Epsilon-Delta Proof of a Lebesgue Integral Let f be a non-negative Lebesgue measurable function with $\int_\mu\ f \,d\lambda&lt; \infty$. Show that for each $\epsilon&gt;0$, there exists a $\delta&gt;0$ such that $\lambda(E)&lt;\delta$ implies $\int_E\ f \,d\lambda&lt; \epsilon$.  --I think I need to use the simple function $\phi=f\chi_E$ such that $\phi&lt;f$ but I do not see how to tie that into the standard epsilon-delta proof with a finite l Lebesgue integral.  --I also know that $\lambda(A_c)\leqslant(1/c)\int_E\ f \,d\lambda$ from the previous problem so I am assuming I will need to use that.
Prove that to each $\epsilon >0$, there exists a $\delta >0$ so that the Lebesgue integral... Suppose $f$ is in $L^1$ space of $\mu$, where $\mu$ is the Lebesgue measure. Prove that to each $\epsilon &gt;0$, there exists a $\delta &gt;0$ so that the Lebesgue integral of the absolute value of $f$ is less than $\epsilon$ (over the set $E$) whenever the measure of $E$ is less than $\delta$.  I know that the integral of $f$ is finite. So for arbitrary $f$ we can find the integral of $f$ to be less than some $\epsilon$, but how would I connect that with the measure of $E$ less than $\delta$. Any help would be welcome.
apt-get build-dep is unable to find a source package What is the difference between "owing to" and "due to"? The  was observed, giving us the frequency and amplitude of the event. Because LIGO has two detectors a  could be derived.  But how do these three factors allow for the mass of the black holes and their distance to be calculcated? If the wave strengths are in a square relationship to distance, then couldn't there be an infinite number of other masses and distances which would give the same signals? Chain length. how long or how many links I measured from the middle of each sprocket and it's 17 inches. Not sure what length of chain I should get Assume that $X$ and $Y$ are independent random variables with $X \sim \exp(\lambda)$ and $Y \sim \exp(\mu)$. It is impossible to obtain direct observations of $X$ and $Y$. Instead, we observe the random variables $Z$ and $W$, where $Z = \min\{X,Y\}$ and $W =1$ if $Z=X$ and $W=0$ if $Z=Y$.  How to find the joint distribution of $Z$ and $W$? OpenLayers 3: Cross-Origin Request Blocked: The Same Origin Policy disallows Showing a set is closed under inverses Let $G$ be a finite group, say a group with $n$ elements, and let $\emptyset \neq S \subseteq G$. Suppose $e \in S$, and that $S$ is closed with respect to multiplication. Prove that $S$ is a subgroup of $G$.    Let $G = \{a_1, \ldots, a_n\}.$ Let $a_i \in G.$ Then any number of $a_ia_1, a_ia_2, \ldots, a_ia_n \in S,$ since $S$ is closed under multiplication. One of those $a_ia_j$ is $e$ since $e \in S.$ Then either $a_i$ or $a_j$ is $a^{-1}.$ Would this argument work? Expectation inequality for a set and its subset. Given that a function $g(x)$ is a monotone increasing function. Its domain is an interval [c,d]. There are two sets $a_j \in A, j= 1,...,J$ and $b_k \in B,k=1,...,K$  on this interval, which satisfy $c_z \in C = B$\A and $c_z \geq a_j$ for all $a_j \in A$. For example, the interval $[c,d]=[0,8]$,$A=\{0,1,2,3,4,5\}$ and $B=\{0,1,2,3,4,5,6,7\}$. In this case, does the following inequality hold? If yes, could you give me some hints to prove that?  $$\frac{1}{K} \sum_{k=1}^{K}g(b_k) \geq \frac{1}{J} \sum_{j=1}^{J}g(a_j)$$ My program cannot run with "command not found" error I just switched from Windows to Ubuntu 11.10.  I wrote the following code in C with the text editor and saved it as HelloWorld.c in Documents.  #include &lt;stdio.h&gt; int main() {     printf("Hello World!