Mixing
Riemann itegrability of functions continuous with the exception of a zero content set
$begingroup$
I'm looking for a proof of this statement
Let $E ⊆ ℝ^n$ be a set of Jordan content zero (it means that for each $ε > 0$ there are compact rectangles $E_1,…,E_m$ such that $E ⊆ ∪_jE_j$ and $∑_jvol(E_j) < ε$). If $I = [a_1,b_1]×…×[a_n,b_n]$ ($a_i < b_i$) and $f : I → ℝ$ is bounded on $I$ and continuous on $I – E$, then $f$ is Riemann integrable on $I$.
I know the proof of the version in which f is supposed continuous, but here the presence of the zero content set complicates the matters.
integration
asked Jan 15 at 6:22
sawesawe
345
$endgroup$
add a comment |
$begingroup$
I'm looking for a proof of this statement
Let $E ⊆ ℝ^n$ be a set of Jordan content zero (it means that for each $ε > 0$ there are compact rectangles $E_1,…,E_m$ such that $E ⊆ ∪_jE_j$ and $∑_jvol(E_j) < ε$). If $I = [a_1,b_1]×…×[a_n,b_n]$ ($a_i < b_i$) and $f : I → ℝ$ is bounded on $I$ and continuous on $I – E$, then $f$ is Riemann integrable on $I$.
I know the proof of the version in which f is supposed continuous, but here the presence of the zero content set complicates the matters.
integration
asked Jan 15 at 6:22
sawesawe
345
$endgroup$
add a comment |
$begingroup$
I'm looking for a proof of this statement
Let $E ⊆ ℝ^n$ be a set of Jordan content zero (it means that for each $ε > 0$ there are compact rectangles $E_1,…,E_m$ such that $E ⊆ ∪_jE_j$ and $∑_jvol(E_j) < ε$). If $I = [a_1,b_1]×…×[a_n,b_n]$ ($a_i < b_i$) and $f : I → ℝ$ is bounded on $I$ and continuous on $I – E$, then $f$ is Riemann integrable on $I$.
I know the proof of the version in which f is supposed continuous, but here the presence of the zero content set complicates the matters.
integration
asked Jan 15 at 6:22
sawesawe
345
$endgroup$
I'm looking for a proof of this statement
Let $E ⊆ ℝ^n$ be a set of Jordan content zero (it means that for each $ε > 0$ there are compact rectangles $E_1,…,E_m$ such that $E ⊆ ∪_jE_j$ and $∑_jvol(E_j) < ε$). If $I = [a_1,b_1]×…×[a_n,b_n]$ ($a_i < b_i$) and $f : I → ℝ$ is bounded on $I$ and continuous on $I – E$, then $f$ is Riemann integrable on $I$.
I know the proof of the version in which f is supposed continuous, but here the presence of the zero content set complicates the matters.
integration
integration
asked Jan 15 at 6:22
sawesawe
345
asked Jan 15 at 6:22
sawesawe
345
asked Jan 15 at 6:22
sawesawe
345
asked Jan 15 at 6:22
sawesawe
345
asked Jan 15 at 6:22
sawesawe
345
345
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The proof is not difficult, but the notation can become heavy when dealing with multidimensional intervals. I'm assuming you know about partitions of intervals, Darboux sums etc. If not you can find a complete treatment of Riemann n-dimensional integral in Munkres book "Analysis on manifolds".
For each $ε > 0$ exist $E_1,…,E_m ⊆ ℝ^n$, with $E_j = [a_1^j,b_1^j]×…×[a_n^j,b_n^j]$, such that $∑_{j=1}^m vol(E_j) < ε$ and $f$ is continuous on $I - D$, with $D = ∪_{j=1}^m E_j$.
Notice that you can suppose that the $E_j$s are non degenerate subrectangles of $I$.
In fact if $E_j$ and $I$ are compact rectangles then $E_j∩I$ is also a compact rectangle and $vol(E_j∩I) ≤ min{vol(E_j),vol(I)}$. Beside if $I$ is non degenerate (i.e. $a_i < b_i$) and $E_j ⊆ I$ is degenerate, we can find a non degenerate rectangle $E_j'$, of arbitrarily small volume, such that $E_j ⊆ E_j' ⊆ I$.
