What are good intuition for understanding that the pre image of a mesurable function on an element of a borel set is mesurableWhat is the intuition behind the Borel Cantelli Lemma?Prove that if E is a measurable set, then the set whose elements are the squares of elements in E is also measurableIs the pre-image (through a measurable function) of a Lebesgue-measurable set also measurable?Is it true or false that every Lebesgue measurable set of a finite measure is contained in a Borel set of the same measure?For a Borel function when does there exist a set of full measure with measurable image$f:mathbb Rlongrightarrow mathbb R$ is measurable $iff$ $f^-1(B)$ is measurable for all Borel set $B$Is the pre-image of a Borel set by a continuous function a Borel set? Is it an open set?Cantor set to show that the Borel measure is not completeA function is a Borel function if and only if for any c, the set $f^-1((c,infty))$ is a Borel set(T/F) If $mu$ is a Borel measure on $mathbbR$ and $A$ is a Borel set such that $mu(A cap K) = 0$ for all compact sets $K$, then $mu(A) = 0$.

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What are good intuition for understanding that the pre image of a mesurable function on an element of a borel set is mesurable


What is the intuition behind the Borel Cantelli Lemma?Prove that if E is a measurable set, then the set whose elements are the squares of elements in E is also measurableIs the pre-image (through a measurable function) of a Lebesgue-measurable set also measurable?Is it true or false that every Lebesgue measurable set of a finite measure is contained in a Borel set of the same measure?For a Borel function when does there exist a set of full measure with measurable image$f:mathbb Rlongrightarrow mathbb R$ is measurable $iff$ $f^-1(B)$ is measurable for all Borel set $B$Is the pre-image of a Borel set by a continuous function a Borel set? Is it an open set?Cantor set to show that the Borel measure is not completeA function is a Borel function if and only if for any c, the set $f^-1((c,infty))$ is a Borel set(T/F) If $mu$ is a Borel measure on $mathbbR$ and $A$ is a Borel set such that $mu(A cap K) = 0$ for all compact sets $K$, then $mu(A) = 0$.













0












$begingroup$


I'm not convinced by the proof I saw in the lecture.



The statement is the following :




If $f$ is measurable, and if $B$ is in the borel set, then $f^-1 (B) in $ the set of measurables.




Can you please give me some intuition about this lemma and abiut the proof? I think it would be better to prove it using definition of the set of measurables.










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    This is a definition of measurability, (at least for Borel measures) (not a lemma), and as such, requires no proof.
    $endgroup$
    – uniquesolution
    Mar 11 at 9:42











  • $begingroup$
    I have another definition of measurability. I have this one : en.wikipedia.org/wiki/Measurable_function
    $endgroup$
    – Marine Galantin
    Mar 11 at 22:42















0












$begingroup$


I'm not convinced by the proof I saw in the lecture.



The statement is the following :




If $f$ is measurable, and if $B$ is in the borel set, then $f^-1 (B) in $ the set of measurables.




Can you please give me some intuition about this lemma and abiut the proof? I think it would be better to prove it using definition of the set of measurables.










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    This is a definition of measurability, (at least for Borel measures) (not a lemma), and as such, requires no proof.
    $endgroup$
    – uniquesolution
    Mar 11 at 9:42











  • $begingroup$
    I have another definition of measurability. I have this one : en.wikipedia.org/wiki/Measurable_function
    $endgroup$
    – Marine Galantin
    Mar 11 at 22:42













0












0








0





$begingroup$


I'm not convinced by the proof I saw in the lecture.



The statement is the following :




If $f$ is measurable, and if $B$ is in the borel set, then $f^-1 (B) in $ the set of measurables.




Can you please give me some intuition about this lemma and abiut the proof? I think it would be better to prove it using definition of the set of measurables.










share|cite|improve this question









$endgroup$




I'm not convinced by the proof I saw in the lecture.



The statement is the following :




If $f$ is measurable, and if $B$ is in the borel set, then $f^-1 (B) in $ the set of measurables.




Can you please give me some intuition about this lemma and abiut the proof? I think it would be better to prove it using definition of the set of measurables.







measure-theory lebesgue-measure outer-measure






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 11 at 9:26









Marine GalantinMarine Galantin

875319




875319







  • 1




    $begingroup$
    This is a definition of measurability, (at least for Borel measures) (not a lemma), and as such, requires no proof.
    $endgroup$
    – uniquesolution
    Mar 11 at 9:42











  • $begingroup$
    I have another definition of measurability. I have this one : en.wikipedia.org/wiki/Measurable_function
    $endgroup$
    – Marine Galantin
    Mar 11 at 22:42












  • 1




    $begingroup$
    This is a definition of measurability, (at least for Borel measures) (not a lemma), and as such, requires no proof.
    $endgroup$
    – uniquesolution
    Mar 11 at 9:42











  • $begingroup$
    I have another definition of measurability. I have this one : en.wikipedia.org/wiki/Measurable_function
    $endgroup$
    – Marine Galantin
    Mar 11 at 22:42







1




1




$begingroup$
This is a definition of measurability, (at least for Borel measures) (not a lemma), and as such, requires no proof.
$endgroup$
– uniquesolution
Mar 11 at 9:42





$begingroup$
This is a definition of measurability, (at least for Borel measures) (not a lemma), and as such, requires no proof.
$endgroup$
– uniquesolution
Mar 11 at 9:42













$begingroup$
I have another definition of measurability. I have this one : en.wikipedia.org/wiki/Measurable_function
$endgroup$
– Marine Galantin
Mar 11 at 22:42




$begingroup$
I have another definition of measurability. I have this one : en.wikipedia.org/wiki/Measurable_function
$endgroup$
– Marine Galantin
Mar 11 at 22:42










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