Mixing
$ mu $ premeasure and continousness - proof verification
$begingroup$
Let be $ mu $ a content on a Ring $R$
I want to prove following
$$ (a) leftrightarrow (b) rightarrow (c) leftrightarrow (d) $$
with
a) $ mu $ is a premeasure
b) $mu $ is continous "from below " , means $ A_n uparrow A, A in R$ and $mu (A_n) rightarrow mu (A) $
c)$mu $ is continous "from above " , means $ A_n downarrow A, A in R$ and $mu (A_n) rightarrow mu (A) $
d) $ mu $ is $ emptyset $- continous, means there is a sequence $ (A_n) in mathbb{R}$ with $ A_n downarrow emptyset $
and $mu (A_n) < infty forall n $ with $ mu (A_n) rightarrow 0 $
__
here my proof:
$ (a) rightarrow (b) $
Let be $ A_n uparrow A $ and define $A_0= emptyset $
Then $B_n= A_n backslash A_{n-1} $ is pairwise disjoint.
Then you can write
$ A_n = cup_{k=1}^n B_k$ and $ A= cup_{k in mathbb{N}} B_k $
so is
$mu (A_n)= mu ( cup_{k=1}^n B_k ) = sum_{k=1}^n mu (B_k) rightarrow sum_{k=1}^{ infty} mu (B_k)= mu ( cup_{k=1}^{ infty } B_k) = mu (A) $
$(a) leftarrow (b) $
here the $ sigma $- additivity needs to be show.
So let be $ B_1,...,B_n $ pairwise disjoint and $ A_n = cup_{k=1}^n B_k $
then $ A_n uparrow = cup_{k=1} ^{ infty} B_n $ hold
and you can write
$mu (A_n) = mu ( cup_{k=1}^n B_k) = sum_{k=1}^n mu (B_k) rightarrow sum_{k=1}^{ infty } mu (B_n) $
and because of continousness from below it follows that :
$ mu (A_n) rightarrow mu (A) $
$ (c) rightarrow (d) $
Set $ A= emptyset $ then $ A_n downarrow emptyset $ and $ mu (A_n) < infty $
and
$ mu (A_n) rightarrow mu (A) = mu ( emptyset) = 0 $
$ (c) leftarrow (d) $
let be $ mu $ finite.
Then $ A backslash A_n uparrow 0 $ holds and
$ mu (A) - lim_{n rightarrow infty} mu (A_n) = lim_{ n rightarrow infty} mu (A backslash A_n ) = mu ( emptyset)= 0 $
$(b) rightarrow (c) $
Is $ A_n downarrow A $ then $ A_1 backslash A_n uparrow A_1 backslash A $
and $ mu (A) < mu (A_n) < infty $
so
$ mu (A_1)- mu (A_n) = mu (A_1 backslash A_n) rightarrow mu ( A_1 backslash A) = mu (A_1 ) - mu (A) $
Is this proof correct so far? Can I leave it like that or woud you add some adjustments? especially i'm not sure with case b) to c)
Thanks in advance !!
measure-theory proof-verification
asked Jan 23 at 15:03
constant94constant94
6310
$endgroup$
add a comment |
$begingroup$
Let be $ mu $ a content on a Ring $R$
I want to prove following
$$ (a) leftrightarrow (b) rightarrow (c) leftrightarrow (d) $$
with
a) $ mu $ is a premeasure
b) $mu $ is continous "from below " , means $ A_n uparrow A, A in R$ and $mu (A_n) rightarrow mu (A) $
c)$mu $ is continous "from above " , means $ A_n downarrow A, A in R$ and $mu (A_n) rightarrow mu (A) $
d) $ mu $ is $ emptyset $- continous, means there is a sequence $ (A_n) in mathbb{R}$ with $ A_n downarrow emptyset $
and $mu (A_n) < infty forall n $ with $ mu (A_n) rightarrow 0 $
__
here my proof:
$ (a) rightarrow (b) $
Let be $ A_n uparrow A $ and define $A_0= emptyset $
Then $B_n= A_n backslash A_{n-1} $ is pairwise disjoint.
