Mixing
Measure on a sigma-algebra with integral
Let $mu$ be a measure on $(X, mathcal{A})$ and a measurable function $f:X to mathbb{R}, f geq 0$.
Define $mu_f(E): mathcal{A} to mathbb{R}, mu_f(E):=int_E f dmu$ for $E in mathcal{A}$.
How to prove that $mu_f$ is a measure on the sigma-algebra $mathcal{A}$?
I tried it with:
$mu_f(emptyset)=int_emptyset f dmu = 0$.
I'm not sure if this is right.
For the countable additivity I don't know how to show that
$mu_f(cup^{i=1}_{infty}E_i)=sum_{i in I}{mu_f(E_i)}$.
measure-theory lebesgue-integral
asked Nov 18 '18 at 16:03
Tartulop
656
add a comment |
Let $mu$ be a measure on $(X, mathcal{A})$ and a measurable function $f:X to mathbb{R}, f geq 0$.
Define $mu_f(E): mathcal{A} to mathbb{R}, mu_f(E):=int_E f dmu$ for $E in mathcal{A}$.
How to prove that $mu_f$ is a measure on the sigma-algebra $mathcal{A}$?
I tried it with:
$mu_f(emptyset)=int_emptyset f dmu = 0$.
I'm not sure if this is right.
For the countable additivity I don't know how to show that
$mu_f(cup^{i=1}_{infty}E_i)=sum_{i in I}{mu_f(E_i)}$.
measure-theory lebesgue-integral
asked Nov 18 '18 at 16:03
Tartulop
656
add a comment |
Let $mu$ be a measure on $(X, mathcal{A})$ and a measurable function $f:X to mathbb{R}, f geq 0$.
Define $mu_f(E): mathcal{A} to mathbb{R}, mu_f(E):=int_E f dmu$ for $E in mathcal{A}$.
How to prove that $mu_f$ is a measure on the sigma-algebra $mathcal{A}$?
I tried it with:
$mu_f(emptyset)=int_emptyset f dmu = 0$.
I'm not sure if this is right.
For the countable additivity I don't know how to show that
$mu_f(cup^{i=1}_{infty}E_i)=sum_{i in I}{mu_f(E_i)}$.
measure-theory lebesgue-integral
asked Nov 18 '18 at 16:03
Tartulop
656
Let $mu$ be a measure on $(X, mathcal{A})$ and a measurable function $f:X to mathbb{R}, f geq 0$.
Define $mu_f(E): mathcal{A} to mathbb{R}, mu_f(E):=int_E f dmu$ for $E in mathcal{A}$.
How to prove that $mu_f$ is a measure on the sigma-algebra $mathcal{A}$?
I tried it with:
$mu_f(emptyset)=int_emptyset f dmu = 0$.
I'm not sure if this is right.
For the countable additivity I don't know how to show that
$mu_f(cup^{i=1}_{infty}E_i)=sum_{i in I}{mu_f(E_i)}$.
measure-theory lebesgue-integral
measure-theory lebesgue-integral
asked Nov 18 '18 at 16:03
Tartulop
656
asked Nov 18 '18 at 16:03
Tartulop
656
edited Nov 20 '18 at 15:19
asked Nov 18 '18 at 16:03
Tartulop
656
asked Nov 18 '18 at 16:03
Tartulop
656
asked Nov 18 '18 at 16:03
Tartulop
656
656
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$$mu_f(varnothing)=int_{varnothing}f;dmu=intmathbf1_{varnothing}f;dmu=int0;dmu=0$$
Further be aware that we always have $intsum_{i=1}^{infty}g_i;dmu=sum_{i=1}^{infty}int g_i;dmu$ if the $g_i$ are measurable and nonnegative.
By disjoint and measurable $E_i$ moreover we have $mathbf1_{bigcup_{i=1}^{infty}E_i}=sum_{i=1}^{infty}intmathbf1_{E_i}$ so that:
$$mu_f(bigcup_{i=1}^{infty}E_i)=int_{bigcup_{i=1}^{infty}E_i}f;dmu=intmathbf1_{bigcup_{i=1}^{infty}E_i}f;dmu=intsum_{i=1}^{infty}mathbf1_{E_i}f;dmu=sum_{i=1}^{infty}intmathbf1_{E_i}f;dmu=$$$$sum_{i=1}^{infty}mu_f(E_i)$$
answered Nov 18 '18 at 16:16
drhab
97.9k544129
add a comment |
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%2f3003719%2fmeasure-on-a-sigma-algebra-with-integral%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
$$mu_f(varnothing)=int_{varnothing}f;dmu=intmathbf1_{varnothing}f;dmu=int0;dmu=0$$
Further be aware that we always have $intsum_{i=1}^{infty}g_i;dmu=sum_{i=1}^{infty}int g_i;dmu$ if the $g_i$ are measurable and nonnegative.
