Dominated convergence theorem - what sequence? The Next CEO of Stack OverflowWhat are some...
How to avoid supervisors with prejudiced views?
What connection does MS Office have to Netscape Navigator?
0 rank tensor vs 1D vector
Make solar eclipses exceedingly rare, but still have new moons
What steps are necessary to read a Modern SSD in Medieval Europe?
Is there always a complete, orthogonal set of unitary matrices?
Is it ever safe to open a suspicious HTML file (e.g. email attachment)?
Would a grinding machine be a simple and workable propulsion system for an interplanetary spacecraft?
The past simple of "gaslight" – "gaslighted" or "gaslit"?
Find non-case sensitive string in a mixed list of elements?
Powershell. How to parse gci Name?
What does "Its cash flow is deeply negative" mean?
How to count occurrences of text in a file?
Why doesn't UK go for the same deal Japan has with EU to resolve Brexit?
How to check if all elements of 1 list are in the *same quantity* and in any order, in the list2?
Are police here, aren't itthey?
How to get from Geneva Airport to Metabief?
The exact meaning of 'Mom made me a sandwich'
Does Germany produce more waste than the US?
Can we say or write : "No, it'sn't"?
Does increasing your ability score affect your main stat?
What flight has the highest ratio of time difference to flight time?
Is a distribution that is normal, but highly skewed considered Gaussian?
Reference request: Grassmannian and Plucker coordinates in type B, C, D
Dominated convergence theorem - what sequence?
The Next CEO of Stack OverflowWhat are some good integration problems where you can use some of the function convergence theorem of Lesbegue integrals?Find Limit Using Lebesgue Dominated ConvergenceSolving these types of integrals, using Monotone convergence theorem and Dominated convergence theorem.Applications of Dominated/Monotone convergence theoremLebesgue Dominated Convergence Theorem exampleDominated convergence theorem for log-integrable rational functionsuniform or dominated convergence of sequence of functions which are boundedBartle's proof of Lebesgue Dominated Convergence TheoremCalculate the limit using dominated or monotone convergence theoremUsing dominated convergence theorem to move limit inside the integral
$begingroup$
Simple question. When are we allowed to exchange limits and integrals? I'm talking about situations like
$$lim_{varepsilonto0^+} int_{-infty}^infty dk f(k,varepsilon) overset{?}{=} int_{-infty}^infty dklim_{varepsilonto0^+} f(k,varepsilon).$$
Everyone refers to either dominated convergence theorem or monotone convergence theorem but I'm not sure if I understand how exactly one should go about applying it. Both theorems are about sequences and I don't see how that relates to integration in practice. Help a physicist out :)
P.S. Before someone marks it as a duplicate, please take a minute to understand (not saying that you won't) what it is that I'm asking here. Thank you!
integration limits
$endgroup$
add a comment |
$begingroup$
Simple question. When are we allowed to exchange limits and integrals? I'm talking about situations like
$$lim_{varepsilonto0^+} int_{-infty}^infty dk f(k,varepsilon) overset{?}{=} int_{-infty}^infty dklim_{varepsilonto0^+} f(k,varepsilon).$$
Everyone refers to either dominated convergence theorem or monotone convergence theorem but I'm not sure if I understand how exactly one should go about applying it. Both theorems are about sequences and I don't see how that relates to integration in practice. Help a physicist out :)
P.S. Before someone marks it as a duplicate, please take a minute to understand (not saying that you won't) what it is that I'm asking here. Thank you!
integration limits
$endgroup$
add a comment |
$begingroup$
Simple question. When are we allowed to exchange limits and integrals? I'm talking about situations like
$$lim_{varepsilonto0^+} int_{-infty}^infty dk f(k,varepsilon) overset{?}{=} int_{-infty}^infty dklim_{varepsilonto0^+} f(k,varepsilon).$$
Everyone refers to either dominated convergence theorem or monotone convergence theorem but I'm not sure if I understand how exactly one should go about applying it. Both theorems are about sequences and I don't see how that relates to integration in practice. Help a physicist out :)
P.S. Before someone marks it as a duplicate, please take a minute to understand (not saying that you won't) what it is that I'm asking here. Thank you!
