Measure of a compact subset of $mathbb{R}^d$












0














Let $K subset mathbb{R}^d$ be closed and bounded.



Define $E_n = {x in mathbb{R}^d vert exists y in K$, $ vert x - y vert < frac{1}{n} }$



wish to show lim$_{n rightarrow infty}$ $m(E_n)$ = $m(K)$.



ATTEMPT:



So I wanted to use double sided inequality to show the equality holds.



NOTE: $E_1 supset E_2 supset E_2 ..$



I have a lemma stating: if additionally $m(E_1) < + infty$ then



lim$_{n rightarrow infty}$ $m(E_n)$ = $m(bigcap_{n = 1}^{infty} E_n$) **



(is ** lemma applicable her?)



using this, If I can make a claim stating $K = bigcap_{n=1}^{infty} E_n $ ( I doubt this is true) then I am done. But I feel like I am missing a subtlety.
Sorry for being a noob, any help is greatly appreciated, I am new to measure theory and trying to learn!



Also, I have an observation, I would like to know is true:



Do all compact subsets of $mathbb{R}^d$ have finite (Lebesgue) measure.






real-analysis measure-theory lebesgue-measure







share|cite|improve this question


















  • 2




    If you have $x in bigcap_{n=1}^{infty}E_{n}$, then $d(x, K) = 0$, right? So since $K$ is closed, $x in K$. So $bigcap_{n=1}^{infty}E_{n} subseteq K$. Also, since $K subseteq E_{n}$ for all $n in mathbb{N}$, you also have $K subseteq bigcap_{n=1}^{infty}E_{n}$, so $K = bigcap_{n=1}^{infty}E_{n}$. Try to work out the details for yourself.
    – Matija Sreckovic
    Jan 3 at 22:19












  • oh yes true, this I can workout, its a simple analysis/point set topology problem. thank you so much!!
    – eyeheartmath
    Jan 3 at 22:26
















0














Let $K subset mathbb{R}^d$ be closed and bounded.



Define $E_n = {x in mathbb{R}^d vert exists y in K$, $ vert x - y vert < frac{1}{n} }$



wish to show lim$_{n rightarrow infty}$ $m(E_n)$ = $m(K)$.



ATTEMPT:



So I wanted to use double sided inequality to show the equality holds.



NOTE: $E_1 supset E_2 supset E_2 ..$



I have a lemma stating: if additionally $m(E_1) < + infty$ then



lim$_{n rightarrow infty}$ $m(E_n)$ = $m(bigcap_{n = 1}^{infty} E_n$) **



(is ** lemma applicable her?)



using this, If I can make a claim stating $K = bigcap_{n=1}^{infty} E_n $ ( I doubt this is true) then I am done. But I feel like I am missing a subtlety.
Sorry for being a noob, any help is greatly appreciated, I am new to measure theory and trying to learn!



Also, I have an observation, I would like to know is true:



Do all compact subsets of $mathbb{R}^d$ have finite (Lebesgue) measure.






real-analysis measure-theory lebesgue-measure







share|cite|improve this question


















  • 2




    If you have $x in bigcap_{n=1}^{infty}E_{n}$, then $d(x, K) = 0$, right? So since $K$ is closed, $x in K$. So $bigcap_{n=1}^{infty}E_{n} subseteq K$. Also, since $K subseteq E_{n}$ for all $n in mathbb{N}$, you also have $K subseteq bigcap_{n=1}^{infty}E_{n}$, so $K = bigcap_{n=1}^{infty}E_{n}$. Try to work out the details for yourself.
    – Matija Sreckovic
    Jan 3 at 22:19












  • oh yes true, this I can workout, its a simple analysis/point set topology problem. thank you so much!!
    – eyeheartmath
    Jan 3 at 22:26














0












0








0







Let $K subset mathbb{R}^d$ be closed and bounded.



Define $E_n = {x in mathbb{R}^d vert exists y in K$, $ vert x - y vert < frac{1}{n} }$



wish to show lim$_{n rightarrow infty}$ $m(E_n)$ = $m(K)$.



ATTEMPT:



So I wanted to use double sided inequality to show the equality holds.



NOTE: $E_1 supset E_2 supset E_2 ..$



I have a lemma stating: if additionally $m(E_1) < + infty$ then



lim$_{n rightarrow infty}$ $m(E_n)$ = $m(bigcap_{n = 1}^{infty} E_n$) **



(is ** lemma applicable her?)



using this, If I can make a claim stating $K = bigcap_{n=1}^{infty} E_n $ ( I doubt this is true) then I am done. But I feel like I am missing a subtlety.
Sorry for being a noob, any help is greatly appreciated, I am new to measure theory and trying to learn!



