Mixing
Proving that a non-negative additive set function is monotonous
$begingroup$
Let $mu:mathscr{C}tomathbb{R}=[-infty,+infty]$ is monotonous if $mu(emptyset)=0$ adnd for $E,Finmathscr{C}$, $Esubset F$ implies that $mu(E)leqslantmu(F)$.
If $mathscr{C}$ is a ring, prove that a non-negative additive set function is monotonous.
I have devised a proof that was wrong but I do not know the reason for that.
Let $A,Binmathscr{C}$ then as $mathscr{C}$ is a ring then $Acup Binmathscr{C}$. Taking $H=Acup B$ then as the measure is non-negative $mu(A)>0$ and $mu(B)>0$.
Then it is obviously seen that:
$Asubset(Acup B)=H$ and $mu(A)leqslant mu(Acup B)=mu(A)+mu(B)=mu(H)$
As $A,B$ are arbitrary sets of $mathscr{C}$ we conclude $mu$ is monotonous.
Questions:
Why is my proof wrong? Which are the alternatives?
Thanks in advance!
measure-theory proof-verification proof-writing
asked Jan 29 at 16:14
Pedro GomesPedro Gomes
1,9822721
$endgroup$
add a comment |
$begingroup$
Let $mu:mathscr{C}tomathbb{R}=[-infty,+infty]$ is monotonous if $mu(emptyset)=0$ adnd for $E,Finmathscr{C}$, $Esubset F$ implies that $mu(E)leqslantmu(F)$.
If $mathscr{C}$ is a ring, prove that a non-negative additive set function is monotonous.
I have devised a proof that was wrong but I do not know the reason for that.
Let $A,Binmathscr{C}$ then as $mathscr{C}$ is a ring then $Acup Binmathscr{C}$. Taking $H=Acup B$ then as the measure is non-negative $mu(A)>0$ and $mu(B)>0$.
Then it is obviously seen that:
$Asubset(Acup B)=H$ and $mu(A)leqslant mu(Acup B)=mu(A)+mu(B)=mu(H)$
As $A,B$ are arbitrary sets of $mathscr{C}$ we conclude $mu$ is monotonous.
Questions:
Why is my proof wrong? Which are the alternatives?
Thanks in advance!
measure-theory proof-verification proof-writing
asked Jan 29 at 16:14
Pedro GomesPedro Gomes
1,9822721
$endgroup$
$begingroup$
Are $A$ and $B$ disjoint?
$endgroup$
– d.k.o.
Jan 29 at 16:38
add a comment |
$begingroup$
Let $mu:mathscr{C}tomathbb{R}=[-infty,+infty]$ is monotonous if $mu(emptyset)=0$ adnd for $E,Finmathscr{C}$, $Esubset F$ implies that $mu(E)leqslantmu(F)$.
If $mathscr{C}$ is a ring, prove that a non-negative additive set function is monotonous.
I have devised a proof that was wrong but I do not know the reason for that.
Let $A,Binmathscr{C}$ then as $mathscr{C}$ is a ring then $Acup Binmathscr{C}$. Taking $H=Acup B$ then as the measure is non-negative $mu(A)>0$ and $mu(B)>0$.
Then it is obviously seen that:
$Asubset(Acup B)=H$ and $mu(A)leqslant mu(Acup B)=mu(A)+mu(B)=mu(H)$
As $A,B$ are arbitrary sets of $mathscr{C}$ we conclude $mu$ is monotonous.
Questions:
Why is my proof wrong? Which are the alternatives?
Thanks in advance!
measure-theory proof-verification proof-writing
asked Jan 29 at 16:14
Pedro GomesPedro Gomes
1,9822721
$endgroup$
Let $mu:mathscr{C}tomathbb{R}=[-infty,+infty]$ is monotonous if $mu(emptyset)=0$ adnd for $E,Finmathscr{C}$, $Esubset F$ implies that $mu(E)leqslantmu(F)$.
If $mathscr{C}$ is a ring, prove that a non-negative additive set function is monotonous.
I have devised a proof that was wrong but I do not know the reason for that.
Let $A,Binmathscr{C}$ then as $mathscr{C}$ is a ring then $Acup Binmathscr{C}$. Taking $H=Acup B$ then as the measure is non-negative $mu(A)>0$ and $mu(B)>0$.
Then it is obviously seen that:
$Asubset(Acup B)=H$ and $mu(A)leqslant mu(Acup B)=mu(A)+mu(B)=mu(H)$
As $A,B$ are arbitrary sets of $mathscr{C}$ we conclude $mu$ is monotonous.
Questions:
Why is my proof wrong? Which are the alternatives?
Thanks in advance!
measure-theory proof-verification proof-writing
measure-theory proof-verification proof-writing
asked Jan 29 at 16:14
Pedro GomesPedro Gomes
1,9822721
asked Jan 29 at 16:14
Pedro GomesPedro Gomes
1,9822721
asked Jan 29 at 16:14
Pedro GomesPedro Gomes
1,9822721
asked Jan 29 at 16:14
Pedro GomesPedro Gomes
1,9822721
asked Jan 29 at 16:14
Pedro GomesPedro Gomes
1,9822721
1,9822721
$begingroup$
Are $A$ and $B$ disjoint?