\n");     return 0; }   And I started the Terminal and enter the following commands:  cd Documents  gcc HelloWorld.c   A file called a.out, which, after some search on Google, is the executable. I entered this command:  a.out   But I get   a.out: command not found   Which step did I do wrongly? Proof that $V = \sum_k U_k$ and $\dim V=\sum_k \dim U_k$ implies $V=\oplus_k U_k$ The limit of reciprocal is the reciprocal of the limit While I was reading the book Exploratory examples for real analysis, I came across this:     to show that : $$\lim_{x\rightarrow x_o}\frac{1}{h(x)} = \frac{1}{\lim_{x\rightarrow x_o}{h(x)}} = \frac{1}{M}$$      Discussion:Let $\epsilon \gt 0$.we want to find $\delta \gt 0$ such that       if  $0 \lt |x-x_0| \lt \delta $,then $\Big|\frac{1}{h(x)}-\frac{1}{M}\Big|\lt \epsilon$.      In order to get expression involving $|h(x)-M|$,we first find the common denominator:        $\Big|\frac{1}{h(x)}-\frac{1}{M}\Big|=\Big|\frac{M-h(x)}{M~h(x)}\Big|$      $~=\Big|\frac{h(x)-M}{|M|~|h(x)|}\Big|$.      As we have to deal with variable term $|h(x)|$ in denominator.We can construct bound .Let $\epsilon =\frac{|M|}{2}$ ,since lim$_{x\to x_o}h(x)=M$,there exists $\delta_1\gt 0~~$ s.t.       if $0 \lt |x-x_0| \lt \delta_1 $,then $|h(x)-{M}|\lt \frac{|M|}{2}$.      As you study steps below,identify that step where $h(x)\neq 0$ for all $0 \lt |x-x_0| \lt \delta_1 $ plays a critical role:      $|M|=|M-h(x)+h(x)|$      $~~~~~~\leq |M-h(x)|+|h(x)|$      $~~~~~~= |h(x)-M|+|h(x)|$      $~~~~~~\lt \frac{M}{2}+|h(x)| \implies$      $\frac{|M|}{2} \lt |h(x)| \implies$      $\frac{1}{h(x)}\lt \frac{2}{|M|}$,where $0 \lt |x-x_0| \lt \delta_1 $.      Now that we've bound for $1/h(x)$,we can apply the definition to term $|h(x)-M|$.      Since     lim$_{x\rightarrow x_o}h(x)=M$,we can find $\delta_2 \gt 0$ such that,       $~~~~~~~~~~~~~~~~~~~~~~~~~~~~$if  $0 \lt |x-x_0| \lt \delta_2$, then $|h(x)-M|\lt \frac{|M|^2 \epsilon }{2}$   The thing I can't understand is for what purpose did the author work with such complicated expression involving $\epsilon $ in the end of the discussion? Subtracting 2 lists in Python Right now I have vector3 values represented as lists. is there a way to subtract 2 of these like vector3 values, like  [2,2,2] - [1,1,1] = [1,1,1]   Should I use tuples?  If none of them defines these operands on these types, can I define it instead?  If not, should I create a new vector3 class? So, I thought I had a good understanding of this, but just ran a test (in response to a conversation where I disagreed with someone) and found that my understanding is flawed...  In as much detail as possible what exactly happens when I execute a file in my shell? What I mean is, if I type: ./somefile some arguments into my shell and press return (and somefile exists in the cwd, and I have read+execute permissions on somefile) then what happens under the hood?  I thought the answer was:   The shell make a syscall to exec, passing the path to somefile The kernel examines somefile and looks at  of the file to determine if it is a format the processor can handle If the magic number indicates that the file is in a format the processor can execute, then   a new process is created (with an entry in the process table) somefile is read/mapped to memory. A stack is created and execution jumps to the entry point of the code of somefile, with ARGV initialized to an array of the parameters (a char**, ["some","arguments"])  If the magic number is a  then exec() spawns a new process as above, but the executable used is the interpreter referenced by the shebang (e.g. /bin/bash or /bin/perl) and somefile is passed to STDIN If the file doesn't have a valid magic number, then an error like "invalid file (bad magic number): Exec format error" occurs   However someone told me that if the file is plain text, then the shell tries to execute the commands (as if I had typed bash somefile). I didn't believe this, but I just tried it, and it was correct. So I clearly have some misconceptions about what actually happens here, and would like to understand the mechanics.  