Regularity of measure in Lemma 7.2.6 of Bogachev
$begingroup$
In the book "Measure Theory" of Bogachev, vol. 2, Lemma 7.2.6 states the following.
Let $mu$ be a $tau$-additive, regular Borel measure on a topological space $X$, and let ${f_alpha}$ be an increasing net of lower-semicontinuous non-negative functions such that $f:= lim_alpha f_alpha$ is bounded. Then
$$
lim_alpha int_X f_alpha , dmu quad = quad int_X f , dmu .
$$
In the proof it is quite clear where the $tau$-additivity comes into play. However, I cannot see why we need $mu$ to be regular. Bogachev defines this property in the following way (paraphrasing its Definition 7.1.5):
A measure $mu$ on a topological space $X$ is called regular if for every measurable set $Asubseteq X$ and for every $varepsilon > 0$ there exists a closed set $C_varepsilon subseteq A$ such that $mu(A) - mu(C_varepsilon) < varepsilon$.
Is regularity of $mu$ in Lemma 7.2.6 then really necessary?
measure-theory lebesgue-integral borel-measures
asked Jan 3 at 15:48
geodudegeodude
4,1761344
$endgroup$
|
show 1 more comment
$begingroup$
In the book "Measure Theory" of Bogachev, vol. 2, Lemma 7.2.6 states the following.
Let $mu$ be a $tau$-additive, regular Borel measure on a topological space $X$, and let ${f_alpha}$ be an increasing net of lower-semicontinuous non-negative functions such that $f:= lim_alpha f_alpha$ is bounded. Then
$$
lim_alpha int_X f_alpha , dmu quad = quad int_X f , dmu .
$$
In the proof it is quite clear where the $tau$-additivity comes into play. However, I cannot see why we need $mu$ to be regular. Bogachev defines this property in the following way (paraphrasing its Definition 7.1.5):
A measure $mu$ on a topological space $X$ is called regular if for every measurable set $Asubseteq X$ and for every $varepsilon > 0$ there exists a closed set $C_varepsilon subseteq A$ such that $mu(A) - mu(C_varepsilon) < varepsilon$.
Is regularity of $mu$ in Lemma 7.2.6 then really necessary?
measure-theory lebesgue-integral borel-measures
asked Jan 3 at 15:48
geodudegeodude
4,1761344
$endgroup$
$begingroup$
Might it be related to the limit over $alpha$?
$endgroup$
– Fede Poncio
Jan 3 at 15:57
$begingroup$
maybe it would be possible that a lower-semicontinuous function would be not measurable if $mu$ is not regular
$endgroup$
– Masacroso
Jan 3 at 16:01
$begingroup$
@FedePoncio I don't think so, since that's what $tau$-additivity takes care of.
$endgroup$
– geodude
Jan 4 at 11:20
$begingroup$
@Masacroso I don't think so, measurability of a function does not depend on the measure (but only on the $sigma$-algebra, which here is fixed - it is the Borel one).
$endgroup$
– geodude
Jan 4 at 11:21
2
$begingroup$
@geodude: right, I had missed the $tau$-additive condition. For what its worth, I read the proof and agree it doesn't need regularity anywhere.
$endgroup$
– Dap
Jan 6 at 19:01
|
show 1 more comment
$begingroup$
In the book "Measure Theory" of Bogachev, vol. 2, Lemma 7.2.6 states the following.
Let $mu$ be a $tau$-additive, regular Borel measure on a topological space $X$, and let ${f_alpha}$ be an increasing net of lower-semicontinuous non-negative functions such that $f:= lim_alpha f_alpha$ is bounded. Then
$$
lim_alpha int_X f_alpha , dmu quad = quad int_X f , dmu .
$$
In the proof it is quite clear where the $tau$-additivity comes into play. However, I cannot see why we need $mu$ to be regular. Bogachev defines this property in the following way (paraphrasing its Definition 7.1.5):
A measure $mu$ on a topological space $X$ is called regular if for every measurable set $Asubseteq X$ and for every $varepsilon > 0$ there exists a closed set $C_varepsilon subseteq A$ such that $mu(A) - mu(C_varepsilon) < varepsilon$.
Is regularity of $mu$ in Lemma 7.2.6 then really necessary?
measure-theory lebesgue-integral borel-measures
asked Jan 3 at 15:48
geodudegeodude
4,1761344
$endgroup$
In the book "Measure Theory" of Bogachev, vol. 2, Lemma 7.2.6 states the following.