Then you can write
$ A_n = cup_{k=1}^n B_k$ and $ A= cup_{k in mathbb{N}} B_k $
so is
$mu (A_n)= mu ( cup_{k=1}^n B_k ) = sum_{k=1}^n mu (B_k) rightarrow sum_{k=1}^{ infty} mu (B_k)= mu ( cup_{k=1}^{ infty } B_k) = mu (A) $
$(a) leftarrow (b) $
here the $ sigma $- additivity needs to be show.
So let be $ B_1,...,B_n $ pairwise disjoint and $ A_n = cup_{k=1}^n B_k $
then $ A_n uparrow = cup_{k=1} ^{ infty} B_n $ hold
and you can write
$mu (A_n) = mu ( cup_{k=1}^n B_k) = sum_{k=1}^n mu (B_k) rightarrow sum_{k=1}^{ infty } mu (B_n) $
and because of continousness from below it follows that :
$ mu (A_n) rightarrow mu (A) $
$ (c) rightarrow (d) $
Set $ A= emptyset $ then $ A_n downarrow emptyset $ and $ mu (A_n) < infty $
and
$ mu (A_n) rightarrow mu (A) = mu ( emptyset) = 0 $
$ (c) leftarrow (d) $
let be $ mu $ finite.
Then $ A backslash A_n uparrow 0 $ holds and
$ mu (A) - lim_{n rightarrow infty} mu (A_n) = lim_{ n rightarrow infty} mu (A backslash A_n ) = mu ( emptyset)= 0 $
$(b) rightarrow (c) $
Is $ A_n downarrow A $ then $ A_1 backslash A_n uparrow A_1 backslash A $
and $ mu (A) < mu (A_n) < infty $
so
$ mu (A_1)- mu (A_n) = mu (A_1 backslash A_n) rightarrow mu ( A_1 backslash A) = mu (A_1 ) - mu (A) $
Is this proof correct so far? Can I leave it like that or woud you add some adjustments? especially i'm not sure with case b) to c)
Thanks in advance !!
measure-theory proof-verification
asked Jan 23 at 15:03
constant94constant94
6310
$endgroup$
add a comment |
$begingroup$
Let be $ mu $ a content on a Ring $R$
I want to prove following
$$ (a) leftrightarrow (b) rightarrow (c) leftrightarrow (d) $$
with
a) $ mu $ is a premeasure
b) $mu $ is continous "from below " , means $ A_n uparrow A, A in R$ and $mu (A_n) rightarrow mu (A) $
c)$mu $ is continous "from above " , means $ A_n downarrow A, A in R$ and $mu (A_n) rightarrow mu (A) $
d) $ mu $ is $ emptyset $- continous, means there is a sequence $ (A_n) in mathbb{R}$ with $ A_n downarrow emptyset $
and $mu (A_n) < infty forall n $ with $ mu (A_n) rightarrow 0 $
__
here my proof:
$ (a) rightarrow (b) $
Let be $ A_n uparrow A $ and define $A_0= emptyset $
Then $B_n= A_n backslash A_{n-1} $ is pairwise disjoint.
Then you can write
$ A_n = cup_{k=1}^n B_k$ and $ A= cup_{k in mathbb{N}} B_k $
so is
$mu (A_n)= mu ( cup_{k=1}^n B_k ) = sum_{k=1}^n mu (B_k) rightarrow sum_{k=1}^{ infty} mu (B_k)= mu ( cup_{k=1}^{ infty } B_k) = mu (A) $
$(a) leftarrow (b) $
here the $ sigma $- additivity needs to be show.
So let be $ B_1,...,B_n $ pairwise disjoint and $ A_n = cup_{k=1}^n B_k $
then $ A_n uparrow = cup_{k=1} ^{ infty} B_n $ hold
and you can write
$mu (A_n) = mu ( cup_{k=1}^n B_k) = sum_{k=1}^n mu (B_k) rightarrow sum_{k=1}^{ infty } mu (B_n) $
and because of continousness from below it follows that :
$ mu (A_n) rightarrow mu (A) $
$ (c) rightarrow (d) $
Set $ A= emptyset $ then $ A_n downarrow emptyset $ and $ mu (A_n) < infty $
and
$ mu (A_n) rightarrow mu (A) = mu ( emptyset) = 0 $
$ (c) leftarrow (d) $
let be $ mu $ finite.