What exactly happens when I execute a file in my shell? (in as much detail is reasonable...) Prove inequality using Schwarz's Lemma Would time freeze if you could travel at the speed of light? Tufte advocates to "minimize the ink-to-data ratio" for charts. I am trying to style a multi data series bar chart, that can have a grid and axes as shown below (chart at the left is from the Economist and on the right my attempt).      Here is the code:  \documentclass[justified]{tufte-book} \usepackage{pgfplots,lipsum} \pgfplotsset{compat=newest} \begin{document} \section{Bar Charts}  \lipsum[1-2]\vfill   \begin{figure*}[htbp]   \fbox{  \begin{tikzpicture}   \centering   \begin{axis}[         ybar,           title={Cumulative Progress of Works},         height=8cm,         width=15.5cm,         bar width=0.4cm,         ymajorgrids, tick align=inside,         enlarge  y limits={value=.2,auto}, % see the upper          ymin=0,         ymax=100,         axis x line=bottom,          axis y line=right,         %enlarge x limits =e,         legend style={at={(0.5,-0.2)},         anchor=north,legend columns=-1},         ylabel={Percentage (\%)},         symbolic x coords={            Sep-11,Oct-11,Nov-11,Dec-11,            Jan-12,Feb-12,            Mar-12,           Apr-12},        xtick=data,        nodes near coords,        every node near coord/.append style={         anchor=mid west,         rotate=70       }     ]     \addplot coordinates {       (Sep-11,75.4064)       (Oct-11, 72.7961)        (Nov-11,94.4597)       (Dec-11,66.6786)        (Jan-12,67.5600)        (Feb-12,88.2339)       (Mar-12,78.6138)        (Apr-12,58.9129) };    \addplot coordinates {       (Sep-11,75.4064)       (Oct-11, 89.7961)        (Nov-11,94.4597)       (Dec-11,76.6786)        (Jan-12,77.5600)        (Feb-12,78.2339)       (Mar-12,88.6138)        (Apr-12,78.9129) };    \addplot coordinates {       (Sep-11,75.4064)       (Oct-11, 89.7961)        (Nov-11,94.4597)       (Dec-11,76.6786)        (Jan-12,77.5600)        (Feb-12,78.2339)       (Mar-12,88.6138)        (Apr-12,78.9129) };      \legend{First Fix,Second Fix,Third Fix}   \end{axis}   \end{tikzpicture} }  \caption{\protect\raggedright Cumulative progress for all MEP works. Notice the slower rate of production during the last three months.} \end{figure*}  \lipsum[1-3] \end{document}   I have moved the y-axis to the right, but cannot get rid of the axis line. The plot is also not fitting within the plot width, despite using \enlarge.   Edit  And thanks to Jake the ugly duckling has been transformed to the following, with minor variations to the code as suggested by Jake. I have only added a stronger color for one of the bars in order to highlight it: Finite $G$ with involutory automorphism $\alpha$ with no nontrivial fixed points. Proving properties of $\alpha$. The question at hand is:     Let G be a finite group and $\alpha$ an involutory automorphism of G, which doesn't fixate any element aside from the trivial one.    1) Prove that $ g \mapsto g^{-1}g^{\alpha} $ is an injection    2) Prove that $\alpha$ maps every element to its inverse    3) Prove that G is abelian    I think I've found 1).  I assume $ g_1^{-1}g_1^{\alpha} = g_2^{-1}g_2^{\alpha} $ and from this I get $ (g_2g_1^{-1})^\alpha = g_2g_1^{-1} $, so $g_2 = g_1$ (is this correct?)   