Also, without loss of generality, you can assume f continuous on $C = I – D° = I - ∪_{j=1}^m(a_1^j,b_1^j)×…×(a_n^j,b_n^j)$. In fact, if $0 < h < min{(b_i^j - a_i^j)/2 | 1 ≤ i ≤ n, 1 ≤ j ≤ m}$ and $F_j = [a_1^j-h,b_1^j+h]×…×[a_n^j-h,b_n^j+h]$, then $vol(F_j) < 2^nvol(E_j) ⇒ ∑_{j=1}^m vol(Fj) < 2^n∑_{j=1}^m vol(E_j) < 2^nε$ with $ε$ arbitrarily small and $f$ continuous on $I - ∪_{j=1}^m E_j ⊇ I - ∪_{j=1}^mF_j°$.
Now, $C$ is compact, so, by Heine-Cantor theorem, we can find $δ > 0$ such that $x, x' ∈ C$ and $∥x'-x∥ < δ$ $⇒ |f(x')-f(x)| < ε$. Let's suppose $|f| ≤ K$, and let $P = (P_1,…,P_n)$ be a (multidimensional) partition of $I$ such that $mesh(P) < δ$. If $P_i^* = P_i ∪ (∪_{j=1}^m{a_i^j,b_i^j})$, $P^* = (P_1^*,…,P_n^*)$ and $J$ is the generic subrectangle of $I$ determined by $P^*$, then:
$$U(f,P^*) - L(f,P^*) = ∑_J osc(f,J)⋅vol(J) = ∑_{J⊆C}osc(f,J)⋅vol(J) + ∑_{J⊆D}osc(f,J)⋅vol(J) ≤ ∑_{J⊆C}ε⋅vol(J) + ∑_{J⊆D}2K⋅vol(J) ≤ ε⋅vol(I) + 2K⋅(∑_{J⊆E_1 }vol(J) +…+ ∑_{J⊆E_m}vol(J)) = ε⋅vol(I) + 2K⋅∑_{j=1}^m vol(E_j) < ε⋅[vol(I)+2K]$$
Because $ε$ is arbitrarily small, this inequality proves that $f$ is integrable on $I$.
As a side note, keep in mind that this result is not reversible, there are integrable function which are not continuous with the exception of a zero-content set. But it will become reversible if we switch from Peano-Jordan to Lebesgue measure. In fact there is a theorem (due to Lebesgue) which affirms that a function is (Riemann) integrable if and only if is continuous (Lebesgue) almost everywhere. You can find the (non trivial) proof of this result in Munkres book I mentioned earlier.
answered Jan 15 at 6:55
Alex123Alex123
775
$endgroup$
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%2f3074132%2friemann-itegrability-of-functions-continuous-with-the-exception-of-a-zero-conten%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
$begingroup$
The proof is not difficult, but the notation can become heavy when dealing with multidimensional intervals. I'm assuming you know about partitions of intervals, Darboux sums etc. If not you can find a complete treatment of Riemann n-dimensional integral in Munkres book "Analysis on manifolds".
For each $ε > 0$ exist $E_1,…,E_m ⊆ ℝ^n$, with $E_j = [a_1^j,b_1^j]×…×[a_n^j,b_n^j]$, such that $∑_{j=1}^m vol(E_j) < ε$ and $f$ is continuous on $I - D$, with $D = ∪_{j=1}^m E_j$.
Notice that you can suppose that the $E_j$s are non degenerate subrectangles of $I$.
In fact if $E_j$ and $I$ are compact rectangles then $E_j∩I$ is also a compact rectangle and $vol(E_j∩I) ≤ min{vol(E_j),vol(I)}$. Beside if $I$ is non degenerate (i.e. $a_i < b_i$) and $E_j ⊆ I$ is degenerate, we can find a non degenerate rectangle $E_j'$, of arbitrarily small volume, such that $E_j ⊆ E_j' ⊆ I$.
Also, without loss of generality, you can assume f continuous on $C = I – D° = I - ∪_{j=1}^m(a_1^j,b_1^j)×…×(a_n^j,b_n^j)$. In fact, if $0 < h < min{(b_i^j - a_i^j)/2 | 1 ≤ i ≤ n, 1 ≤ j ≤ m}$ and $F_j = [a_1^j-h,b_1^j+h]×…×[a_n^j-h,b_n^j+h]$, then $vol(F_j) < 2^nvol(E_j) ⇒ ∑_{j=1}^m vol(Fj) < 2^n∑_{j=1}^m vol(E_j) < 2^nε$ with $ε$ arbitrarily small and $f$ continuous on $I - ∪_{j=1}^m E_j ⊇ I - ∪_{j=1}^mF_j°$.
