If $m(E)=m_*(E_1)+m_*(E_2)$ , then $E_1$ and $E_2$ are measurable
$begingroup$
Suppose $E$ is measurable with $m(E) lt infty$ , and $$E=E_1 cup E_2 , E_1 cap E_2 text{is empty}$$ Show that if $m(E)=m_*(E_1)+m_*(E_2)$ , then $E_1$ and $E_2$ are measurable.
My attempt:
Since $E_2=E-E_1$ , it suffice to prove $E_1$ is measurable . And for $E_1$ , since $m_*(E_1)le m(E) lt infty$ , for any $epsilon$ , we can find an open set $O$ such taht $E_1 subset O$ and $m(O)-m_*(E_1) lt epsilon$ , but how to show that $m_*(O-E_1)lt epsilon$ ?
real-analysis lebesgue-measure
asked Jan 8 at 16:43
J.GuoJ.Guo
3979
$endgroup$
add a comment |
$begingroup$
Suppose $E$ is measurable with $m(E) lt infty$ , and $$E=E_1 cup E_2 , E_1 cap E_2 text{is empty}$$ Show that if $m(E)=m_*(E_1)+m_*(E_2)$ , then $E_1$ and $E_2$ are measurable.
My attempt:
Since $E_2=E-E_1$ , it suffice to prove $E_1$ is measurable . And for $E_1$ , since $m_*(E_1)le m(E) lt infty$ , for any $epsilon$ , we can find an open set $O$ such taht $E_1 subset O$ and $m(O)-m_*(E_1) lt epsilon$ , but how to show that $m_*(O-E_1)lt epsilon$ ?
real-analysis lebesgue-measure
asked Jan 8 at 16:43
J.GuoJ.Guo
3979
$endgroup$
add a comment |
$begingroup$
Suppose $E$ is measurable with $m(E) lt infty$ , and $$E=E_1 cup E_2 , E_1 cap E_2 text{is empty}$$ Show that if $m(E)=m_*(E_1)+m_*(E_2)$ , then $E_1$ and $E_2$ are measurable.
My attempt:
Since $E_2=E-E_1$ , it suffice to prove $E_1$ is measurable . And for $E_1$ , since $m_*(E_1)le m(E) lt infty$ , for any $epsilon$ , we can find an open set $O$ such taht $E_1 subset O$ and $m(O)-m_*(E_1) lt epsilon$ , but how to show that $m_*(O-E_1)lt epsilon$ ?
real-analysis lebesgue-measure
asked Jan 8 at 16:43
J.GuoJ.Guo
3979
$endgroup$
Suppose $E$ is measurable with $m(E) lt infty$ , and $$E=E_1 cup E_2 , E_1 cap E_2 text{is empty}$$ Show that if $m(E)=m_*(E_1)+m_*(E_2)$ , then $E_1$ and $E_2$ are measurable.
My attempt:
Since $E_2=E-E_1$ , it suffice to prove $E_1$ is measurable . And for $E_1$ , since $m_*(E_1)le m(E) lt infty$ , for any $epsilon$ , we can find an open set $O$ such taht $E_1 subset O$ and $m(O)-m_*(E_1) lt epsilon$ , but how to show that $m_*(O-E_1)lt epsilon$ ?
real-analysis lebesgue-measure
real-analysis lebesgue-measure
asked Jan 8 at 16:43
J.GuoJ.Guo
3979
asked Jan 8 at 16:43
J.GuoJ.Guo
3979
asked Jan 8 at 16:43
J.GuoJ.Guo
3979
asked Jan 8 at 16:43
J.GuoJ.Guo
3979
asked Jan 8 at 16:43
J.GuoJ.Guo
3979
3979
add a comment |
add a comment |
1 Answer
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$begingroup$
Let $epsilon>0$ be arbitrarily given and $U_i$, $i=1,2$ be open sets such that $E_isubset U_i$ and $$
m_*E_ile mU_i<m_*E_i +epsilon
$$ for $i=1,2$. Since $
Esubset U_1cup U_2,
$ we have $$
mE leq m(U_1cup U_2)=mU_1+mU_2-m(U_1cap U_2).
$$ This gives $m(U_1cap U_2)le mU_1+mU_2-mE<2epsilon$ by the assumption that $mE=m_*E_1+m_* E_2$. Now, observe that
$$
U_isetminus E_i subset left(U_1cap U_2right)cup left(left(U_1cup U_2right)setminus Eright).
$$ This implies
$$begin{eqnarray}
m_*(U_isetminus E_i)&le& m_*left( left(U_1cap U_2right)cup left(left(U_1cup U_2right)setminus Eright)right)\
&le& m(U_1cap U_2)+m(left(U_1cup U_2right)setminus E)\
&le& 2epsilon +m(U_1cup U_2)-mE\
&le& 2epsilon + mU_1+mU_2-mE<4epsilon.
end{eqnarray}$$ Since $epsilon>0$ was arbitrary, it says that $E_i$ are measurable for $i=1,2$.
answered Jan 8 at 17:47
SongSong
14.2k1634
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1 Answer
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1 Answer
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$begingroup$
Let $epsilon>0$ be arbitrarily given and $U_i$, $i=1,2$ be open sets such that $E_isubset U_i$ and $$
m_*E_ile mU_i<m_*E_i +epsilon
$$ for $i=1,2$. Since $
Esubset U_1cup U_2,
$ we have $$
mE leq m(U_1cup U_2)=mU_1+mU_2-m(U_1cap U_2).