Also, I have an observation, I would like to know is true:



Do all compact subsets of $mathbb{R}^d$ have finite (Lebesgue) measure.






real-analysis measure-theory lebesgue-measure







share|cite|improve this question













Let $K subset mathbb{R}^d$ be closed and bounded.



Define $E_n = {x in mathbb{R}^d vert exists y in K$, $ vert x - y vert < frac{1}{n} }$



wish to show lim$_{n rightarrow infty}$ $m(E_n)$ = $m(K)$.



ATTEMPT:



So I wanted to use double sided inequality to show the equality holds.



NOTE: $E_1 supset E_2 supset E_2 ..$



I have a lemma stating: if additionally $m(E_1) < + infty$ then



lim$_{n rightarrow infty}$ $m(E_n)$ = $m(bigcap_{n = 1}^{infty} E_n$) **



(is ** lemma applicable her?)



using this, If I can make a claim stating $K = bigcap_{n=1}^{infty} E_n $ ( I doubt this is true) then I am done. But I feel like I am missing a subtlety.
Sorry for being a noob, any help is greatly appreciated, I am new to measure theory and trying to learn!



Also, I have an observation, I would like to know is true:



Do all compact subsets of $mathbb{R}^d$ have finite (Lebesgue) measure.






real-analysis measure-theory lebesgue-measure




real-analysis measure-theory lebesgue-measure






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 3 at 22:09









eyeheartmath

657




657








  • 2




    If you have $x in bigcap_{n=1}^{infty}E_{n}$, then $d(x, K) = 0$, right? So since $K$ is closed, $x in K$. So $bigcap_{n=1}^{infty}E_{n} subseteq K$. Also, since $K subseteq E_{n}$ for all $n in mathbb{N}$, you also have $K subseteq bigcap_{n=1}^{infty}E_{n}$, so $K = bigcap_{n=1}^{infty}E_{n}$. Try to work out the details for yourself.
    – Matija Sreckovic
    Jan 3 at 22:19












  • oh yes true, this I can workout, its a simple analysis/point set topology problem. thank you so much!!
    – eyeheartmath
    Jan 3 at 22:26














  • 2




    If you have $x in bigcap_{n=1}^{infty}E_{n}$, then $d(x, K) = 0$, right? So since $K$ is closed, $x in K$. So $bigcap_{n=1}^{infty}E_{n} subseteq K$. Also, since $K subseteq E_{n}$ for all $n in mathbb{N}$, you also have $K subseteq bigcap_{n=1}^{infty}E_{n}$, so $K = bigcap_{n=1}^{infty}E_{n}$. Try to work out the details for yourself.
    – Matija Sreckovic
    Jan 3 at 22:19












  • oh yes true, this I can workout, its a simple analysis/point set topology problem. thank you so much!!
    – eyeheartmath
    Jan 3 at 22:26








2




2




If you have $x in bigcap_{n=1}^{infty}E_{n}$, then $d(x, K) = 0$, right? So since $K$ is closed, $x in K$. So $bigcap_{n=1}^{infty}E_{n} subseteq K$. Also, since $K subseteq E_{n}$ for all $n in mathbb{N}$, you also have $K subseteq bigcap_{n=1}^{infty}E_{n}$, so $K = bigcap_{n=1}^{infty}E_{n}$. Try to work out the details for yourself.
– Matija Sreckovic
Jan 3 at 22:19






If you have $x in bigcap_{n=1}^{infty}E_{n}$, then $d(x, K) = 0$, right? So since $K$ is closed, $x in K$. So $bigcap_{n=1}^{infty}E_{n} subseteq K$. Also, since $K subseteq E_{n}$ for all $n in mathbb{N}$, you also have $K subseteq bigcap_{n=1}^{infty}E_{n}$, so $K = bigcap_{n=1}^{infty}E_{n}$. Try to work out the details for yourself.
– Matija Sreckovic
Jan 3 at 22:19














oh yes true, this I can workout, its a simple analysis/point set topology problem. thank you so much!!
– eyeheartmath
Jan 3 at 22:26




oh yes true, this I can workout, its a simple analysis/point set topology problem. thank you so much!!
– eyeheartmath
Jan 3 at 22:26










2 Answers
2






active

oldest

votes


















2














For each $xinmathbb{R}^{d}$ and $r>0$, let $B(x,r)={ymid d(x,y)<r}$
be the open ball centered at $x$ with radius $r$. Note that
$$
E_{n}=bigcup_{xin K}B(x,frac{1}{n})
$$
which is a bounded (because $K$ is bounded) open set. In particular
$m(E_{n})<infty$.



