Mixing
Convergence in measure and almost everywhere proof
$begingroup$
I'm trying to prove the following :
Let $(X,mathcal A,mu)$ be a measure space. Let $f_n in L^1$ converge in measure to $f in L^1$ and that
$$ sum_k int vert f_k - fvert d mu < infty$$
Then $f_n$ converges almost everywhere to $f$.
Attempt & ideas :
By contradiction
Suppose $f_n$ doesn't converge almost everywhere to $f$. Then $ not exists A in mathcal A ; s.t. ; mu(A) = 0$ and
$$ N = { x : f_k not rightarrow f } subset A.$$
Let x $in N,exists epsilon_x > 0,$ s.t. $forall k$ :
$$ vert f_k(x) - f(x) vert > epsilon_x.$$
Now the rough idea of rest of my "proof" is the following :
take $epsilon$ the smallest of all $epsilon_x$ then if $epsilon neq 0$ then
$$ 0 < mu{x :vert f_k(x) - f(x)vert > epsilon } < frac{1}{epsilon} int vert f_k - f vert d mu$$
and get a contradiction since $sum_k int vert f_k - fvert d mu < infty Rightarrow int vert f_k - f vert d mu$ $rightarrow 0$.
measure-theory convergence proof-writing
asked Jan 12 at 22:21
DigitalisDigitalis
528216
$endgroup$
add a comment |
$begingroup$
I'm trying to prove the following :
Let $(X,mathcal A,mu)$ be a measure space. Let $f_n in L^1$ converge in measure to $f in L^1$ and that
$$ sum_k int vert f_k - fvert d mu < infty$$
Then $f_n$ converges almost everywhere to $f$.
Attempt & ideas :
By contradiction
Suppose $f_n$ doesn't converge almost everywhere to $f$. Then $ not exists A in mathcal A ; s.t. ; mu(A) = 0$ and
$$ N = { x : f_k not rightarrow f } subset A.$$
Let x $in N,exists epsilon_x > 0,$ s.t. $forall k$ :
$$ vert f_k(x) - f(x) vert > epsilon_x.$$
Now the rough idea of rest of my "proof" is the following :
take $epsilon$ the smallest of all $epsilon_x$ then if $epsilon neq 0$ then
$$ 0 < mu{x :vert f_k(x) - f(x)vert > epsilon } < frac{1}{epsilon} int vert f_k - f vert d mu$$
and get a contradiction since $sum_k int vert f_k - fvert d mu < infty Rightarrow int vert f_k - f vert d mu$ $rightarrow 0$.
measure-theory convergence proof-writing
asked Jan 12 at 22:21
DigitalisDigitalis
528216
$endgroup$
add a comment |
$begingroup$
I'm trying to prove the following :
Let $(X,mathcal A,mu)$ be a measure space. Let $f_n in L^1$ converge in measure to $f in L^1$ and that
$$ sum_k int vert f_k - fvert d mu < infty$$
Then $f_n$ converges almost everywhere to $f$.
Attempt & ideas :
By contradiction
Suppose $f_n$ doesn't converge almost everywhere to $f$. Then $ not exists A in mathcal A ; s.t. ; mu(A) = 0$ and
$$ N = { x : f_k not rightarrow f } subset A.$$
Let x $in N,exists epsilon_x > 0,$ s.t. $forall k$ :
$$ vert f_k(x) - f(x) vert > epsilon_x.$$
Now the rough idea of rest of my "proof" is the following :
take $epsilon$ the smallest of all $epsilon_x$ then if $epsilon neq 0$ then
$$ 0 < mu{x :vert f_k(x) - f(x)vert > epsilon } < frac{1}{epsilon} int vert f_k - f vert d mu$$
and get a contradiction since $sum_k int vert f_k - fvert d mu < infty Rightarrow int vert f_k - f vert d mu$ $rightarrow 0$.
measure-theory convergence proof-writing
asked Jan 12 at 22:21
DigitalisDigitalis
528216
$endgroup$
I'm trying to prove the following :
Let $(X,mathcal A,mu)$ be a measure space. Let $f_n in L^1$ converge in measure to $f in L^1$ and that
$$ sum_k int vert f_k - fvert d mu < infty$$
Then $f_n$ converges almost everywhere to $f$.
