Mixing
$f$ measurable iff $f^2$ measurable and ${f > 0}$ measurable
$begingroup$
I know for sure that if $f^2$ is measurable, that doesn't imply that $f$ is measurable, but how does the condition:
$$ {f > 0} text{ measurable }$$
play into making $f$ automatically measurable?
measure-theory lebesgue-measure borel-sets borel-measures
edited Jan 4 at 19:24
Umberto P.
38.8k13064
asked Jan 4 at 19:09
sashasasha
133
$endgroup$
add a comment |
$begingroup$
I know for sure that if $f^2$ is measurable, that doesn't imply that $f$ is measurable, but how does the condition:
$$ {f > 0} text{ measurable }$$
play into making $f$ automatically measurable?
measure-theory lebesgue-measure borel-sets borel-measures
edited Jan 4 at 19:24
Umberto P.
38.8k13064
asked Jan 4 at 19:09
sashasasha
133
$endgroup$
add a comment |
$begingroup$
I know for sure that if $f^2$ is measurable, that doesn't imply that $f$ is measurable, but how does the condition:
$$ {f > 0} text{ measurable }$$
play into making $f$ automatically measurable?
measure-theory lebesgue-measure borel-sets borel-measures
edited Jan 4 at 19:24
Umberto P.
38.8k13064
asked Jan 4 at 19:09
sashasasha
133
$endgroup$
I know for sure that if $f^2$ is measurable, that doesn't imply that $f$ is measurable, but how does the condition:
$$ {f > 0} text{ measurable }$$
play into making $f$ automatically measurable?
measure-theory lebesgue-measure borel-sets borel-measures
measure-theory lebesgue-measure borel-sets borel-measures
edited Jan 4 at 19:24
Umberto P.
38.8k13064
asked Jan 4 at 19:09
sashasasha
133
edited Jan 4 at 19:24
Umberto P.
38.8k13064
asked Jan 4 at 19:09
sashasasha
133
edited Jan 4 at 19:24
Umberto P.
38.8k13064
edited Jan 4 at 19:24
Umberto P.
38.8k13064
edited Jan 4 at 19:24
Umberto P.
38.8k13064
38.8k13064
asked Jan 4 at 19:09
sashasasha
133
asked Jan 4 at 19:09
sashasasha
133
asked Jan 4 at 19:09
sashasasha
133
133
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: For $a>0$, we have ${f>a} ={f^2 >a^2} cap{f>0}$.
answered Jan 4 at 19:13
BerciBerci
60.1k23672
$endgroup$
$begingroup$
Ohhh, I knew it had to do with a way of rewriting the sets in some sort of way.
$endgroup$
– sasha
Jan 4 at 19:25
add a comment |
$begingroup$
Another Hint: It holds
$$f =sqrt{f^2}1_{{f>0}}-sqrt{f^2}1_{{f>0}^c}.
$$
answered Jan 4 at 19:17
SongSong
9,211627
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
Hint: For $a>0$, we have ${f>a} ={f^2 >a^2} cap{f>0}$.
answered Jan 4 at 19:13
BerciBerci
60.1k23672
$endgroup$
$begingroup$
Ohhh, I knew it had to do with a way of rewriting the sets in some sort of way.
$endgroup$
– sasha
Jan 4 at 19:25
add a comment |
$begingroup$
Hint: For $a>0$, we have ${f>a} ={f^2 >a^2} cap{f>0}$.
answered Jan 4 at 19:13
BerciBerci
60.1k23672
$endgroup$
$begingroup$
Ohhh, I knew it had to do with a way of rewriting the sets in some sort of way.
$endgroup$
– sasha
Jan 4 at 19:25
add a comment |
$begingroup$
Hint: For $a>0$, we have ${f>a} ={f^2 >a^2} cap{f>0}$.
answered Jan 4 at 19:13
BerciBerci
60.1k23672
$endgroup$
Hint: For $a>0$, we have ${f>a} ={f^2 >a^2} cap{f>0}$.
answered Jan 4 at 19:13
BerciBerci
60.1k23672
answered Jan 4 at 19:13
BerciBerci
60.1k23672
answered Jan 4 at 19:13
BerciBerci
60.1k23672
answered Jan 4 at 19:13
BerciBerci
60.1k23672
60.1k23672
$begingroup$
Ohhh, I knew it had to do with a way of rewriting the sets in some sort of way.
$endgroup$
– sasha
Jan 4 at 19:25
add a comment |
$begingroup$
Ohhh, I knew it had to do with a way of rewriting the sets in some sort of way.
$endgroup$
– sasha
Jan 4 at 19:25
$begingroup$
Ohhh, I knew it had to do with a way of rewriting the sets in some sort of way.
$endgroup$
– sasha
Jan 4 at 19:25
$begingroup$
Ohhh, I knew it had to do with a way of rewriting the sets in some sort of way.
$endgroup$
– sasha
Jan 4 at 19:25
add a comment |
$begingroup$
Another Hint: It holds
$$f =sqrt{f^2}1_{{f>0}}-sqrt{f^2}1_{{f>0}^c}.
$$
answered Jan 4 at 19:17
SongSong
9,211627
$endgroup$
add a comment |
$begingroup$
Another Hint: It holds
$$f =sqrt{f^2}1_{{f>0}}-sqrt{f^2}1_{{f>0}^c}.
$$
answered Jan 4 at 19:17
SongSong
9,211627
$endgroup$
add a comment |
$begingroup$
Another Hint: It holds
$$f =sqrt{f^2}1_{{f>0}}-sqrt{f^2}1_{{f>0}^c}.
$$
answered Jan 4 at 19:17
SongSong
9,211627
$endgroup$
Another Hint: It holds
$$f =sqrt{f^2}1_{{f>0}}-sqrt{f^2}1_{{f>0}^c}.
$$
answered Jan 4 at 19:17
SongSong
9,211627
answered Jan 4 at 19:17
SongSong
9,211627
answered Jan 4 at 19:17
SongSong
9,211627
answered Jan 4 at 19:17
SongSong
9,211627
9,211627
add a comment |
add a comment |
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