Now, $C$ is compact, so, by Heine-Cantor theorem, we can find $δ > 0$ such that $x, x' ∈ C$ and $∥x'-x∥ < δ$ $⇒ |f(x')-f(x)| < ε$. Let's suppose $|f| ≤ K$, and let $P = (P_1,…,P_n)$ be a (multidimensional) partition of $I$ such that $mesh(P) < δ$. If $P_i^* = P_i ∪ (∪_{j=1}^m{a_i^j,b_i^j})$, $P^* = (P_1^*,…,P_n^*)$ and $J$ is the generic subrectangle of $I$ determined by $P^*$, then:
$$U(f,P^*) - L(f,P^*) = ∑_J osc(f,J)⋅vol(J) = ∑_{J⊆C}osc(f,J)⋅vol(J) + ∑_{J⊆D}osc(f,J)⋅vol(J) ≤ ∑_{J⊆C}ε⋅vol(J) + ∑_{J⊆D}2K⋅vol(J) ≤ ε⋅vol(I) + 2K⋅(∑_{J⊆E_1 }vol(J) +…+ ∑_{J⊆E_m}vol(J)) = ε⋅vol(I) + 2K⋅∑_{j=1}^m vol(E_j) < ε⋅[vol(I)+2K]$$
Because $ε$ is arbitrarily small, this inequality proves that $f$ is integrable on $I$.
As a side note, keep in mind that this result is not reversible, there are integrable function which are not continuous with the exception of a zero-content set. But it will become reversible if we switch from Peano-Jordan to Lebesgue measure. In fact there is a theorem (due to Lebesgue) which affirms that a function is (Riemann) integrable if and only if is continuous (Lebesgue) almost everywhere. You can find the (non trivial) proof of this result in Munkres book I mentioned earlier.
answered Jan 15 at 6:55
Alex123Alex123
775
$endgroup$
add a comment |
$begingroup$
The proof is not difficult, but the notation can become heavy when dealing with multidimensional intervals. I'm assuming you know about partitions of intervals, Darboux sums etc. If not you can find a complete treatment of Riemann n-dimensional integral in Munkres book "Analysis on manifolds".
For each $ε > 0$ exist $E_1,…,E_m ⊆ ℝ^n$, with $E_j = [a_1^j,b_1^j]×…×[a_n^j,b_n^j]$, such that $∑_{j=1}^m vol(E_j) < ε$ and $f$ is continuous on $I - D$, with $D = ∪_{j=1}^m E_j$.
Notice that you can suppose that the $E_j$s are non degenerate subrectangles of $I$.
In fact if $E_j$ and $I$ are compact rectangles then $E_j∩I$ is also a compact rectangle and $vol(E_j∩I) ≤ min{vol(E_j),vol(I)}$. Beside if $I$ is non degenerate (i.e. $a_i < b_i$) and $E_j ⊆ I$ is degenerate, we can find a non degenerate rectangle $E_j'$, of arbitrarily small volume, such that $E_j ⊆ E_j' ⊆ I$.
Also, without loss of generality, you can assume f continuous on $C = I – D° = I - ∪_{j=1}^m(a_1^j,b_1^j)×…×(a_n^j,b_n^j)$. In fact, if $0 < h < min{(b_i^j - a_i^j)/2 | 1 ≤ i ≤ n, 1 ≤ j ≤ m}$ and $F_j = [a_1^j-h,b_1^j+h]×…×[a_n^j-h,b_n^j+h]$, then $vol(F_j) < 2^nvol(E_j) ⇒ ∑_{j=1}^m vol(Fj) < 2^n∑_{j=1}^m vol(E_j) < 2^nε$ with $ε$ arbitrarily small and $f$ continuous on $I - ∪_{j=1}^m E_j ⊇ I - ∪_{j=1}^mF_j°$.
Now, $C$ is compact, so, by Heine-Cantor theorem, we can find $δ > 0$ such that $x, x' ∈ C$ and $∥x'-x∥ < δ$ $⇒ |f(x')-f(x)| < ε$. Let's suppose $|f| ≤ K$, and let $P = (P_1,…,P_n)$ be a (multidimensional) partition of $I$ such that $mesh(P) < δ$. If $P_i^* = P_i ∪ (∪_{j=1}^m{a_i^j,b_i^j})$, $P^* = (P_1^*,…,P_n^*)$ and $J$ is the generic subrectangle of $I$ determined by $P^*$, then:
$$U(f,P^*) - L(f,P^*) = ∑_J osc(f,J)⋅vol(J) = ∑_{J⊆C}osc(f,J)⋅vol(J) + ∑_{J⊆D}osc(f,J)⋅vol(J) ≤ ∑_{J⊆C}ε⋅vol(J) + ∑_{J⊆D}2K⋅vol(J) ≤ ε⋅vol(I) + 2K⋅(∑_{J⊆E_1 }vol(J) +…+ ∑_{J⊆E_m}vol(J)) = ε⋅vol(I) + 2K⋅∑_{j=1}^m vol(E_j) < ε⋅[vol(I)+2K]$$
Because $ε$ is arbitrarily small, this inequality proves that $f$ is integrable on $I$.
