Regularity of measure in Lemma 7.2.6 of Bogachev
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 2 days ago
geodude
4,0641141
add a comment |
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 2 days ago
geodude
4,0641141
Might it be related to the limit over $alpha$?
– Fede Poncio
2 days ago
maybe it would be possible that a lower-semicontinuous function would be not measurable if $mu$ is not regular
– Masacroso
2 days ago
@FedePoncio I don't think so, since that's what $tau$-additivity takes care of.
– geodude
yesterday
@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).
– geodude
yesterday
add a comment |
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 2 days ago
geodude
4,0641141
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 2 days ago
geodude
4,0641141
asked 2 days ago
geodude
4,0641141
edited yesterday
asked 2 days ago
geodude
4,0641141
asked 2 days ago
geodude
4,0641141
asked 2 days ago
geodude
4,0641141
4,0641141
Might it be related to the limit over $alpha$?
– Fede Poncio
2 days ago
maybe it would be possible that a lower-semicontinuous function would be not measurable if $mu$ is not regular
– Masacroso
2 days ago
@FedePoncio I don't think so, since that's what $tau$-additivity takes care of.
– geodude
yesterday
@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).
– geodude
yesterday
add a comment |
Might it be related to the limit over $alpha$?
– Fede Poncio
2 days ago
maybe it would be possible that a lower-semicontinuous function would be not measurable if $mu$ is not regular
– Masacroso
2 days ago
@FedePoncio I don't think so, since that's what $tau$-additivity takes care of.
– geodude
yesterday
@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).
– geodude
yesterday
Might it be related to the limit over $alpha$?
– Fede Poncio
2 days ago
Might it be related to the limit over $alpha$?
– Fede Poncio
2 days ago
maybe it would be possible that a lower-semicontinuous function would be not measurable if $mu$ is not regular
– Masacroso
2 days ago
maybe it would be possible that a lower-semicontinuous function would be not measurable if $mu$ is not regular
– Masacroso
2 days ago
@FedePoncio I don't think so, since that's what $tau$-additivity takes care of.
– geodude
yesterday
@FedePoncio I don't think so, since that's what $tau$-additivity takes care of.
– geodude
yesterday
@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).
– geodude
yesterday
@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).
– geodude
yesterday
add a comment |
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Might it be related to the limit over $alpha$?
– Fede Poncio
2 days ago
maybe it would be possible that a lower-semicontinuous function would be not measurable if $mu$ is not regular
– Masacroso
2 days ago
@FedePoncio I don't think so, since that's what $tau$-additivity takes care of.
– geodude
yesterday
@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).
– geodude
yesterday