.
True or False? $f$ is measurable if and only if there is a sequence ${E_n}$ (finite or infinite) of measurable subsets of $[0,1]$ and a sequence of non-negative constants ${c_n}$ such that $f(x) = sum_n c_n chi_{E_n}(x)$ for every $xin[0,1]$, where $chi_{E_n}$ is the the characteristic function of $E_n$ for each $n$.
I think the statement is true. For the direction $(Longleftarrow$), since $f(x) = sum_n c_n chi_{E_n}(x)$ for every $xin[0,1]$, we see that ${sum_n^k c_n chi_{E_n}}_k$ is a sequence of simple functions on $[0,1]$ converging pointwise to $f$ and $|sum_n^k c_n chi_{E_n}|le |f|$ on $[0,1]$ for all $k$. By simple approximation theorem, $f$ is measurable.
But I got stuck on the forward ($Longrightarrow$) direction. Also, if the above proof is not right, please point it out. Thank you!
real-analysis measure-theory lebesgue-measure
asked Jan 7 at 3:45
AlexAlex
567
$endgroup$
add a comment |
$begingroup$
I could use some help for the problem:
Let $f$ be a non-negative real-valued function defined on $[0,1]$.
True or False? $f$ is measurable if and only if there is a sequence ${E_n}$ (finite or infinite) of measurable subsets of $[0,1]$ and a sequence of non-negative constants ${c_n}$ such that $f(x) = sum_n c_n chi_{E_n}(x)$ for every $xin[0,1]$, where $chi_{E_n}$ is the the characteristic function of $E_n$ for each $n$.
I think the statement is true. For the direction $(Longleftarrow$), since $f(x) = sum_n c_n chi_{E_n}(x)$ for every $xin[0,1]$, we see that ${sum_n^k c_n chi_{E_n}}_k$ is a sequence of simple functions on $[0,1]$ converging pointwise to $f$ and $|sum_n^k c_n chi_{E_n}|le |f|$ on $[0,1]$ for all $k$. By simple approximation theorem, $f$ is measurable.
But I got stuck on the forward ($Longrightarrow$) direction. Also, if the above proof is not right, please point it out. Thank you!
real-analysis measure-theory lebesgue-measure
asked Jan 7 at 3:45
AlexAlex
567
$endgroup$
I could use some help for the problem:
Let $f$ be a non-negative real-valued function defined on $[0,1]$.
True or False? $f$ is measurable if and only if there is a sequence ${E_n}$ (finite or infinite) of measurable subsets of $[0,1]$ and a sequence of non-negative constants ${c_n}$ such that $f(x) = sum_n c_n chi_{E_n}(x)$ for every $xin[0,1]$, where $chi_{E_n}$ is the the characteristic function of $E_n$ for each $n$.
I think the statement is true. For the direction $(Longleftarrow$), since $f(x) = sum_n c_n chi_{E_n}(x)$ for every $xin[0,1]$, we see that ${sum_n^k c_n chi_{E_n}}_k$ is a sequence of simple functions on $[0,1]$ converging pointwise to $f$ and $|sum_n^k c_n chi_{E_n}|le |f|$ on $[0,1]$ for all $k$. By simple approximation theorem, $f$ is measurable.
But I got stuck on the forward ($Longrightarrow$) direction. Also, if the above proof is not right, please point it out. Thank you!
real-analysis measure-theory lebesgue-measure
real-analysis measure-theory lebesgue-measure
asked Jan 7 at 3:45
AlexAlex
567
asked Jan 7 at 3:45
AlexAlex
567
edited Jan 7 at 3:54
Alex
asked Jan 7 at 3:45
AlexAlex
567
asked Jan 7 at 3:45
AlexAlex
567
asked Jan 7 at 3:45
AlexAlex
567
567
add a comment |
add a comment |
1 Answer
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The forward part is also true. Let $f_n$ be simple functions increasing to $f$. Then $f=f_1+(f_2-f_1)+(f_3-f_2)+cdots$. Writing $f_{n+1}-f_n=sum_{j=1}^{k_n} a_{j,n} I_{E_{jn}}$ (with $a_{j,n}$'s non-negative) we
see that $f$ is a countable sum of terms which are constant times characteristic functions.
answered Jan 7 at 5:34
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
$begingroup$
Thanks for your answer. Slick! I would never have come up with something like this.
$endgroup$
– Alex
Jan 7 at 8:18
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The forward part is also true. Let $f_n$ be simple functions increasing to $f$. Then $f=f_1+(f_2-f_1)+(f_3-f_2)+cdots$. Writing $f_{n+1}-f_n=sum_{j=1}^{k_n} a_{j,n} I_{E_{jn}}$ (with $a_{j,n}$'s non-negative) we
see that $f$ is a countable sum of terms which are constant times characteristic functions.
answered Jan 7 at 5:34
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
$begingroup$
Thanks for your answer. Slick! I would never have come up with something like this.
$endgroup$
– Alex
Jan 7 at 8:18
add a comment |
$begingroup$
The forward part is also true. Let $f_n$ be simple functions increasing to $f$. Then $f=f_1+(f_2-f_1)+(f_3-f_2)+cdots$. Writing $f_{n+1}-f_n=sum_{j=1}^{k_n} a_{j,n} I_{E_{jn}}$ (with $a_{j,n}$'s non-negative) we
see that $f$ is a countable sum of terms which are constant times characteristic functions.
answered Jan 7 at 5:34
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
$begingroup$
Thanks for your answer. Slick! I would never have come up with something like this.
$endgroup$
– Alex
Jan 7 at 8:18
add a comment |
$begingroup$
The forward part is also true. Let $f_n$ be simple functions increasing to $f$. Then $f=f_1+(f_2-f_1)+(f_3-f_2)+cdots$. Writing $f_{n+1}-f_n=sum_{j=1}^{k_n} a_{j,n} I_{E_{jn}}$ (with $a_{j,n}$'s non-negative) we
see that $f$ is a countable sum of terms which are constant times characteristic functions.
answered Jan 7 at 5:34
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
The forward part is also true. Let $f_n$ be simple functions increasing to $f$. Then $f=f_1+(f_2-f_1)+(f_3-f_2)+cdots$. Writing $f_{n+1}-f_n=sum_{j=1}^{k_n} a_{j,n} I_{E_{jn}}$ (with $a_{j,n}$'s non-negative) we
see that $f$ is a countable sum of terms which are constant times characteristic functions.
answered Jan 7 at 5:34
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Jan 7 at 5:34
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Jan 7 at 5:34
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Jan 7 at 5:34
Kavi Rama MurthyKavi Rama Murthy
56k42158
56k42158
$begingroup$
Thanks for your answer. Slick! I would never have come up with something like this.
$endgroup$
– Alex
Jan 7 at 8:18
add a comment |
$begingroup$
Thanks for your answer. Slick! I would never have come up with something like this.
$endgroup$
– Alex
Jan 7 at 8:18
$begingroup$
Thanks for your answer. Slick! I would never have come up with something like this.
$endgroup$
– Alex
Jan 7 at 8:18
$begingroup$
Thanks for your answer. Slick! I would never have come up with something like this.
$endgroup$
– Alex
Jan 7 at 8:18
add a comment |
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