Then $ A backslash A_n uparrow 0 $ holds and
$ mu (A) - lim_{n rightarrow infty} mu (A_n) = lim_{ n rightarrow infty} mu (A backslash A_n ) = mu ( emptyset)= 0 $
$(b) rightarrow (c) $
Is $ A_n downarrow A $ then $ A_1 backslash A_n uparrow A_1 backslash A $
and $ mu (A) < mu (A_n) < infty $
so
$ mu (A_1)- mu (A_n) = mu (A_1 backslash A_n) rightarrow mu ( A_1 backslash A) = mu (A_1 ) - mu (A) $
Is this proof correct so far? Can I leave it like that or woud you add some adjustments? especially i'm not sure with case b) to c)
Thanks in advance !!
measure-theory proof-verification
asked Jan 23 at 15:03
constant94constant94
6310
$endgroup$
Let be $ mu $ a content on a Ring $R$
I want to prove following
$$ (a) leftrightarrow (b) rightarrow (c) leftrightarrow (d) $$
with
a) $ mu $ is a premeasure
b) $mu $ is continous "from below " , means $ A_n uparrow A, A in R$ and $mu (A_n) rightarrow mu (A) $
c)$mu $ is continous "from above " , means $ A_n downarrow A, A in R$ and $mu (A_n) rightarrow mu (A) $
d) $ mu $ is $ emptyset $- continous, means there is a sequence $ (A_n) in mathbb{R}$ with $ A_n downarrow emptyset $
and $mu (A_n) < infty forall n $ with $ mu (A_n) rightarrow 0 $
__
here my proof:
$ (a) rightarrow (b) $
Let be $ A_n uparrow A $ and define $A_0= emptyset $
Then $B_n= A_n backslash A_{n-1} $ is pairwise disjoint.
Then you can write
$ A_n = cup_{k=1}^n B_k$ and $ A= cup_{k in mathbb{N}} B_k $
so is
$mu (A_n)= mu ( cup_{k=1}^n B_k ) = sum_{k=1}^n mu (B_k) rightarrow sum_{k=1}^{ infty} mu (B_k)= mu ( cup_{k=1}^{ infty } B_k) = mu (A) $
$(a) leftarrow (b) $
here the $ sigma $- additivity needs to be show.
So let be $ B_1,...,B_n $ pairwise disjoint and $ A_n = cup_{k=1}^n B_k $
then $ A_n uparrow = cup_{k=1} ^{ infty} B_n $ hold
and you can write
$mu (A_n) = mu ( cup_{k=1}^n B_k) = sum_{k=1}^n mu (B_k) rightarrow sum_{k=1}^{ infty } mu (B_n) $
and because of continousness from below it follows that :
$ mu (A_n) rightarrow mu (A) $
$ (c) rightarrow (d) $
Set $ A= emptyset $ then $ A_n downarrow emptyset $ and $ mu (A_n) < infty $
and
$ mu (A_n) rightarrow mu (A) = mu ( emptyset) = 0 $
$ (c) leftarrow (d) $
let be $ mu $ finite.
Then $ A backslash A_n uparrow 0 $ holds and
$ mu (A) - lim_{n rightarrow infty} mu (A_n) = lim_{ n rightarrow infty} mu (A backslash A_n ) = mu ( emptyset)= 0 $
$(b) rightarrow (c) $
Is $ A_n downarrow A $ then $ A_1 backslash A_n uparrow A_1 backslash A $
and $ mu (A) < mu (A_n) < infty $
so
$ mu (A_1)- mu (A_n) = mu (A_1 backslash A_n) rightarrow mu ( A_1 backslash A) = mu (A_1 ) - mu (A) $
Is this proof correct so far? Can I leave it like that or woud you add some adjustments? especially i'm not sure with case b) to c)
Thanks in advance !!
measure-theory proof-verification
measure-theory proof-verification
asked Jan 23 at 15:03
constant94constant94
6310
asked Jan 23 at 15:03
constant94constant94
6310
edited Jan 23 at 16:43
constant94
asked Jan 23 at 15:03
constant94constant94
6310
asked Jan 23 at 15:03
constant94constant94
6310
asked Jan 23 at 15:03
constant94constant94
6310
6310
add a comment |
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%2f3084582%2fmu-premeasure-and-continousness-proof-verification%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.
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%2f3084582%2fmu-premeasure-and-continousness-proof-verification%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