$$ This gives $m(U_1cap U_2)le mU_1+mU_2-mE<2epsilon$ by the assumption that $mE=m_*E_1+m_* E_2$. Now, observe that
$$
U_isetminus E_i subset left(U_1cap U_2right)cup left(left(U_1cup U_2right)setminus Eright).
$$ This implies
$$begin{eqnarray}
m_*(U_isetminus E_i)&le& m_*left( left(U_1cap U_2right)cup left(left(U_1cup U_2right)setminus Eright)right)\
&le& m(U_1cap U_2)+m(left(U_1cup U_2right)setminus E)\
&le& 2epsilon +m(U_1cup U_2)-mE\
&le& 2epsilon + mU_1+mU_2-mE<4epsilon.
end{eqnarray}$$ Since $epsilon>0$ was arbitrary, it says that $E_i$ are measurable for $i=1,2$.
answered Jan 8 at 17:47
SongSong
14.2k1634
$endgroup$
add a comment |
$begingroup$
Let $epsilon>0$ be arbitrarily given and $U_i$, $i=1,2$ be open sets such that $E_isubset U_i$ and $$
m_*E_ile mU_i<m_*E_i +epsilon
$$ for $i=1,2$. Since $
Esubset U_1cup U_2,
$ we have $$
mE leq m(U_1cup U_2)=mU_1+mU_2-m(U_1cap U_2).
$$ This gives $m(U_1cap U_2)le mU_1+mU_2-mE<2epsilon$ by the assumption that $mE=m_*E_1+m_* E_2$. Now, observe that
$$
U_isetminus E_i subset left(U_1cap U_2right)cup left(left(U_1cup U_2right)setminus Eright).
$$ This implies
$$begin{eqnarray}
m_*(U_isetminus E_i)&le& m_*left( left(U_1cap U_2right)cup left(left(U_1cup U_2right)setminus Eright)right)\
&le& m(U_1cap U_2)+m(left(U_1cup U_2right)setminus E)\
&le& 2epsilon +m(U_1cup U_2)-mE\
&le& 2epsilon + mU_1+mU_2-mE<4epsilon.
end{eqnarray}$$ Since $epsilon>0$ was arbitrary, it says that $E_i$ are measurable for $i=1,2$.
answered Jan 8 at 17:47
SongSong
14.2k1634
$endgroup$
add a comment |
$begingroup$
Let $epsilon>0$ be arbitrarily given and $U_i$, $i=1,2$ be open sets such that $E_isubset U_i$ and $$
m_*E_ile mU_i<m_*E_i +epsilon
$$ for $i=1,2$. Since $
Esubset U_1cup U_2,
$ we have $$
mE leq m(U_1cup U_2)=mU_1+mU_2-m(U_1cap U_2).
$$ This gives $m(U_1cap U_2)le mU_1+mU_2-mE<2epsilon$ by the assumption that $mE=m_*E_1+m_* E_2$. Now, observe that
$$
U_isetminus E_i subset left(U_1cap U_2right)cup left(left(U_1cup U_2right)setminus Eright).
$$ This implies
$$begin{eqnarray}
m_*(U_isetminus E_i)&le& m_*left( left(U_1cap U_2right)cup left(left(U_1cup U_2right)setminus Eright)right)\
&le& m(U_1cap U_2)+m(left(U_1cup U_2right)setminus E)\
&le& 2epsilon +m(U_1cup U_2)-mE\
&le& 2epsilon + mU_1+mU_2-mE<4epsilon.
end{eqnarray}$$ Since $epsilon>0$ was arbitrary, it says that $E_i$ are measurable for $i=1,2$.
answered Jan 8 at 17:47
SongSong
14.2k1634
$endgroup$
Let $epsilon>0$ be arbitrarily given and $U_i$, $i=1,2$ be open sets such that $E_isubset U_i$ and $$
m_*E_ile mU_i<m_*E_i +epsilon
$$ for $i=1,2$. Since $
Esubset U_1cup U_2,
$ we have $$
mE leq m(U_1cup U_2)=mU_1+mU_2-m(U_1cap U_2).
$$ This gives $m(U_1cap U_2)le mU_1+mU_2-mE<2epsilon$ by the assumption that $mE=m_*E_1+m_* E_2$. Now, observe that
$$
U_isetminus E_i subset left(U_1cap U_2right)cup left(left(U_1cup U_2right)setminus Eright).
$$ This implies
$$begin{eqnarray}
m_*(U_isetminus E_i)&le& m_*left( left(U_1cap U_2right)cup left(left(U_1cup U_2right)setminus Eright)right)\
&le& m(U_1cap U_2)+m(left(U_1cup U_2right)setminus E)\
&le& 2epsilon +m(U_1cup U_2)-mE\
&le& 2epsilon + mU_1+mU_2-mE<4epsilon.
end{eqnarray}$$ Since $epsilon>0$ was arbitrary, it says that $E_i$ are measurable for $i=1,2$.
answered Jan 8 at 17:47
SongSong
14.2k1634
answered Jan 8 at 17:47
SongSong
14.2k1634
answered Jan 8 at 17:47
SongSong
14.2k1634
answered Jan 8 at 17:47
SongSong
14.2k1634
14.2k1634
add a comment |
add a comment |
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