By disjoint and measurable $E_i$ moreover we have $mathbf1_{bigcup_{i=1}^{infty}E_i}=sum_{i=1}^{infty}intmathbf1_{E_i}$ so that:
$$mu_f(bigcup_{i=1}^{infty}E_i)=int_{bigcup_{i=1}^{infty}E_i}f;dmu=intmathbf1_{bigcup_{i=1}^{infty}E_i}f;dmu=intsum_{i=1}^{infty}mathbf1_{E_i}f;dmu=sum_{i=1}^{infty}intmathbf1_{E_i}f;dmu=$$$$sum_{i=1}^{infty}mu_f(E_i)$$
answered Nov 18 '18 at 16:16
drhab
97.9k544129
add a comment |
$$mu_f(varnothing)=int_{varnothing}f;dmu=intmathbf1_{varnothing}f;dmu=int0;dmu=0$$
Further be aware that we always have $intsum_{i=1}^{infty}g_i;dmu=sum_{i=1}^{infty}int g_i;dmu$ if the $g_i$ are measurable and nonnegative.
By disjoint and measurable $E_i$ moreover we have $mathbf1_{bigcup_{i=1}^{infty}E_i}=sum_{i=1}^{infty}intmathbf1_{E_i}$ so that:
$$mu_f(bigcup_{i=1}^{infty}E_i)=int_{bigcup_{i=1}^{infty}E_i}f;dmu=intmathbf1_{bigcup_{i=1}^{infty}E_i}f;dmu=intsum_{i=1}^{infty}mathbf1_{E_i}f;dmu=sum_{i=1}^{infty}intmathbf1_{E_i}f;dmu=$$$$sum_{i=1}^{infty}mu_f(E_i)$$
answered Nov 18 '18 at 16:16
drhab
97.9k544129
add a comment |
$$mu_f(varnothing)=int_{varnothing}f;dmu=intmathbf1_{varnothing}f;dmu=int0;dmu=0$$
Further be aware that we always have $intsum_{i=1}^{infty}g_i;dmu=sum_{i=1}^{infty}int g_i;dmu$ if the $g_i$ are measurable and nonnegative.
By disjoint and measurable $E_i$ moreover we have $mathbf1_{bigcup_{i=1}^{infty}E_i}=sum_{i=1}^{infty}intmathbf1_{E_i}$ so that:
$$mu_f(bigcup_{i=1}^{infty}E_i)=int_{bigcup_{i=1}^{infty}E_i}f;dmu=intmathbf1_{bigcup_{i=1}^{infty}E_i}f;dmu=intsum_{i=1}^{infty}mathbf1_{E_i}f;dmu=sum_{i=1}^{infty}intmathbf1_{E_i}f;dmu=$$$$sum_{i=1}^{infty}mu_f(E_i)$$
answered Nov 18 '18 at 16:16
drhab
97.9k544129
$$mu_f(varnothing)=int_{varnothing}f;dmu=intmathbf1_{varnothing}f;dmu=int0;dmu=0$$
Further be aware that we always have $intsum_{i=1}^{infty}g_i;dmu=sum_{i=1}^{infty}int g_i;dmu$ if the $g_i$ are measurable and nonnegative.
By disjoint and measurable $E_i$ moreover we have $mathbf1_{bigcup_{i=1}^{infty}E_i}=sum_{i=1}^{infty}intmathbf1_{E_i}$ so that:
$$mu_f(bigcup_{i=1}^{infty}E_i)=int_{bigcup_{i=1}^{infty}E_i}f;dmu=intmathbf1_{bigcup_{i=1}^{infty}E_i}f;dmu=intsum_{i=1}^{infty}mathbf1_{E_i}f;dmu=sum_{i=1}^{infty}intmathbf1_{E_i}f;dmu=$$$$sum_{i=1}^{infty}mu_f(E_i)$$
answered Nov 18 '18 at 16:16
drhab
97.9k544129
answered Nov 18 '18 at 16:16
drhab
97.9k544129
answered Nov 18 '18 at 16:16
drhab
97.9k544129
answered Nov 18 '18 at 16:16
drhab
97.9k544129
97.9k544129
add a comment |
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.
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%2f3003719%2fmeasure-on-a-sigma-algebra-with-integral%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
