I am trying to solve Exercise 8, Chapter 2 of Rudin's Real and Complex Analysis, but I am not sure of what I have done and I am not able to answer to the final part of the question. The requirement is as follows  Construct a Borel set $E \in \Bbb R^1$ such that $$0&lt;m(E\cap I) &lt; m(I)$$ for every nonempty segment $I$. Is it possibile to have $m(E)&lt; \infty$?  My try on the construction: Let $E_0 =\emptyset$, and let $r_1, r_2, \dots$ be an enumeration of the rationals. For $i\geq 1$, do the following.  If there exists a nonempty segment $I_i=(a_i,b_i)$, with $a_i &lt; r_i &lt; b_i$, such that $m(E_{i-1} \cap I_i) = 0$, then let $m_i = \min\left(b_i-r_i,r_i-a_i\right)$ and $$E_i=E_{i-1} \cup \left(r_i-\frac{m_i}2,r_i+\frac{m_i}2\right).$$ Otherwise, if there is a nonempty segment $I_i=(a_i,b_i)$ with $a_i&lt;r_i&lt;b_i$ such that $m(E_{i-1}\cap I_i) = m(I_i)$, let \begin{eqnarray*} E_i &amp;=&amp; E_{i-1} - \left[r_i-\frac{m_i}2,r_i+\frac{m_i}2\right]=\\ &amp;=&amp; E_{i-1} \cap \left[r_i-\frac{m_i}2,r_i+\frac{m_i}2\right]^c,\end{eqnarray*} with $m_i$ defined as above.  $E = E_1 \cup E_2\cup E_3\dots$ is a Borel set, since it is obtained with a countable number of unions, intersections, and complements of open sets. Since any nonempty segment $I$ contains rationals, both $m(E \cap I)=0$ and $m(E\cap I) = m(I)$ would contradict the construction.  Question 1: Is my construction correct? Question 2: I would like to show that $m(E) &lt; \infty$ cannot hold, in general. I would appreciate a hint on how to do that, if possible.
I was wondering how one can construct a Borel set that doesn't have full measure on any interval of the real line but does have positive measure everywhere.   To be precise, if $\mu$ denotes Lebesgue measure, how would one construct a Borel set $A \subset \mathbb{R}$ such that $$0 &lt; \mu(A \cap I) &lt; \mu(I)$$ for every interval $I$ in $\mathbb{R}$?   Moreover, would such a set necessarily have to contain infinite measure?
I have installed a package using snap and I need to modify one of the files but when I try to change its ownership or permissions, I always get the following message:  sudo chmod +x ./my_file.js chmod: changing permissions of '/snap/my_app/my_file.js': Read-only file system   How can I modify files that are installed via snap? How to use xcolor to color a pgfplot How do I use xcolor to color a specific cell from a table in pgfplots? Is xcolor the best tool to use here? I'm trying to loop over a list of colors and color the cells in the column individually.  \documentclass{article} \usepackage{pgfplotstable,filecontents} \RequirePackage{listofitems} \usepackage{}  \begin{filecontents*}{test.csv}     ColA, ColB, ColC     0.17, 0.91, 0.67     0.15, 0.17, 0.92     0.48, 0.1, 0.28 \end{filecontents*}     \begin{document}     \pgfplotstabletypeset[col sep=comma,     ]{test.csv}     \def\mylistOne{green, blue, orange}     \readlist*\mylistTwo{\mylistOne}     \newcounter{myCounter}     \setcounter{myCounter}{-1}     \foreachitem\x\in\mylistTwo[]{         \stepcounter{myCounter}         \pgfplotstableset{             every row \themyCounter column 1/.style={                 \cellcolor{\x}         }     } } \end{document} How to Make A button do more than one action xcode How do I increase the render distance? I have created a simple model here   When rendering it the end of it is cut off   How can I get the camera to render a longer distance? Easy is to beginner as medium is to I want to display three levels in my game...but I don't know what the text should say.  easy -&gt; Beginner  medium -&gt; ?  hard -&gt; Expert   Basically, I want the form that describes the person playing the game, instead of the game itself. What adjective is best?  I have considered adept.  I doubt my users will know what adept means. Also, I need something in between expert and beginner. Adept falls into the expert category. Is there anything more common that I can use, that has the same impact as the other two words I mentioned?  