Attempt & ideas :
By contradiction
Suppose $f_n$ doesn't converge almost everywhere to $f$. Then $ not exists A in mathcal A ; s.t. ; mu(A) = 0$ and
$$ N = { x : f_k not rightarrow f } subset A.$$
Let x $in N,exists epsilon_x > 0,$ s.t. $forall k$ :
$$ vert f_k(x) - f(x) vert > epsilon_x.$$
Now the rough idea of rest of my "proof" is the following :
take $epsilon$ the smallest of all $epsilon_x$ then if $epsilon neq 0$ then
$$ 0 < mu{x :vert f_k(x) - f(x)vert > epsilon } < frac{1}{epsilon} int vert f_k - f vert d mu$$
and get a contradiction since $sum_k int vert f_k - fvert d mu < infty Rightarrow int vert f_k - f vert d mu$ $rightarrow 0$.
measure-theory convergence proof-writing
measure-theory convergence proof-writing
asked Jan 12 at 22:21
DigitalisDigitalis
528216
asked Jan 12 at 22:21
DigitalisDigitalis
528216
asked Jan 12 at 22:21
DigitalisDigitalis
528216
asked Jan 12 at 22:21
DigitalisDigitalis
528216
asked Jan 12 at 22:21
DigitalisDigitalis
528216
528216
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Fix $varepsilon>0$. Note that
$$
sum_{n=1}^inftymu(|f_n-f|>varepsilon)leqsum_{n=1}^inftyfrac{1}{varepsilon}int|f_n-f|, dmu<infty
$$
by Markov's inequality. It follows by Borel Cantelli that
$$
mu(|f_n-f|>varepsilon, text{i.o})=0.tag{0}
$$
Note that the set $A$ on which the sequence $(f_n)$ does not converge can be written as
$$
A=bigcup_{varepsilon>0}bigcap_{Ngeq1}bigcup_{ngeq N}(|f_n-f|>varepsilon).tag{1}
$$
where we take $varepsilon$ through a countable sequence decreasing towards zero in the union above (like $1/n$). A union bound applied to $(1)$ together with $(0)$ yield that
$$
mu(A)=0
$$
as desired.
answered Jan 12 at 23:02
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
1
$begingroup$
what does the i.o. stand for in $mu (vert f_n -f vert > epsilon$ i.o.$) = 0.$ ?
$endgroup$
– Digitalis
Jan 12 at 23:30
$begingroup$
It means "infinitely often". The set ${A_n$ i.o$}$ is the set of elements that belong to infinitely many of the sets $A_n$.
$endgroup$
– Mark
Jan 12 at 23:33
add a comment |
$begingroup$
Alright, but what if $epsilon=0$? Usually it is hard to take a smallest element in an infinite set.
Here is what you can do. Let any $ninmathbb{N}$. For each $kinmathbb{N}$ we have $mu{x: |f_k(x)-f(x)|geqfrac{1}{n}}leq nint|f_k-f|dmu$. Hence using the fact that the sum of integrals converges we get $sum_{k=1}^infty mu{x: |f_k(x)-f(x)|geqfrac{1}{n}}<infty$. And now from the Borel-Cantelli lemma we know that there exists a null set $E_n$ such that if $xin Xsetminus E_n$ then $x$ belongs only to a finite number of the sets $({x: |f_k(x)-f(x)|geqfrac{1}{n}})_{k=1}^infty$.
Now do that for every $ninmathbb{N}$ and then define $E=cup_{n=1}^infty E_n$. It is a null set as a countable union of null sets. Now show that there is pointwise convergence in $Xsetminus E$.
answered Jan 12 at 22:50
MarkMark
7,293417
$endgroup$
add a comment |
$begingroup$
Using the dominated convergence theorem
$$ sum_k int vert f_k - fvert d mu = int sum_k vert f_k - f vert dmu < infty.$$
Therefore $sum_k vert f_k - f vert $ is integrable and is finite almost everywhere and
$$ lim_{k to infty} vert f_k - f vert = 0 qquad mu.a.e. $$
and the sequence $f_k$ converges to $f$ almost everywhere.