Let $mu$ be a $tau$-additive, regular Borel measure on a topological space $X$, and let ${f_alpha}$ be an increasing net of lower-semicontinuous non-negative functions such that $f:= lim_alpha f_alpha$ is bounded. Then
$$
lim_alpha int_X f_alpha , dmu quad = quad int_X f , dmu .
$$
In the proof it is quite clear where the $tau$-additivity comes into play. However, I cannot see why we need $mu$ to be regular. Bogachev defines this property in the following way (paraphrasing its Definition 7.1.5):
A measure $mu$ on a topological space $X$ is called regular if for every measurable set $Asubseteq X$ and for every $varepsilon > 0$ there exists a closed set $C_varepsilon subseteq A$ such that $mu(A) - mu(C_varepsilon) < varepsilon$.
Is regularity of $mu$ in Lemma 7.2.6 then really necessary?
measure-theory lebesgue-integral borel-measures
measure-theory lebesgue-integral borel-measures
asked Jan 3 at 15:48
geodudegeodude
4,1761344
asked Jan 3 at 15:48
geodudegeodude
4,1761344
edited Jan 4 at 14:39
geodude
asked Jan 3 at 15:48
geodudegeodude
4,1761344
asked Jan 3 at 15:48
geodudegeodude
4,1761344
asked Jan 3 at 15:48
geodudegeodude
4,1761344
4,1761344
$begingroup$
Might it be related to the limit over $alpha$?
$endgroup$
– Fede Poncio
Jan 3 at 15:57
$begingroup$
maybe it would be possible that a lower-semicontinuous function would be not measurable if $mu$ is not regular
$endgroup$
– Masacroso
Jan 3 at 16:01
$begingroup$
@FedePoncio I don't think so, since that's what $tau$-additivity takes care of.
$endgroup$
– geodude
Jan 4 at 11:20
$begingroup$
@Masacroso I don't think so, measurability of a function does not depend on the measure (but only on the $sigma$-algebra, which here is fixed - it is the Borel one).
$endgroup$
– geodude
Jan 4 at 11:21
2
$begingroup$
@geodude: right, I had missed the $tau$-additive condition. For what its worth, I read the proof and agree it doesn't need regularity anywhere.
$endgroup$
– Dap
Jan 6 at 19:01
|
show 1 more comment
$begingroup$
Might it be related to the limit over $alpha$?
$endgroup$
– Fede Poncio
Jan 3 at 15:57
$begingroup$
maybe it would be possible that a lower-semicontinuous function would be not measurable if $mu$ is not regular
$endgroup$
– Masacroso
Jan 3 at 16:01
$begingroup$
@FedePoncio I don't think so, since that's what $tau$-additivity takes care of.
$endgroup$
– geodude
Jan 4 at 11:20
$begingroup$
@Masacroso I don't think so, measurability of a function does not depend on the measure (but only on the $sigma$-algebra, which here is fixed - it is the Borel one).
$endgroup$
– geodude
Jan 4 at 11:21
2
$begingroup$
@geodude: right, I had missed the $tau$-additive condition. For what its worth, I read the proof and agree it doesn't need regularity anywhere.
$endgroup$
– Dap
Jan 6 at 19:01
$begingroup$
Might it be related to the limit over $alpha$?
$endgroup$
– Fede Poncio
Jan 3 at 15:57
$begingroup$
Might it be related to the limit over $alpha$?
$endgroup$
– Fede Poncio
Jan 3 at 15:57
$begingroup$
maybe it would be possible that a lower-semicontinuous function would be not measurable if $mu$ is not regular
$endgroup$
– Masacroso
Jan 3 at 16:01
$begingroup$
maybe it would be possible that a lower-semicontinuous function would be not measurable if $mu$ is not regular
$endgroup$
– Masacroso
Jan 3 at 16:01
$begingroup$
@FedePoncio I don't think so, since that's what $tau$-additivity takes care of.
$endgroup$
– geodude
Jan 4 at 11:20
$begingroup$
@FedePoncio I don't think so, since that's what $tau$-additivity takes care of.
$endgroup$
– geodude
Jan 4 at 11:20
$begingroup$
@Masacroso I don't think so, measurability of a function does not depend on the measure (but only on the $sigma$-algebra, which here is fixed - it is the Borel one).
$endgroup$
– geodude
Jan 4 at 11:21
$begingroup$
@Masacroso I don't think so, measurability of a function does not depend on the measure (but only on the $sigma$-algebra, which here is fixed - it is the Borel one).
$endgroup$
– geodude
Jan 4 at 11:21
2
2
$begingroup$
@geodude: right, I had missed the $tau$-additive condition. For what its worth, I read the proof and agree it doesn't need regularity anywhere.
$endgroup$
– Dap
Jan 6 at 19:01
$begingroup$
@geodude: right, I had missed the $tau$-additive condition. For what its worth, I read the proof and agree it doesn't need regularity anywhere.
$endgroup$
– Dap
Jan 6 at 19:01
|
show 1 more comment
1 Answer
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I found an authoritative reference: "Measure Theory" by D. H. Fremlin, vol. 4.
