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 yesterday
geodude
4,0541141
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 yesterday
geodude
4,0541141
Might it be related to the limit over $alpha$?
– Fede Poncio
yesterday
maybe it would be possible that a lower-semicontinuous function would be not measurable if $mu$ is not regular
– Masacroso
yesterday
@FedePoncio I don't think so, since that's what $tau$-additivity takes care of.
– geodude
22 hours ago
@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
22 hours ago
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 yesterday
geodude
4,0541141
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 yesterday
geodude
4,0541141
asked yesterday
geodude
4,0541141
edited 19 hours ago
asked yesterday
geodude
4,0541141
asked yesterday
geodude
4,0541141
asked yesterday
geodude
4,0541141
4,0541141
Might it be related to the limit over $alpha$?
– Fede Poncio
yesterday
maybe it would be possible that a lower-semicontinuous function would be not measurable if $mu$ is not regular
– Masacroso
yesterday
@FedePoncio I don't think so, since that's what $tau$-additivity takes care of.
– geodude
22 hours ago
@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
22 hours ago
add a comment |
Might it be related to the limit over $alpha$?
– Fede Poncio
yesterday
maybe it would be possible that a lower-semicontinuous function would be not measurable if $mu$ is not regular
– Masacroso
yesterday
@FedePoncio I don't think so, since that's what $tau$-additivity takes care of.
– geodude
22 hours ago
@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
22 hours ago
Might it be related to the limit over $alpha$?
– Fede Poncio
yesterday
Might it be related to the limit over $alpha$?
– Fede Poncio
yesterday
maybe it would be possible that a lower-semicontinuous function would be not measurable if $mu$ is not regular
– Masacroso
yesterday
maybe it would be possible that a lower-semicontinuous function would be not measurable if $mu$ is not regular
– Masacroso
yesterday
@FedePoncio I don't think so, since that's what $tau$-additivity takes care of.
– geodude
22 hours ago
@FedePoncio I don't think so, since that's what $tau$-additivity takes care of.
– geodude
22 hours ago
@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
22 hours ago
@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
22 hours ago
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060701%2fregularity-of-measure-in-lemma-7-2-6-of-bogachev%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060701%2fregularity-of-measure-in-lemma-7-2-6-of-bogachev%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Might it be related to the limit over $alpha$?
– Fede Poncio
yesterday
maybe it would be possible that a lower-semicontinuous function would be not measurable if $mu$ is not regular
– Masacroso
yesterday
@FedePoncio I don't think so, since that's what $tau$-additivity takes care of.
– geodude
22 hours ago
@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
22 hours ago