answered Jan 13 at 15:56
DigitalisDigitalis
528216
$endgroup$
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%2f3071455%2fconvergence-in-measure-and-almost-everywhere-proof%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Fix $varepsilon>0$. Note that
$$
sum_{n=1}^inftymu(|f_n-f|>varepsilon)leqsum_{n=1}^inftyfrac{1}{varepsilon}int|f_n-f|, dmu<infty
$$
by Markov's inequality. It follows by Borel Cantelli that
$$
mu(|f_n-f|>varepsilon, text{i.o})=0.tag{0}
$$
Note that the set $A$ on which the sequence $(f_n)$ does not converge can be written as
$$
A=bigcup_{varepsilon>0}bigcap_{Ngeq1}bigcup_{ngeq N}(|f_n-f|>varepsilon).tag{1}
$$
where we take $varepsilon$ through a countable sequence decreasing towards zero in the union above (like $1/n$). A union bound applied to $(1)$ together with $(0)$ yield that
$$
mu(A)=0
$$
as desired.
answered Jan 12 at 23:02
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
1
$begingroup$
what does the i.o. stand for in $mu (vert f_n -f vert > epsilon$ i.o.$) = 0.$ ?
$endgroup$
– Digitalis
Jan 12 at 23:30
$begingroup$
It means "infinitely often". The set ${A_n$ i.o$}$ is the set of elements that belong to infinitely many of the sets $A_n$.
$endgroup$
– Mark
Jan 12 at 23:33
add a comment |
$begingroup$
Fix $varepsilon>0$. Note that
$$
sum_{n=1}^inftymu(|f_n-f|>varepsilon)leqsum_{n=1}^inftyfrac{1}{varepsilon}int|f_n-f|, dmu<infty
$$
by Markov's inequality. It follows by Borel Cantelli that
$$
mu(|f_n-f|>varepsilon, text{i.o})=0.tag{0}
$$
Note that the set $A$ on which the sequence $(f_n)$ does not converge can be written as
$$
A=bigcup_{varepsilon>0}bigcap_{Ngeq1}bigcup_{ngeq N}(|f_n-f|>varepsilon).tag{1}
$$
where we take $varepsilon$ through a countable sequence decreasing towards zero in the union above (like $1/n$). A union bound applied to $(1)$ together with $(0)$ yield that
$$
mu(A)=0
$$
as desired.
answered Jan 12 at 23:02
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
1
$begingroup$
what does the i.o. stand for in $mu (vert f_n -f vert > epsilon$ i.o.$) = 0.$ ?
$endgroup$
– Digitalis
Jan 12 at 23:30
$begingroup$
It means "infinitely often". The set ${A_n$ i.o$}$ is the set of elements that belong to infinitely many of the sets $A_n$.
$endgroup$
– Mark
Jan 12 at 23:33
add a comment |
$begingroup$
Fix $varepsilon>0$. Note that
$$
sum_{n=1}^inftymu(|f_n-f|>varepsilon)leqsum_{n=1}^inftyfrac{1}{varepsilon}int|f_n-f|, dmu<infty
$$
by Markov's inequality. It follows by Borel Cantelli that
$$
mu(|f_n-f|>varepsilon, text{i.o})=0.tag{0}
$$
Note that the set $A$ on which the sequence $(f_n)$ does not converge can be written as
$$
A=bigcup_{varepsilon>0}bigcap_{Ngeq1}bigcup_{ngeq N}(|f_n-f|>varepsilon).tag{1}
$$
where we take $varepsilon$ through a countable sequence decreasing towards zero in the union above (like $1/n$). A union bound applied to $(1)$ together with $(0)$ yield that
$$
mu(A)=0
$$
as desired.
answered Jan 12 at 23:02
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
Fix $varepsilon>0$. Note that
$$
sum_{n=1}^inftymu(|f_n-f|>varepsilon)leqsum_{n=1}^inftyfrac{1}{varepsilon}int|f_n-f|, dmu<infty
$$
by Markov's inequality. It follows by Borel Cantelli that
$$
mu(|f_n-f|>varepsilon, text{i.o})=0.tag{0}
$$
Note that the set $A$ on which the sequence $(f_n)$ does not converge can be written as
$$
A=bigcup_{varepsilon>0}bigcap_{Ngeq1}bigcup_{ngeq N}(|f_n-f|>varepsilon).tag{1}
$$
where we take $varepsilon$ through a countable sequence decreasing towards zero in the union above (like $1/n$). A union bound applied to $(1)$ together with $(0)$ yield that
$$
mu(A)=0
$$
as desired.
answered Jan 12 at 23:02
Foobaz JohnFoobaz John
22.1k41352
edited Jan 12 at 23:11
answered Jan 12 at 23:02
Foobaz JohnFoobaz John
22.1k41352
answered Jan 12 at 23:02
Foobaz JohnFoobaz John
22.1k41352
answered Jan 12 at 23:02
Foobaz JohnFoobaz John
22.1k41352
22.1k41352
1
$begingroup$
what does the i.o. stand for in $mu (vert f_n -f vert > epsilon$ i.o.$) = 0.$ ?