Point a) of Corollary 414B therein says the following (the notes in the brackets are mine):
Let $X$ be a topological space and $mu$ an effectively locally finite (for example, finite) $tau$-additive topological (for example, Borel) measure on $X$. Suppose that $A$ is a non-empty upwards-directed family of lower-semicontinuous functions from $X$ to $[0.infty]$. Set
$$
g(x) := sup_{fin A} f(x)
$$
for all $xin X$. Then
$$
int g , dmu = sup_{fin A} int f,dmu .
$$
Modulo the difference in the notation, this implies the assert in the question.
answered Jan 7 at 17:37
geodudegeodude
4,1761344
$endgroup$
add a comment |
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$begingroup$
I found an authoritative reference: "Measure Theory" by D. H. Fremlin, vol. 4.
Point a) of Corollary 414B therein says the following (the notes in the brackets are mine):
Let $X$ be a topological space and $mu$ an effectively locally finite (for example, finite) $tau$-additive topological (for example, Borel) measure on $X$. Suppose that $A$ is a non-empty upwards-directed family of lower-semicontinuous functions from $X$ to $[0.infty]$. Set
$$
g(x) := sup_{fin A} f(x)
$$
for all $xin X$. Then
$$
int g , dmu = sup_{fin A} int f,dmu .
$$
Modulo the difference in the notation, this implies the assert in the question.
answered Jan 7 at 17:37
geodudegeodude
4,1761344
$endgroup$
add a comment |
$begingroup$
I found an authoritative reference: "Measure Theory" by D. H. Fremlin, vol. 4.
Point a) of Corollary 414B therein says the following (the notes in the brackets are mine):
Let $X$ be a topological space and $mu$ an effectively locally finite (for example, finite) $tau$-additive topological (for example, Borel) measure on $X$. Suppose that $A$ is a non-empty upwards-directed family of lower-semicontinuous functions from $X$ to $[0.infty]$. Set
$$
g(x) := sup_{fin A} f(x)
$$
for all $xin X$. Then
$$
int g , dmu = sup_{fin A} int f,dmu .
$$
Modulo the difference in the notation, this implies the assert in the question.
answered Jan 7 at 17:37
geodudegeodude
4,1761344
$endgroup$
add a comment |
$begingroup$
I found an authoritative reference: "Measure Theory" by D. H. Fremlin, vol. 4.
Point a) of Corollary 414B therein says the following (the notes in the brackets are mine):
Let $X$ be a topological space and $mu$ an effectively locally finite (for example, finite) $tau$-additive topological (for example, Borel) measure on $X$. Suppose that $A$ is a non-empty upwards-directed family of lower-semicontinuous functions from $X$ to $[0.infty]$. Set
$$
g(x) := sup_{fin A} f(x)
$$
for all $xin X$. Then
$$
int g , dmu = sup_{fin A} int f,dmu .
$$
Modulo the difference in the notation, this implies the assert in the question.
answered Jan 7 at 17:37
geodudegeodude
4,1761344
$endgroup$
I found an authoritative reference: "Measure Theory" by D. H. Fremlin, vol. 4.
Point a) of Corollary 414B therein says the following (the notes in the brackets are mine):
Let $X$ be a topological space and $mu$ an effectively locally finite (for example, finite) $tau$-additive topological (for example, Borel) measure on $X$. Suppose that $A$ is a non-empty upwards-directed family of lower-semicontinuous functions from $X$ to $[0.infty]$. Set
$$
g(x) := sup_{fin A} f(x)
$$
for all $xin X$. Then
$$
int g , dmu = sup_{fin A} int f,dmu .
$$
Modulo the difference in the notation, this implies the assert in the question.
answered Jan 7 at 17:37
geodudegeodude
4,1761344
answered Jan 7 at 17:37
geodudegeodude
4,1761344
answered Jan 7 at 17:37
geodudegeodude
4,1761344
answered Jan 7 at 17:37
geodudegeodude
4,1761344
4,1761344
add a comment |
add a comment |
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$begingroup$
Might it be related to the limit over $alpha$?
$endgroup$
– Fede Poncio
Jan 3 at 15:57
$begingroup$
maybe it would be possible that a lower-semicontinuous function would be not measurable if $mu$ is not regular
$endgroup$
– Masacroso
Jan 3 at 16:01
$begingroup$
@FedePoncio I don't think so, since that's what $tau$-additivity takes care of.
$endgroup$
– geodude
Jan 4 at 11:20
$begingroup$
@Masacroso I don't think so, measurability of a function does not depend on the measure (but only on the $sigma$-algebra, which here is fixed - it is the Borel one).
$endgroup$
– geodude
Jan 4 at 11:21
2
$begingroup$
@geodude: right, I had missed the $tau$-additive condition. For what its worth, I read the proof and agree it doesn't need regularity anywhere.
$endgroup$
– Dap
Jan 6 at 19:01