integration limits
$endgroup$
Simple question. When are we allowed to exchange limits and integrals? I'm talking about situations like
$$lim_{varepsilonto0^+} int_{-infty}^infty dk f(k,varepsilon) overset{?}{=} int_{-infty}^infty dklim_{varepsilonto0^+} f(k,varepsilon).$$
Everyone refers to either dominated convergence theorem or monotone convergence theorem but I'm not sure if I understand how exactly one should go about applying it. Both theorems are about sequences and I don't see how that relates to integration in practice. Help a physicist out :)
P.S. Before someone marks it as a duplicate, please take a minute to understand (not saying that you won't) what it is that I'm asking here. Thank you!
integration limits
integration limits
asked 4 hours ago
Ivan V.Ivan V.
911216
911216
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let's look at it in a silly case. We want to prove by DCT that $$lim_{varepsilonto0^+} int_0^infty e^{-y/varepsilon},dy=0$$
This is the case if and only if for all sequence $varepsilon_nto 0^+$ it holds $$lim_{ntoinfty}int_0^infty e^{-y/varepsilon_n},dy=0$$
And now you can use DCT on each of these sequences. Of course, the limiting function will always be the zero function and you may consider the dominating function $e^{-x}$.
$endgroup$
add a comment |
$begingroup$
The statement of the dominated convergence theorem (DCT) is as follows:
"Discrete" DCT. Suppose ${f_n}_{n=1}^infty$ is a sequence of (measurable) functions such that $|f_n| le g$ for some integrable function $g$ and all $n$, and $lim_{ntoinfty}f_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_{ntoinfty}int f_n = int f$ (by the triangle inequality). This can be written as
$$ lim_{ntoinfty}int f_n = int lim_{ntoinfty} f_n.$$
(The statement and conclusion of the monotone convergence theorem are similar, but it has a somewhat different set of hypotheses.)
As you note, the statements of these theorems involve sequences of functions, i.e., a $1$-discrete-parameter family of functions ${f_n}_{n=1}^infty$. To apply these theorems to a $1$-continuous-parameter family of functions, say ${f_epsilon}_{0<epsilon<epsilon_0}$, one typically uses a characterization of limits involving a continuous parameter in terms of sequences:
Proposition. If $f$ is a function, then
$$lim_{epsilonto0^+}f(epsilon) = L iff lim_{ntoinfty}f(a_n) = Lquad text{for $mathbf{all}$ sequences $a_nto 0^+$.}$$
With this characterization, we can formulate a version of the dominated convergence theorem involving continuous-parameter families of functions (note that I use quotations to title these versions of the DCT because these names are not standard as far as I know):
"Continuous" DCT. Suppose ${f_epsilon}_{0<epsilon<epsilon_0}$ is a $1$-continuous-parameter family of (measurable) functions such that $|f_epsilon| le g$ for some integrable function $g$ and all $0<epsilon<epsilon_0$, and $lim_{epsilonto0^+}f_epsilon=f$ pointwise almost everywhere. Then, $f$ is an integrable function and $lim_{epsilonto 0^+}int f_epsilon = int f$. This can be written as
$$ lim_{epsilonto0^+}int f_epsilon = int lim_{epsilonto0^+} f_epsilon.$$
The way we use the continuous DCT in practice is by picking an arbitrary sequence $pmb{a_nto 0^+}$ and showing that the hypotheses of the "discrete" DCT are satisfied for this arbitrary sequence $a_n$, using only the assumption that $a_nto 0^+$ and properties of the family ${f_epsilon}$ that are known to us.
$endgroup$
$begingroup$
Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
$endgroup$
– Ivan V.