$endgroup$
– Digitalis
Jan 12 at 23:30
$begingroup$
It means "infinitely often". The set ${A_n$ i.o$}$ is the set of elements that belong to infinitely many of the sets $A_n$.
$endgroup$
– Mark
Jan 12 at 23:33
add a comment |
1
$begingroup$
what does the i.o. stand for in $mu (vert f_n -f vert > epsilon$ i.o.$) = 0.$ ?
$endgroup$
– Digitalis
Jan 12 at 23:30
$begingroup$
It means "infinitely often". The set ${A_n$ i.o$}$ is the set of elements that belong to infinitely many of the sets $A_n$.
$endgroup$
– Mark
Jan 12 at 23:33
1
1
$begingroup$
what does the i.o. stand for in $mu (vert f_n -f vert > epsilon$ i.o.$) = 0.$ ?
$endgroup$
– Digitalis
Jan 12 at 23:30
$begingroup$
what does the i.o. stand for in $mu (vert f_n -f vert > epsilon$ i.o.$) = 0.$ ?
$endgroup$
– Digitalis
Jan 12 at 23:30
$begingroup$
It means "infinitely often". The set ${A_n$ i.o$}$ is the set of elements that belong to infinitely many of the sets $A_n$.
$endgroup$
– Mark
Jan 12 at 23:33
$begingroup$
It means "infinitely often". The set ${A_n$ i.o$}$ is the set of elements that belong to infinitely many of the sets $A_n$.
$endgroup$
– Mark
Jan 12 at 23:33
add a comment |
$begingroup$
Alright, but what if $epsilon=0$? Usually it is hard to take a smallest element in an infinite set.
Here is what you can do. Let any $ninmathbb{N}$. For each $kinmathbb{N}$ we have $mu{x: |f_k(x)-f(x)|geqfrac{1}{n}}leq nint|f_k-f|dmu$. Hence using the fact that the sum of integrals converges we get $sum_{k=1}^infty mu{x: |f_k(x)-f(x)|geqfrac{1}{n}}<infty$. And now from the Borel-Cantelli lemma we know that there exists a null set $E_n$ such that if $xin Xsetminus E_n$ then $x$ belongs only to a finite number of the sets $({x: |f_k(x)-f(x)|geqfrac{1}{n}})_{k=1}^infty$.
Now do that for every $ninmathbb{N}$ and then define $E=cup_{n=1}^infty E_n$. It is a null set as a countable union of null sets. Now show that there is pointwise convergence in $Xsetminus E$.
answered Jan 12 at 22:50
MarkMark
7,293417
$endgroup$
add a comment |
$begingroup$
Alright, but what if $epsilon=0$? Usually it is hard to take a smallest element in an infinite set.
Here is what you can do. Let any $ninmathbb{N}$. For each $kinmathbb{N}$ we have $mu{x: |f_k(x)-f(x)|geqfrac{1}{n}}leq nint|f_k-f|dmu$. Hence using the fact that the sum of integrals converges we get $sum_{k=1}^infty mu{x: |f_k(x)-f(x)|geqfrac{1}{n}}<infty$. And now from the Borel-Cantelli lemma we know that there exists a null set $E_n$ such that if $xin Xsetminus E_n$ then $x$ belongs only to a finite number of the sets $({x: |f_k(x)-f(x)|geqfrac{1}{n}})_{k=1}^infty$.
Now do that for every $ninmathbb{N}$ and then define $E=cup_{n=1}^infty E_n$. It is a null set as a countable union of null sets. Now show that there is pointwise convergence in $Xsetminus E$.
answered Jan 12 at 22:50
MarkMark
7,293417
$endgroup$
add a comment |
$begingroup$
Alright, but what if $epsilon=0$? Usually it is hard to take a smallest element in an infinite set.