$endgroup$
– d.k.o.
Jan 29 at 16:38
add a comment |
$begingroup$
Are $A$ and $B$ disjoint?
$endgroup$
– d.k.o.
Jan 29 at 16:38
$begingroup$
Are $A$ and $B$ disjoint?
$endgroup$
– d.k.o.
Jan 29 at 16:38
$begingroup$
Are $A$ and $B$ disjoint?
$endgroup$
– d.k.o.
Jan 29 at 16:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $E,Fin mathscr{C}$ s.t. $Esubset F$,
$$
mu(F)=mu(E)+mu(Fsetminus E)ge mu(E).
$$
answered Jan 29 at 16:31
d.k.o.d.k.o.
10.5k630
$endgroup$
$begingroup$
Thanks for your answer! That is better. But I still do not understand why my answer is wrong. Could you explain me please?
$endgroup$
– Pedro Gomes
Jan 29 at 17:00
$begingroup$
You argue that $ mu(Acup B)=mu(A)+mu(B)$. Are these sets disjoint?
$endgroup$
– d.k.o.
Jan 29 at 17:46
add a comment |
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%2f3092362%2fproving-that-a-non-negative-additive-set-function-is-monotonous%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For $E,Fin mathscr{C}$ s.t. $Esubset F$,
$$
mu(F)=mu(E)+mu(Fsetminus E)ge mu(E).
$$
answered Jan 29 at 16:31
d.k.o.d.k.o.
10.5k630
$endgroup$
$begingroup$
Thanks for your answer! That is better. But I still do not understand why my answer is wrong. Could you explain me please?
$endgroup$
– Pedro Gomes
Jan 29 at 17:00
$begingroup$
You argue that $ mu(Acup B)=mu(A)+mu(B)$. Are these sets disjoint?
$endgroup$
– d.k.o.
Jan 29 at 17:46
add a comment |
$begingroup$
For $E,Fin mathscr{C}$ s.t. $Esubset F$,
$$
mu(F)=mu(E)+mu(Fsetminus E)ge mu(E).
$$
answered Jan 29 at 16:31
d.k.o.d.k.o.
10.5k630
$endgroup$
$begingroup$
Thanks for your answer! That is better. But I still do not understand why my answer is wrong. Could you explain me please?
$endgroup$
– Pedro Gomes
Jan 29 at 17:00
$begingroup$
You argue that $ mu(Acup B)=mu(A)+mu(B)$. Are these sets disjoint?
$endgroup$
– d.k.o.
Jan 29 at 17:46
add a comment |
$begingroup$
For $E,Fin mathscr{C}$ s.t. $Esubset F$,
$$
mu(F)=mu(E)+mu(Fsetminus E)ge mu(E).
$$
answered Jan 29 at 16:31
d.k.o.d.k.o.
10.5k630
$endgroup$
For $E,Fin mathscr{C}$ s.t. $Esubset F$,
$$
mu(F)=mu(E)+mu(Fsetminus E)ge mu(E).
$$
answered Jan 29 at 16:31
d.k.o.d.k.o.
10.5k630
answered Jan 29 at 16:31
d.k.o.d.k.o.
10.5k630
answered Jan 29 at 16:31
d.k.o.d.k.o.
10.5k630
answered Jan 29 at 16:31
d.k.o.d.k.o.
10.5k630
10.5k630
$begingroup$
Thanks for your answer! That is better. But I still do not understand why my answer is wrong. Could you explain me please?
$endgroup$
– Pedro Gomes
Jan 29 at 17:00
$begingroup$
You argue that $ mu(Acup B)=mu(A)+mu(B)$. Are these sets disjoint?
$endgroup$
– d.k.o.
Jan 29 at 17:46
add a comment |
$begingroup$
Thanks for your answer! That is better. But I still do not understand why my answer is wrong. Could you explain me please?
$endgroup$
– Pedro Gomes
Jan 29 at 17:00
$begingroup$
You argue that $ mu(Acup B)=mu(A)+mu(B)$. Are these sets disjoint?
$endgroup$
– d.k.o.
Jan 29 at 17:46
$begingroup$
Thanks for your answer! That is better. But I still do not understand why my answer is wrong. Could you explain me please?
$endgroup$
– Pedro Gomes
Jan 29 at 17:00
$begingroup$
Thanks for your answer! That is better. But I still do not understand why my answer is wrong. Could you explain me please?
$endgroup$
– Pedro Gomes
Jan 29 at 17:00
$begingroup$
You argue that $ mu(Acup B)=mu(A)+mu(B)$. Are these sets disjoint?
$endgroup$
– d.k.o.
Jan 29 at 17:46
$begingroup$
You argue that $ mu(Acup B)=mu(A)+mu(B)$. Are these sets disjoint?
$endgroup$
– d.k.o.
Jan 29 at 17:46
add a comment |
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%2f3092362%2fproving-that-a-non-negative-additive-set-function-is-monotonous%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
$begingroup$
Are $A$ and $B$ disjoint?
$endgroup$
– d.k.o.
Jan 29 at 16:38