Mixing
Exact definition of Radon-Nikodym Theorem
$begingroup$
A part of the RND Theorem states for $sigma-$ finite measures $nu, mu$ on $(X,mathcal{A})$
$nu << mu iff exists f$ non-negative and measurable: $nu(A)=int_{A}fdmu$ for all $A in mathcal{A}$
Questions:
$1.$ This may fall under logic but does the "for all $A in mathcal{A}$" mean "$exists f,forall A in mathcal{A}$" or rather $forall A in mathcal{A},exists f"$. In other words, is that $f$ dependent on the choice of $A in mathcal{A}$?
I realize the RND derivative (i.e. in our case, $f = frac{d nu}{d mu}$) is an equivalence class of non-negative, measurable functions that are equal to each other almost everywhere. But I am not sure whether I understand the theorem precisely (regarding $1.$).
But if I have $nu << mu$ and
$int gdnu=int fdmu$, does that mean $g = f, mu-$a.e.?
real-analysis measure-theory radon-nikodym
asked Jan 17 at 17:07
SABOYSABOY
649311
$endgroup$
add a comment |
$begingroup$
A part of the RND Theorem states for $sigma-$ finite measures $nu, mu$ on $(X,mathcal{A})$
$nu << mu iff exists f$ non-negative and measurable: $nu(A)=int_{A}fdmu$ for all $A in mathcal{A}$
Questions:
$1.$ This may fall under logic but does the "for all $A in mathcal{A}$" mean "$exists f,forall A in mathcal{A}$" or rather $forall A in mathcal{A},exists f"$. In other words, is that $f$ dependent on the choice of $A in mathcal{A}$?
I realize the RND derivative (i.e. in our case, $f = frac{d nu}{d mu}$) is an equivalence class of non-negative, measurable functions that are equal to each other almost everywhere. But I am not sure whether I understand the theorem precisely (regarding $1.$).
But if I have $nu << mu$ and
$int gdnu=int fdmu$, does that mean $g = f, mu-$a.e.?
real-analysis measure-theory radon-nikodym
asked Jan 17 at 17:07
SABOYSABOY
649311
$endgroup$
$begingroup$
It means $exists f,forall A$, i.e., one function works for all $Ainmathcal A$ simultaneously. (The other interpretation would be trivial.)
$endgroup$
– Andreas Blass
Jan 17 at 19:46
add a comment |
$begingroup$
A part of the RND Theorem states for $sigma-$ finite measures $nu, mu$ on $(X,mathcal{A})$
$nu << mu iff exists f$ non-negative and measurable: $nu(A)=int_{A}fdmu$ for all $A in mathcal{A}$
Questions:
$1.$ This may fall under logic but does the "for all $A in mathcal{A}$" mean "$exists f,forall A in mathcal{A}$" or rather $forall A in mathcal{A},exists f"$. In other words, is that $f$ dependent on the choice of $A in mathcal{A}$?
I realize the RND derivative (i.e. in our case, $f = frac{d nu}{d mu}$) is an equivalence class of non-negative, measurable functions that are equal to each other almost everywhere. But I am not sure whether I understand the theorem precisely (regarding $1.$).
But if I have $nu << mu$ and
$int gdnu=int fdmu$, does that mean $g = f, mu-$a.e.?
real-analysis measure-theory radon-nikodym
asked Jan 17 at 17:07
SABOYSABOY
649311
$endgroup$
A part of the RND Theorem states for $sigma-$ finite measures $nu, mu$ on $(X,mathcal{A})$
$nu << mu iff exists f$ non-negative and measurable: $nu(A)=int_{A}fdmu$ for all $A in mathcal{A}$
Questions:
$1.$ This may fall under logic but does the "for all $A in mathcal{A}$" mean "$exists f,forall A in mathcal{A}$" or rather $forall A in mathcal{A},exists f"$. In other words, is that $f$ dependent on the choice of $A in mathcal{A}$?
I realize the RND derivative (i.e. in our case, $f = frac{d nu}{d mu}$) is an equivalence class of non-negative, measurable functions that are equal to each other almost everywhere. But I am not sure whether I understand the theorem precisely (regarding $1.$).
But if I have $nu << mu$ and
$int gdnu=int fdmu$, does that mean $g = f, mu-$a.e.?
real-analysis measure-theory radon-nikodym
real-analysis measure-theory radon-nikodym
asked Jan 17 at 17:07
SABOYSABOY
649311
asked Jan 17 at 17:07
SABOYSABOY
649311
edited Jan 17 at 17:16
SABOY
asked Jan 17 at 17:07
SABOYSABOY
649311
asked Jan 17 at 17:07
SABOYSABOY
649311
asked Jan 17 at 17:07
SABOYSABOY
649311
649311
$begingroup$
It means $exists f,forall A$, i.e., one function works for all $Ainmathcal A$ simultaneously. (The other interpretation would be trivial.)
$endgroup$
– Andreas Blass
Jan 17 at 19:46
add a comment |
$begingroup$
It means $exists f,forall A$, i.e., one function works for all $Ainmathcal A$ simultaneously. (The other interpretation would be trivial.)
$endgroup$
– Andreas Blass
Jan 17 at 19:46
$begingroup$
It means $exists f,forall A$, i.e., one function works for all $Ainmathcal A$ simultaneously. (The other interpretation would be trivial.)
$endgroup$
– Andreas Blass
Jan 17 at 19:46
$begingroup$
It means $exists f,forall A$, i.e., one function works for all $Ainmathcal A$ simultaneously. (The other interpretation would be trivial.)
$endgroup$
– Andreas Blass
Jan 17 at 19:46
add a comment |
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$begingroup$
It means $exists f,forall A$, i.e., one function works for all $Ainmathcal A$ simultaneously. (The other interpretation would be trivial.)
$endgroup$
– Andreas Blass
Jan 17 at 19:46