Here is what you can do. Let any $ninmathbb{N}$. For each $kinmathbb{N}$ we have $mu{x: |f_k(x)-f(x)|geqfrac{1}{n}}leq nint|f_k-f|dmu$. Hence using the fact that the sum of integrals converges we get $sum_{k=1}^infty mu{x: |f_k(x)-f(x)|geqfrac{1}{n}}<infty$. And now from the Borel-Cantelli lemma we know that there exists a null set $E_n$ such that if $xin Xsetminus E_n$ then $x$ belongs only to a finite number of the sets $({x: |f_k(x)-f(x)|geqfrac{1}{n}})_{k=1}^infty$.
Now do that for every $ninmathbb{N}$ and then define $E=cup_{n=1}^infty E_n$. It is a null set as a countable union of null sets. Now show that there is pointwise convergence in $Xsetminus E$.
answered Jan 12 at 22:50
MarkMark
7,293417
$endgroup$
Alright, but what if $epsilon=0$? Usually it is hard to take a smallest element in an infinite set.
Here is what you can do. Let any $ninmathbb{N}$. For each $kinmathbb{N}$ we have $mu{x: |f_k(x)-f(x)|geqfrac{1}{n}}leq nint|f_k-f|dmu$. Hence using the fact that the sum of integrals converges we get $sum_{k=1}^infty mu{x: |f_k(x)-f(x)|geqfrac{1}{n}}<infty$. And now from the Borel-Cantelli lemma we know that there exists a null set $E_n$ such that if $xin Xsetminus E_n$ then $x$ belongs only to a finite number of the sets $({x: |f_k(x)-f(x)|geqfrac{1}{n}})_{k=1}^infty$.
Now do that for every $ninmathbb{N}$ and then define $E=cup_{n=1}^infty E_n$. It is a null set as a countable union of null sets. Now show that there is pointwise convergence in $Xsetminus E$.
answered Jan 12 at 22:50
MarkMark
7,293417
answered Jan 12 at 22:50
MarkMark
7,293417
answered Jan 12 at 22:50
MarkMark
7,293417
answered Jan 12 at 22:50
MarkMark
7,293417
7,293417
add a comment |
add a comment |
$begingroup$
Using the dominated convergence theorem
$$ sum_k int vert f_k - fvert d mu = int sum_k vert f_k - f vert dmu < infty.$$
Therefore $sum_k vert f_k - f vert $ is integrable and is finite almost everywhere and
$$ lim_{k to infty} vert f_k - f vert = 0 qquad mu.a.e. $$
and the sequence $f_k$ converges to $f$ almost everywhere.
answered Jan 13 at 15:56
DigitalisDigitalis
528216
$endgroup$
add a comment |
$begingroup$
Using the dominated convergence theorem
$$ sum_k int vert f_k - fvert d mu = int sum_k vert f_k - f vert dmu < infty.$$
Therefore $sum_k vert f_k - f vert $ is integrable and is finite almost everywhere and
$$ lim_{k to infty} vert f_k - f vert = 0 qquad mu.a.e. $$
and the sequence $f_k$ converges to $f$ almost everywhere.
answered Jan 13 at 15:56
DigitalisDigitalis
528216
$endgroup$
add a comment |
$begingroup$
Using the dominated convergence theorem
$$ sum_k int vert f_k - fvert d mu = int sum_k vert f_k - f vert dmu < infty.$$
Therefore $sum_k vert f_k - f vert $ is integrable and is finite almost everywhere and
$$ lim_{k to infty} vert f_k - f vert = 0 qquad mu.a.e. $$
and the sequence $f_k$ converges to $f$ almost everywhere.
answered Jan 13 at 15:56
DigitalisDigitalis
528216
$endgroup$
Using the dominated convergence theorem
$$ sum_k int vert f_k - fvert d mu = int sum_k vert f_k - f vert dmu < infty.$$
Therefore $sum_k vert f_k - f vert $ is integrable and is finite almost everywhere and
$$ lim_{k to infty} vert f_k - f vert = 0 qquad mu.a.e. $$
and the sequence $f_k$ converges to $f$ almost everywhere.
answered Jan 13 at 15:56
DigitalisDigitalis
528216
answered Jan 13 at 15:56
DigitalisDigitalis
528216
answered Jan 13 at 15:56
DigitalisDigitalis
528216
answered Jan 13 at 15:56
DigitalisDigitalis
528216
528216
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.
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%2f3071455%2fconvergence-in-measure-and-almost-everywhere-proof%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