To show that $K=cap_{n}E_{n}$. Clearly $E_{1}supseteq E_{2}supseteqcdotssupseteq K$.
We only need to show that $cap_{n}E_{n}subseteq K$. Let $xincap_{n}E_{n}$.
For each $n$, $xin E_{n}Rightarrowexists y_{n}in K$ such that
$xin B(y_{n},frac{1}{n})$. Since $K$ is compact, for the sequence
$(y_{n})$, there exists a convergent subsequence $(y_{n_{k}})$ and
$yin K$ such that $y_{n_{k}}rightarrow y$ as $krightarrowinfty$.
Now
begin{eqnarray*}
d(x,y) & leq & d(x,y_{n_{k}})+d(y_{n_{k}},y)\
& < & frac{1}{n_{k}}+d(y_{n_{k}},y)\
& rightarrow & 0,
end{eqnarray*}
as $krightarrowinfty$. Therefore, $d(x,y)=0$ and hence $x=yin K$.
This shows that $cap_{n}E_{n}subseteq K$.






share|cite|improve this answer























  • got it thanks so much!!!
    – eyeheartmath
    Jan 3 at 22:29



















2















I doubt this is true




It is, and your method is fine, once you've also checked that $m(E_1) < +infty$.




Do all compact subsets of $mathbb{R}^d$ have finite (Lebesgue) measure.




Yes. This follows from the Heine-Borel Theorem: all compact subsets of $mathbb{R}^d$ are bounded, so are contained in some product of closed, bounded intervals, hence have measure bounded above by the measure of that product of closed, bounded intervals, which is finite. There's extra work to do to check that all compact subsets of $mathbb{R}^d$ are measurable, but indeed they are.






share|cite|improve this answer





















  • Awesome!! thank you so much!! I am always very very hesitant on making claims unless I'm 100000% certain. I thought the equality was true but wanted to be extra certain, and yes Heine-Borel, luckily enough my point-set topology is relatively strong specially in comparison to my measure theory. again, thanks!!!!
    – eyeheartmath
    Jan 3 at 22:23













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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061067%2fmeasure-of-a-compact-subset-of-mathbbrd%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









2














For each $xinmathbb{R}^{d}$ and $r>0$, let $B(x,r)={ymid d(x,y)<r}$
be the open ball centered at $x$ with radius $r$. Note that
$$
E_{n}=bigcup_{xin K}B(x,frac{1}{n})
$$
which is a bounded (because $K$ is bounded) open set. In particular
$m(E_{n})<infty$.



To show that $K=cap_{n}E_{n}$. Clearly $E_{1}supseteq E_{2}supseteqcdotssupseteq K$.
We only need to show that $cap_{n}E_{n}subseteq K$. Let $xincap_{n}E_{n}$.
For each $n$, $xin E_{n}Rightarrowexists y_{n}in K$ such that
$xin B(y_{n},frac{1}{n})$. Since $K$ is compact, for the sequence
$(y_{n})$, there exists a convergent subsequence $(y_{n_{k}})$ and
$yin K$ such that $y_{n_{k}}rightarrow y$ as $krightarrowinfty$.
Now
begin{eqnarray*}
d(x,y) & leq & d(x,y_{n_{k}})+d(y_{n_{k}},y)\
& < & frac{1}{n_{k}}+d(y_{n_{k}},y)\
& rightarrow & 0,
end{eqnarray*}
as $krightarrowinfty$. Therefore, $d(x,y)=0$ and hence $x=yin K$.
This shows that $cap_{n}E_{n}subseteq K$.






share|cite|improve this answer























  • got it thanks so much!!!
    – eyeheartmath
    Jan 3 at 22:29
















2














For each $xinmathbb{R}^{d}$ and $r>0$, let $B(x,r)={ymid d(x,y)<r}$
be the open ball centered at $x$ with radius $r$. Note that
$$
E_{n}=bigcup_{xin K}B(x,frac{1}{n})
$$
which is a bounded (because $K$ is bounded) open set. In particular
$m(E_{n})<infty$.