As a side note, keep in mind that this result is not reversible, there are integrable function which are not continuous with the exception of a zero-content set. But it will become reversible if we switch from Peano-Jordan to Lebesgue measure. In fact there is a theorem (due to Lebesgue) which affirms that a function is (Riemann) integrable if and only if is continuous (Lebesgue) almost everywhere. You can find the (non trivial) proof of this result in Munkres book I mentioned earlier.
answered Jan 15 at 6:55
Alex123Alex123
775
$endgroup$
add a comment |
$begingroup$
The proof is not difficult, but the notation can become heavy when dealing with multidimensional intervals. I'm assuming you know about partitions of intervals, Darboux sums etc. If not you can find a complete treatment of Riemann n-dimensional integral in Munkres book "Analysis on manifolds".
For each $ε > 0$ exist $E_1,…,E_m ⊆ ℝ^n$, with $E_j = [a_1^j,b_1^j]×…×[a_n^j,b_n^j]$, such that $∑_{j=1}^m vol(E_j) < ε$ and $f$ is continuous on $I - D$, with $D = ∪_{j=1}^m E_j$.
Notice that you can suppose that the $E_j$s are non degenerate subrectangles of $I$.
In fact if $E_j$ and $I$ are compact rectangles then $E_j∩I$ is also a compact rectangle and $vol(E_j∩I) ≤ min{vol(E_j),vol(I)}$. Beside if $I$ is non degenerate (i.e. $a_i < b_i$) and $E_j ⊆ I$ is degenerate, we can find a non degenerate rectangle $E_j'$, of arbitrarily small volume, such that $E_j ⊆ E_j' ⊆ I$.
Also, without loss of generality, you can assume f continuous on $C = I – D° = I - ∪_{j=1}^m(a_1^j,b_1^j)×…×(a_n^j,b_n^j)$. In fact, if $0 < h < min{(b_i^j - a_i^j)/2 | 1 ≤ i ≤ n, 1 ≤ j ≤ m}$ and $F_j = [a_1^j-h,b_1^j+h]×…×[a_n^j-h,b_n^j+h]$, then $vol(F_j) < 2^nvol(E_j) ⇒ ∑_{j=1}^m vol(Fj) < 2^n∑_{j=1}^m vol(E_j) < 2^nε$ with $ε$ arbitrarily small and $f$ continuous on $I - ∪_{j=1}^m E_j ⊇ I - ∪_{j=1}^mF_j°$.
Now, $C$ is compact, so, by Heine-Cantor theorem, we can find $δ > 0$ such that $x, x' ∈ C$ and $∥x'-x∥ < δ$ $⇒ |f(x')-f(x)| < ε$. Let's suppose $|f| ≤ K$, and let $P = (P_1,…,P_n)$ be a (multidimensional) partition of $I$ such that $mesh(P) < δ$. If $P_i^* = P_i ∪ (∪_{j=1}^m{a_i^j,b_i^j})$, $P^* = (P_1^*,…,P_n^*)$ and $J$ is the generic subrectangle of $I$ determined by $P^*$, then:
$$U(f,P^*) - L(f,P^*) = ∑_J osc(f,J)⋅vol(J) = ∑_{J⊆C}osc(f,J)⋅vol(J) + ∑_{J⊆D}osc(f,J)⋅vol(J) ≤ ∑_{J⊆C}ε⋅vol(J) + ∑_{J⊆D}2K⋅vol(J) ≤ ε⋅vol(I) + 2K⋅(∑_{J⊆E_1 }vol(J) +…+ ∑_{J⊆E_m}vol(J)) = ε⋅vol(I) + 2K⋅∑_{j=1}^m vol(E_j) < ε⋅[vol(I)+2K]$$
Because $ε$ is arbitrarily small, this inequality proves that $f$ is integrable on $I$.