2 hours ago
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
1 hour ago
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%2f3168778%2fdominated-convergence-theorem-what-sequence%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's look at it in a silly case. We want to prove by DCT that $$lim_{varepsilonto0^+} int_0^infty e^{-y/varepsilon},dy=0$$
This is the case if and only if for all sequence $varepsilon_nto 0^+$ it holds $$lim_{ntoinfty}int_0^infty e^{-y/varepsilon_n},dy=0$$
And now you can use DCT on each of these sequences. Of course, the limiting function will always be the zero function and you may consider the dominating function $e^{-x}$.
$endgroup$
add a comment |
$begingroup$
Let's look at it in a silly case. We want to prove by DCT that $$lim_{varepsilonto0^+} int_0^infty e^{-y/varepsilon},dy=0$$
This is the case if and only if for all sequence $varepsilon_nto 0^+$ it holds $$lim_{ntoinfty}int_0^infty e^{-y/varepsilon_n},dy=0$$
And now you can use DCT on each of these sequences. Of course, the limiting function will always be the zero function and you may consider the dominating function $e^{-x}$.
$endgroup$
add a comment |
$begingroup$
Let's look at it in a silly case. We want to prove by DCT that $$lim_{varepsilonto0^+} int_0^infty e^{-y/varepsilon},dy=0$$
This is the case if and only if for all sequence $varepsilon_nto 0^+$ it holds $$lim_{ntoinfty}int_0^infty e^{-y/varepsilon_n},dy=0$$
And now you can use DCT on each of these sequences. Of course, the limiting function will always be the zero function and you may consider the dominating function $e^{-x}$.
$endgroup$
Let's look at it in a silly case. We want to prove by DCT that $$lim_{varepsilonto0^+} int_0^infty e^{-y/varepsilon},dy=0$$
This is the case if and only if for all sequence $varepsilon_nto 0^+$ it holds $$lim_{ntoinfty}int_0^infty e^{-y/varepsilon_n},dy=0$$
And now you can use DCT on each of these sequences. Of course, the limiting function will always be the zero function and you may consider the dominating function $e^{-x}$.
answered 4 hours ago
Saucy O'PathSaucy O'Path
6,2141627
6,2141627
add a comment |
add a comment |
$begingroup$
The statement of the dominated convergence theorem (DCT) is as follows:
"Discrete" DCT. Suppose ${f_n}_{n=1}^infty$ is a sequence of (measurable) functions such that $|f_n| le g$ for some integrable function $g$ and all $n$, and $lim_{ntoinfty}f_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_{ntoinfty}int f_n = int f$ (by the triangle inequality). This can be written as
$$ lim_{ntoinfty}int f_n = int lim_{ntoinfty} f_n.$$
(The statement and conclusion of the monotone convergence theorem are similar, but it has a somewhat different set of hypotheses.)
As you note, the statements of these theorems involve sequences of functions, i.e., a $1$-discrete-parameter family of functions ${f_n}_{n=1}^infty$. To apply these theorems to a $1$-continuous-parameter family of functions, say ${f_epsilon}_{0<epsilon<epsilon_0}$, one typically uses a characterization of limits involving a continuous parameter in terms of sequences:
Proposition. If $f$ is a function, then
$$lim_{epsilonto0^+}f(epsilon) = L iff lim_{ntoinfty}f(a_n) = Lquad text{for $mathbf{all}$ sequences $a_nto 0^+$.}$$
With this characterization, we can formulate a version of the dominated convergence theorem involving continuous-parameter families of functions (note that I use quotations to title these versions of the DCT because these names are not standard as far as I know):
"Continuous" DCT. Suppose ${f_epsilon}_{0<epsilon<epsilon_0}$ is a $1$-continuous-parameter family of (measurable) functions such that $|f_epsilon| le g$ for some integrable function $g$ and all $0<epsilon<epsilon_0$, and $lim_{epsilonto0^+}f_epsilon=f$ pointwise almost everywhere. Then, $f$ is an integrable function and $lim_{epsilonto 0^+}int f_epsilon = int f$. This can be written as
$$ lim_{epsilonto0^+}int f_epsilon = int lim_{epsilonto0^+} f_epsilon.$$
The way we use the continuous DCT in practice is by picking an arbitrary sequence $pmb{a_nto 0^+}$ and showing that the hypotheses of the "discrete" DCT are satisfied for this arbitrary sequence $a_n$, using only the assumption that $a_nto 0^+$ and properties of the family ${f_epsilon}$ that are known to us.