Proficient would work, theoretically, but in day-to-day term, the levels "beginner - proficient - expert" won't sound great. If $f(x)$ is differentiable at x, I need to prove that $\lim_{h\to0}\frac{f(x+h)-f(x-h)}{2h}$ exist and is finite.  so if $f(x)$ is differentiable at a $x$, the difference quotient exist for this point, and also $f(x)$ must be continuous at x as well  so that mean that:  $\lim_{h\to0^+}\frac{f(x+h)-f(x)}{h} =  \lim_{h\to0^-}\frac{f(x+h)-f(x)}{h}$  and that: $\lim_{x\to x_0^+}=\lim_{x\to x_0^-}=f(x_0)$  I know I should probably use arithmetic limit laws to prove this but I can't see how what I figured out could help me. any help with that?  Thanks! How to find the mime type of a file in python? How to find the center of an ellipse? How do I permanently allow a website to access my location? In iOS, I only need to be asked once whether I want to allow a website to share my location data. Is it possible to permanently do this in OS X version of Safari as well? What software do you use to accurately draw geometry diagrams? Let $X$ ~ $Bin(n,p)$ and $Y$ ~ $Bin(m,p)$ be two independent random variables. Find the distribution of $Z=X-Y$.   see also   I figured this out:  $$ P(Z=z)=\cases{\sum_{i=o}^{min(m,n)} Bin(k+i,n,p)*Bin(i,m,p), &amp;if $z\ge0$;\cr \sum_{i=0}^{min(m,n)} Bin(i,n,p) * Bin(i-z, m, p),&amp;otherwise. \cr}$$  I also validated it by Monte Carlo simulation. For $n=30$, $m=20$ and $p=0.5$, I get the following distribution, where the circles are the analytical probabilities and the line connects the MC estimates.    Because that looked to me pretty much like a binomial distribution, I gave it a try and figured out that its actually a binomial, just shifted by m to the left. This can be simply written as $P(Y=y) = Bin(y+m, m+n, p)$. Hence, given equal success probabilities, the sum of two independent binomially distributed random variables is binomial, but also their difference, just shifted to the left.  This question here  is actually the same as mine, but received no answer and only the comment that there would be no simple formula. But the above formula looks pretty simple to me.   Is it correct that for the case of equal success probabilities, the above equations actually describes the distribution of $Z=X-Y$? I read in a book that $Z$ could not be binomial distributed because it had negative support. Is it right to call it a shifted binomial? Setting-up SSH-tunnel for VPN? I have a laptop running Ubuntu.  I also rent a VPS running Ubuntu.  I've bought a paid VPN-service, which I use on my laptop.  I was wondering how I can set-up a SSH-tunnel or something similar, from my laptop to the VPS I'm renting, so I then can run VPN (from my laptop) "through" it (to the VPS), to the VPN-server.  The reason I wish to do so, is that a place I often work blocks my VPN-service, but not my SSHing to my VPS...  While I have been using Linux for a while, I'm a novice when it comes to setting-up virtual network-devices, advanced routing, and so on... so please spoonfeed me (using small words and colorful pictures ;-)  I have been googling and looking at various answers on the StackExchange-network, but not seems to be exactly what I'm after.  SSH-tunneling seemed promising, but since I don't have/know an exact domain and port for my VPN, I couldn't figure out how to use it...  I guess I may need to set-up some virtual network-device on my laptop (which ends on the VPS), and connect VPN through it - but how to do this, sadly goes over my head. Prove that the number of $\alpha\in\mathbb{F}_{27}$ such that $|A_\alpha|=26 $ equals 12. Let $\mathbb{F}_{27}$ denote the finite field of size 27. For each $\alpha\in\mathbb{F}_{27}$ , define $ A_{\alpha}$ $=$ { $ 1, 1+\alpha , 1+\alpha+{\alpha}^2 , 1+\alpha+{\alpha}^2+{\alpha}^3 , .... $} . Then  prove that  the number of  $\alpha\in\mathbb{F}_{27}$ such that $|A_\alpha|$=26 equals 12.  $ 0\in A_\alpha $ if and only if $\alpha\neq 0$ For the first proof , the multiplicative group of the field $\mathbb{F}_{27}$ is cyclic and its order is 26  , so there are 12 generators in that group.   