For 2) I'm sort of stumped though, not sure how to start proving that, so any help please? I would like to filter the markers on my (geojson) layer according to two features: year and type of event, ideally using both filters (year and eventType) as check-boxes without using layerGroup at all. I would like to be able to click on a checkbox and update dynamically the markers by filtering them according to feature.properties.eventType too, in addition to the year displayed, something like    &lt;input type="checkbox" id="all" name="all" value="all"&gt;   &lt;label for="all"&gt;All&lt;/label&gt;   &lt;input type="checkbox" id="funfair" name="funfair" value="newsletter"&gt;   &lt;label for="funfair"&gt;Funfair&lt;/label&gt;   &lt;input type="checkbox" id="trade" name="trade" value="trade"&gt;   &lt;label for="trade"&gt;Trade&lt;/label&gt;   &lt;input type="checkbox" id="exhibition" name="exhibition" value="exhibition"&gt;   &lt;label for="exhibition"&gt;Paid exhibition&lt;/label&gt;   etc... which would trigger a  filter: function(feature, layer) {   return feature.properties.eventType == value-of-checkbox; }   I have five types of eventType. So, for instance, if 1890 is checked, the map is showing all 1890 events, no matter what kind of event it is (because by default it would show all events). If the user checks the 'funfair' and the 'trade' checkbox, though, I want the map to show only the 1890 events of the funfair and trade kind, and not other kind of events.  This is my complete script, which so far is working fine with filtering by year.  var oldmap = L.tileLayer('address', {   maxZoom: 12 }),  var modernmap = L.tileLayer('https://api.tiles.mapbox.com/v4/{id}/{z}/{x}/{y}.png?access_token={accessToken}', {   maxZoom: 12,   id: 'mapbox.streets',   accessToken: 'mytoken' });  var map = L.map('map', {   center: [40.1857864,-4.5536861],   zoom: 7,   layers: [oldmap, modernmap] });  var baseMaps = {   "Old map": oldmap,   "Modern map": modernmap };  var jsontest = {"type": "FeatureCollection",   "features": [     {       "type": "Feature",       "geometry": {         "type": "Point",         "coordinates":  [ -16.5471268,28.4136726 ]       },       "properties": {         "Name":"Point1",         "year":'1892',         "eventType":"funfair",         "notes":""       }     },     {       "type": "Feature",       "geometry": {         "type": "Point",         "coordinates":  [ 2.1753895,41.3767925 ]       },       "properties": {         "Name":"Point2",         "year":'1890',         "eventType":"trade demonstration",         "notes":""       }     }   ]};  var mapLayerGroupsYear = [];  function onEachFeature(feature, layer) {   var year = mapLayerGroupsYear[feature.properties.year];   if (year === undefined) {     year = new L.layerGroup();      //add the layer to the map     year.addTo(map);      //store layer     mapLayerGroupsYear[feature.properties.year] = year;   }    //add the feature to the layer   year.addLayer(layer);    }   var myLayer = L.geoJSON(jsontest, {   onEachFeature: onEachFeature })  L.control.layers(baseMaps, mapLayerGroupsYear).addTo(map);   I am a beginner with leaflet so please forgive the dumb question, in case it is. How to add number of days to today's date? How to convert a HEIF/HEIC image to JPEG in El Capitan? So, here’s the thing. I have a new iPhone and an old Mac. iOS 11 (still in beta as of today) and El Capitan (no more updates for this Mac). I use iCloud photo library in both devices. When I take a photo in the new format with my iPhone 7, is there any way to use it right away in my mac, without having to convert it in my iPhone first? I mean, can I get those photos synced through iCloud and converted locally on my Mac?  I have tried to find a third party software that could do that kind of conversion, but was not able to find any that would work on El Capitan. Any idea?