$endgroup$
$begingroup$
Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
$endgroup$
– Ivan V.
2 hours ago
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
1 hour ago
add a comment |
$begingroup$
The statement of the dominated convergence theorem (DCT) is as follows:
"Discrete" DCT. Suppose ${f_n}_{n=1}^infty$ is a sequence of (measurable) functions such that $|f_n| le g$ for some integrable function $g$ and all $n$, and $lim_{ntoinfty}f_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_{ntoinfty}int f_n = int f$ (by the triangle inequality). This can be written as
$$ lim_{ntoinfty}int f_n = int lim_{ntoinfty} f_n.$$
(The statement and conclusion of the monotone convergence theorem are similar, but it has a somewhat different set of hypotheses.)
As you note, the statements of these theorems involve sequences of functions, i.e., a $1$-discrete-parameter family of functions ${f_n}_{n=1}^infty$. To apply these theorems to a $1$-continuous-parameter family of functions, say ${f_epsilon}_{0<epsilon<epsilon_0}$, one typically uses a characterization of limits involving a continuous parameter in terms of sequences:
Proposition. If $f$ is a function, then
$$lim_{epsilonto0^+}f(epsilon) = L iff lim_{ntoinfty}f(a_n) = Lquad text{for $mathbf{all}$ sequences $a_nto 0^+$.}$$
With this characterization, we can formulate a version of the dominated convergence theorem involving continuous-parameter families of functions (note that I use quotations to title these versions of the DCT because these names are not standard as far as I know):
"Continuous" DCT. Suppose ${f_epsilon}_{0<epsilon<epsilon_0}$ is a $1$-continuous-parameter family of (measurable) functions such that $|f_epsilon| le g$ for some integrable function $g$ and all $0<epsilon<epsilon_0$, and $lim_{epsilonto0^+}f_epsilon=f$ pointwise almost everywhere. Then, $f$ is an integrable function and $lim_{epsilonto 0^+}int f_epsilon = int f$. This can be written as
$$ lim_{epsilonto0^+}int f_epsilon = int lim_{epsilonto0^+} f_epsilon.$$
The way we use the continuous DCT in practice is by picking an arbitrary sequence $pmb{a_nto 0^+}$ and showing that the hypotheses of the "discrete" DCT are satisfied for this arbitrary sequence $a_n$, using only the assumption that $a_nto 0^+$ and properties of the family ${f_epsilon}$ that are known to us.
$endgroup$
$begingroup$
Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
$endgroup$
– Ivan V.
2 hours ago
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
1 hour ago
add a comment |
$begingroup$
The statement of the dominated convergence theorem (DCT) is as follows:
"Discrete" DCT. Suppose ${f_n}_{n=1}^infty$ is a sequence of (measurable) functions such that $|f_n| le g$ for some integrable function $g$ and all $n$, and $lim_{ntoinfty}f_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_{ntoinfty}int f_n = int f$ (by the triangle inequality). This can be written as
$$ lim_{ntoinfty}int f_n = int lim_{ntoinfty} f_n.$$
(The statement and conclusion of the monotone convergence theorem are similar, but it has a somewhat different set of hypotheses.)