Now  I need  to  show  that  $ 1+\alpha+{\alpha}^2+.. {\alpha}^{25} $ =0  , where order of $\alpha$=26 . Since   $\alpha ^{26}$ =1 ($\alpha$-1)(1+$\alpha$+${\alpha}^2$+...+${\alpha}^{25}$)=0 , since $\alpha\neq 1$  , So $ 1+\alpha+...+{\alpha}^{25} $ =0 My question is , can I write these last two  lines?  I mean are those two lines alright to write? If those two lines are invalid then I need some help to show  $ 1+\alpha+{\alpha}^2+..+{\alpha}^{25} $=0  .Thanks EDIT: As of Java 8, static methods are now allowed in interfaces.  Here's the example:  public interface IXMLizable&lt;T&gt; {   static T newInstanceFromXML(Element e);   Element toXMLElement(); }   Of course this won't work.  But why not?  One of the possible issues would be, what happens when you call:  IXMLizable.newInstanceFromXML(e);   In this case, I think it should just call an empty method (i.e. {}).  All subclasses would be forced to implement the static method, so they'd all be fine when calling the static method.  So why isn't this possible?  EDIT: I guess I'm looking for answer that's deeper than "because that's the way Java is".   Is there a particular technological reason why static methods can't be overwritten? That is, why did the designers of Java decide to make instance methods overrideable but not static methods?  EDIT: The problem with my design is I'm trying to use interfaces to enforce a coding convention.  That is, the goal of the interface is twofold:   I want the IXMLizable interface to allow me to convert classes that implement it to XML elements (using polymorphism, works fine). If someone wants to make a new instance of a class that implements the IXMLizable interface, they will always know that there will be a newInstanceFromXML(Element e) static constructor.   Is there any other way to ensure this, other than just putting a comment in the interface? I am running out of storage, I usually have bought a lot of my games on disk but if I buy a game on the store then delete it, and want to redownload it do I have to buy it again Arrangements of MISSISSIPPI with all S's and P's separated If the all S,P are separated in the word MISSISSIPPI then the total possible arrangements are $k\frac {10!}{4!.4!} $ then value of k is? Now all S,P are separated so we can have slash method. Thus they can be placed alternatively in $6! $ ways and rest 5 can be placed in $5! $ ways. Thus we equate to get $$6!.5!=k\frac {10!}{4!.4!} $$  to get $k=13.7$ while $k=4/3$ .Where is my mistake Thanks Compile error: package javax.servlet does not exist about://(null) popup? I encountered a popup similar to this one this morning while checking the weather report at  in Safari, though I have not been able to reproduce it.  The title is "about://(null)" and the text "Why you no play Candy Crush!!"  The popup occurred twice.  I tapped Cancel both times.  I suspect the source of this is rogue JavaScript in an ad network, judging from the fact that the screenshot below shows it in an ad-supported application, and when I saw it, I was on an ad-supported website.  My phone is not jailbroken.  I'm wondering if anyone is able to identify the source of the popup definitively.  Speculation is running rampant on Apple's support communities and elsewhere, from push notifications (delete Facebook and all your games!) to carrier notifications (which don't work on iPod Touches…), and my gut tells me little of it is helpful.  But I could also be very wrong.  (This is not my screenshot; it comes from .)    UPDATE 1: The Candy Crush people have this to say, but I'm not (yet) convinced there's actually something installed that is to blame:   UPDATE 2: Just reproduced it in the iOS Simulator while clicking around woodtv.com, but I'm not sure how to track it down further from here. Where To start Learning With Salesforce Development Where should I start learning? There are various topics to learn like Objects, Classes, Apex, Visualforce, SOSL, SOQL, Triggers, etc. There are so many options out there to start learning. I couldn't decide what topic to learn first, and then step by step get proficiency in all aspects of Saleforce. How to reinstall Windows 10 on new hard disk