To show that $K=cap_{n}E_{n}$. Clearly $E_{1}supseteq E_{2}supseteqcdotssupseteq K$.
We only need to show that $cap_{n}E_{n}subseteq K$. Let $xincap_{n}E_{n}$.
For each $n$, $xin E_{n}Rightarrowexists y_{n}in K$ such that
$xin B(y_{n},frac{1}{n})$. Since $K$ is compact, for the sequence
$(y_{n})$, there exists a convergent subsequence $(y_{n_{k}})$ and
$yin K$ such that $y_{n_{k}}rightarrow y$ as $krightarrowinfty$.
Now
begin{eqnarray*}
d(x,y) & leq & d(x,y_{n_{k}})+d(y_{n_{k}},y)\
& < & frac{1}{n_{k}}+d(y_{n_{k}},y)\
& rightarrow & 0,
end{eqnarray*}
as $krightarrowinfty$. Therefore, $d(x,y)=0$ and hence $x=yin K$.
This shows that $cap_{n}E_{n}subseteq K$.






share|cite|improve this answer























  • got it thanks so much!!!
    – eyeheartmath
    Jan 3 at 22:29














2












2








2






For each $xinmathbb{R}^{d}$ and $r>0$, let $B(x,r)={ymid d(x,y)<r}$
be the open ball centered at $x$ with radius $r$. Note that
$$
E_{n}=bigcup_{xin K}B(x,frac{1}{n})
$$
which is a bounded (because $K$ is bounded) open set. In particular
$m(E_{n})<infty$.



To show that $K=cap_{n}E_{n}$. Clearly $E_{1}supseteq E_{2}supseteqcdotssupseteq K$.
We only need to show that $cap_{n}E_{n}subseteq K$. Let $xincap_{n}E_{n}$.
For each $n$, $xin E_{n}Rightarrowexists y_{n}in K$ such that
$xin B(y_{n},frac{1}{n})$. Since $K$ is compact, for the sequence
$(y_{n})$, there exists a convergent subsequence $(y_{n_{k}})$ and
$yin K$ such that $y_{n_{k}}rightarrow y$ as $krightarrowinfty$.
Now
begin{eqnarray*}
d(x,y) & leq & d(x,y_{n_{k}})+d(y_{n_{k}},y)\
& < & frac{1}{n_{k}}+d(y_{n_{k}},y)\
& rightarrow & 0,
end{eqnarray*}
as $krightarrowinfty$. Therefore, $d(x,y)=0$ and hence $x=yin K$.
This shows that $cap_{n}E_{n}subseteq K$.






share|cite|improve this answer














For each $xinmathbb{R}^{d}$ and $r>0$, let $B(x,r)={ymid d(x,y)<r}$
be the open ball centered at $x$ with radius $r$. Note that
$$
E_{n}=bigcup_{xin K}B(x,frac{1}{n})
$$
which is a bounded (because $K$ is bounded) open set. In particular
$m(E_{n})<infty$.



To show that $K=cap_{n}E_{n}$. Clearly $E_{1}supseteq E_{2}supseteqcdotssupseteq K$.
We only need to show that $cap_{n}E_{n}subseteq K$. Let $xincap_{n}E_{n}$.
For each $n$, $xin E_{n}Rightarrowexists y_{n}in K$ such that
$xin B(y_{n},frac{1}{n})$. Since $K$ is compact, for the sequence
$(y_{n})$, there exists a convergent subsequence $(y_{n_{k}})$ and
$yin K$ such that $y_{n_{k}}rightarrow y$ as $krightarrowinfty$.
Now
begin{eqnarray*}
d(x,y) & leq & d(x,y_{n_{k}})+d(y_{n_{k}},y)\
& < & frac{1}{n_{k}}+d(y_{n_{k}},y)\
& rightarrow & 0,
end{eqnarray*}
as $krightarrowinfty$. Therefore, $d(x,y)=0$ and hence $x=yin K$.
This shows that $cap_{n}E_{n}subseteq K$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 3 at 22:30

























answered Jan 3 at 22:28









Danny Pak-Keung Chan

2,19038




2,19038












  • got it thanks so much!!!
    – eyeheartmath
    Jan 3 at 22:29


















  • got it thanks so much!!!
    – eyeheartmath
    Jan 3 at 22:29
















got it thanks so much!!!
– eyeheartmath
Jan 3 at 22:29




got it thanks so much!!!
– eyeheartmath
Jan 3 at 22:29











2















I doubt this is true




It is, and your method is fine, once you've also checked that $m(E_1) < +infty$.