As you note, the statements of these theorems involve sequences of functions, i.e., a $1$-discrete-parameter family of functions ${f_n}_{n=1}^infty$. To apply these theorems to a $1$-continuous-parameter family of functions, say ${f_epsilon}_{0<epsilon<epsilon_0}$, one typically uses a characterization of limits involving a continuous parameter in terms of sequences:
Proposition. If $f$ is a function, then
$$lim_{epsilonto0^+}f(epsilon) = L iff lim_{ntoinfty}f(a_n) = Lquad text{for $mathbf{all}$ sequences $a_nto 0^+$.}$$
With this characterization, we can formulate a version of the dominated convergence theorem involving continuous-parameter families of functions (note that I use quotations to title these versions of the DCT because these names are not standard as far as I know):
"Continuous" DCT. Suppose ${f_epsilon}_{0<epsilon<epsilon_0}$ is a $1$-continuous-parameter family of (measurable) functions such that $|f_epsilon| le g$ for some integrable function $g$ and all $0<epsilon<epsilon_0$, and $lim_{epsilonto0^+}f_epsilon=f$ pointwise almost everywhere. Then, $f$ is an integrable function and $lim_{epsilonto 0^+}int f_epsilon = int f$. This can be written as
$$ lim_{epsilonto0^+}int f_epsilon = int lim_{epsilonto0^+} f_epsilon.$$
The way we use the continuous DCT in practice is by picking an arbitrary sequence $pmb{a_nto 0^+}$ and showing that the hypotheses of the "discrete" DCT are satisfied for this arbitrary sequence $a_n$, using only the assumption that $a_nto 0^+$ and properties of the family ${f_epsilon}$ that are known to us.
$endgroup$
The statement of the dominated convergence theorem (DCT) is as follows:
"Discrete" DCT. Suppose ${f_n}_{n=1}^infty$ is a sequence of (measurable) functions such that $|f_n| le g$ for some integrable function $g$ and all $n$, and $lim_{ntoinfty}f_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_{ntoinfty}int f_n = int f$ (by the triangle inequality). This can be written as
$$ lim_{ntoinfty}int f_n = int lim_{ntoinfty} f_n.$$
(The statement and conclusion of the monotone convergence theorem are similar, but it has a somewhat different set of hypotheses.)
As you note, the statements of these theorems involve sequences of functions, i.e., a $1$-discrete-parameter family of functions ${f_n}_{n=1}^infty$. To apply these theorems to a $1$-continuous-parameter family of functions, say ${f_epsilon}_{0<epsilon<epsilon_0}$, one typically uses a characterization of limits involving a continuous parameter in terms of sequences:
Proposition. If $f$ is a function, then
$$lim_{epsilonto0^+}f(epsilon) = L iff lim_{ntoinfty}f(a_n) = Lquad text{for $mathbf{all}$ sequences $a_nto 0^+$.}$$
With this characterization, we can formulate a version of the dominated convergence theorem involving continuous-parameter families of functions (note that I use quotations to title these versions of the DCT because these names are not standard as far as I know):
"Continuous" DCT. Suppose ${f_epsilon}_{0<epsilon<epsilon_0}$ is a $1$-continuous-parameter family of (measurable) functions such that $|f_epsilon| le g$ for some integrable function $g$ and all $0<epsilon<epsilon_0$, and $lim_{epsilonto0^+}f_epsilon=f$ pointwise almost everywhere. Then, $f$ is an integrable function and $lim_{epsilonto 0^+}int f_epsilon = int f$. This can be written as
$$ lim_{epsilonto0^+}int f_epsilon = int lim_{epsilonto0^+} f_epsilon.$$
The way we use the continuous DCT in practice is by picking an arbitrary sequence $pmb{a_nto 0^+}$ and showing that the hypotheses of the "discrete" DCT are satisfied for this arbitrary sequence $a_n$, using only the assumption that $a_nto 0^+$ and properties of the family ${f_epsilon}$ that are known to us.
edited 4 hours ago
answered 4 hours ago
Alex OrtizAlex Ortiz
11.2k21441
11.2k21441
$begingroup$
Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
$endgroup$
– Ivan V.
2 hours ago
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
1 hour ago
add a comment |
$begingroup$
Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
$endgroup$
– Ivan V.
2 hours ago
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
1 hour ago
$begingroup$
Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
$endgroup$
– Ivan V.
2 hours ago
$begingroup$
Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
$endgroup$
– Ivan V.
2 hours ago
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
1 hour ago
$begingroup$
@IvanV.: Yes, that's correct!
$endgroup$
– Alex Ortiz
1 hour ago
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%2f3168778%2fdominated-convergence-theorem-what-sequence%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