As a side note, keep in mind that this result is not reversible, there are integrable function which are not continuous with the exception of a zero-content set. But it will become reversible if we switch from Peano-Jordan to Lebesgue measure. In fact there is a theorem (due to Lebesgue) which affirms that a function is (Riemann) integrable if and only if is continuous (Lebesgue) almost everywhere. You can find the (non trivial) proof of this result in Munkres book I mentioned earlier.
answered Jan 15 at 6:55
Alex123Alex123
775
$endgroup$
The proof is not difficult, but the notation can become heavy when dealing with multidimensional intervals. I'm assuming you know about partitions of intervals, Darboux sums etc. If not you can find a complete treatment of Riemann n-dimensional integral in Munkres book "Analysis on manifolds".
For each $ε > 0$ exist $E_1,…,E_m ⊆ ℝ^n$, with $E_j = [a_1^j,b_1^j]×…×[a_n^j,b_n^j]$, such that $∑_{j=1}^m vol(E_j) < ε$ and $f$ is continuous on $I - D$, with $D = ∪_{j=1}^m E_j$.
Notice that you can suppose that the $E_j$s are non degenerate subrectangles of $I$.
In fact if $E_j$ and $I$ are compact rectangles then $E_j∩I$ is also a compact rectangle and $vol(E_j∩I) ≤ min{vol(E_j),vol(I)}$. Beside if $I$ is non degenerate (i.e. $a_i < b_i$) and $E_j ⊆ I$ is degenerate, we can find a non degenerate rectangle $E_j'$, of arbitrarily small volume, such that $E_j ⊆ E_j' ⊆ I$.
Also, without loss of generality, you can assume f continuous on $C = I – D° = I - ∪_{j=1}^m(a_1^j,b_1^j)×…×(a_n^j,b_n^j)$. In fact, if $0 < h < min{(b_i^j - a_i^j)/2 | 1 ≤ i ≤ n, 1 ≤ j ≤ m}$ and $F_j = [a_1^j-h,b_1^j+h]×…×[a_n^j-h,b_n^j+h]$, then $vol(F_j) < 2^nvol(E_j) ⇒ ∑_{j=1}^m vol(Fj) < 2^n∑_{j=1}^m vol(E_j) < 2^nε$ with $ε$ arbitrarily small and $f$ continuous on $I - ∪_{j=1}^m E_j ⊇ I - ∪_{j=1}^mF_j°$.
Now, $C$ is compact, so, by Heine-Cantor theorem, we can find $δ > 0$ such that $x, x' ∈ C$ and $∥x'-x∥ < δ$ $⇒ |f(x')-f(x)| < ε$. Let's suppose $|f| ≤ K$, and let $P = (P_1,…,P_n)$ be a (multidimensional) partition of $I$ such that $mesh(P) < δ$. If $P_i^* = P_i ∪ (∪_{j=1}^m{a_i^j,b_i^j})$, $P^* = (P_1^*,…,P_n^*)$ and $J$ is the generic subrectangle of $I$ determined by $P^*$, then:
$$U(f,P^*) - L(f,P^*) = ∑_J osc(f,J)⋅vol(J) = ∑_{J⊆C}osc(f,J)⋅vol(J) + ∑_{J⊆D}osc(f,J)⋅vol(J) ≤ ∑_{J⊆C}ε⋅vol(J) + ∑_{J⊆D}2K⋅vol(J) ≤ ε⋅vol(I) + 2K⋅(∑_{J⊆E_1 }vol(J) +…+ ∑_{J⊆E_m}vol(J)) = ε⋅vol(I) + 2K⋅∑_{j=1}^m vol(E_j) < ε⋅[vol(I)+2K]$$
Because $ε$ is arbitrarily small, this inequality proves that $f$ is integrable on $I$.
As a side note, keep in mind that this result is not reversible, there are integrable function which are not continuous with the exception of a zero-content set. But it will become reversible if we switch from Peano-Jordan to Lebesgue measure. In fact there is a theorem (due to Lebesgue) which affirms that a function is (Riemann) integrable if and only if is continuous (Lebesgue) almost everywhere. You can find the (non trivial) proof of this result in Munkres book I mentioned earlier.
answered Jan 15 at 6:55
Alex123Alex123
775
edited Jan 15 at 7:15
answered Jan 15 at 6:55
Alex123Alex123
775
answered Jan 15 at 6:55
Alex123Alex123
775
answered Jan 15 at 6:55
Alex123Alex123
775
775
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%2f3074132%2friemann-itegrability-of-functions-continuous-with-the-exception-of-a-zero-conten%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