Do all compact subsets of $mathbb{R}^d$ have finite (Lebesgue) measure.




Yes. This follows from the Heine-Borel Theorem: all compact subsets of $mathbb{R}^d$ are bounded, so are contained in some product of closed, bounded intervals, hence have measure bounded above by the measure of that product of closed, bounded intervals, which is finite. There's extra work to do to check that all compact subsets of $mathbb{R}^d$ are measurable, but indeed they are.






share|cite|improve this answer





















  • Awesome!! thank you so much!! I am always very very hesitant on making claims unless I'm 100000% certain. I thought the equality was true but wanted to be extra certain, and yes Heine-Borel, luckily enough my point-set topology is relatively strong specially in comparison to my measure theory. again, thanks!!!!
    – eyeheartmath
    Jan 3 at 22:23


















2















I doubt this is true




It is, and your method is fine, once you've also checked that $m(E_1) < +infty$.




Do all compact subsets of $mathbb{R}^d$ have finite (Lebesgue) measure.




Yes. This follows from the Heine-Borel Theorem: all compact subsets of $mathbb{R}^d$ are bounded, so are contained in some product of closed, bounded intervals, hence have measure bounded above by the measure of that product of closed, bounded intervals, which is finite. There's extra work to do to check that all compact subsets of $mathbb{R}^d$ are measurable, but indeed they are.






share|cite|improve this answer





















  • Awesome!! thank you so much!! I am always very very hesitant on making claims unless I'm 100000% certain. I thought the equality was true but wanted to be extra certain, and yes Heine-Borel, luckily enough my point-set topology is relatively strong specially in comparison to my measure theory. again, thanks!!!!
    – eyeheartmath
    Jan 3 at 22:23
















2












2








2







I doubt this is true




It is, and your method is fine, once you've also checked that $m(E_1) < +infty$.




Do all compact subsets of $mathbb{R}^d$ have finite (Lebesgue) measure.




Yes. This follows from the Heine-Borel Theorem: all compact subsets of $mathbb{R}^d$ are bounded, so are contained in some product of closed, bounded intervals, hence have measure bounded above by the measure of that product of closed, bounded intervals, which is finite. There's extra work to do to check that all compact subsets of $mathbb{R}^d$ are measurable, but indeed they are.






share|cite|improve this answer













I doubt this is true




It is, and your method is fine, once you've also checked that $m(E_1) < +infty$.




Do all compact subsets of $mathbb{R}^d$ have finite (Lebesgue) measure.




Yes. This follows from the Heine-Borel Theorem: all compact subsets of $mathbb{R}^d$ are bounded, so are contained in some product of closed, bounded intervals, hence have measure bounded above by the measure of that product of closed, bounded intervals, which is finite. There's extra work to do to check that all compact subsets of $mathbb{R}^d$ are measurable, but indeed they are.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 3 at 22:20









user3482749

2,738414




2,738414












  • Awesome!! thank you so much!! I am always very very hesitant on making claims unless I'm 100000% certain. I thought the equality was true but wanted to be extra certain, and yes Heine-Borel, luckily enough my point-set topology is relatively strong specially in comparison to my measure theory. again, thanks!!!!
    – eyeheartmath
    Jan 3 at 22:23




















  • Awesome!! thank you so much!! I am always very very hesitant on making claims unless I'm 100000% certain. I thought the equality was true but wanted to be extra certain, and yes Heine-Borel, luckily enough my point-set topology is relatively strong specially in comparison to my measure theory. again, thanks!!!!
    – eyeheartmath
    Jan 3 at 22:23


















Awesome!! thank you so much!! I am always very very hesitant on making claims unless I'm 100000% certain. I thought the equality was true but wanted to be extra certain, and yes Heine-Borel, luckily enough my point-set topology is relatively strong specially in comparison to my measure theory. again, thanks!!!!
– eyeheartmath
Jan 3 at 22:23






Awesome!! thank you so much!! I am always very very hesitant on making claims unless I'm 100000% certain. I thought the equality was true but wanted to be extra certain, and yes Heine-Borel, luckily enough my point-set topology is relatively strong specially in comparison to my measure theory. again, thanks!!!!
– eyeheartmath
Jan 3 at 22:23




















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061067%2fmeasure-of-a-compact-subset-of-mathbbrd%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

1300-talet

Image
Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
Read more

1300-talet

Image
Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
Read more

Display a custom attribute below product name in the front-end Magento 1.9.3.8

Image
0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
Read more