This is Problem 3.1.5 in Cohn's Measure Theory, 2nd edition.  Let $\mu$ be a measure on $(X, \mathcal A)$, and let $f, f_1,f_2, \ldots$ and $g,g_1,g_2,\ldots$ be real-valued $\mathcal A$-measureable functions on $X$. (a) Show that if $\mu$ is finite, if $f_n \to f$ in measure, and $g_n \to g$ in measure, then $f_n g_n \to fg$ in measure.  With some effort, I was able to work out a direct proof, similar to the one outlined in . After re-reading the section, I came up with an alternative argument similar to the one used by Cohn to prove another theorem. It results in a much quicker proof, and almost seems like a magic trick. I want to make sure that my argument is correct. I use the following results, proved by Cohn in this section, for real-valued measurable functions:  Proposition 3.1.3: If $f_n \to f$ in measure, then there is a subsequence such that $f_{n_k} \to f$ almost everywhere. Proposition 3.1.2: If $\mu$ is finite and $f_n \to f$ almost everywhere, then $f_n \to f$ in measure.  Here is my proof: Let $\{n_k\}$ be any subsequence of $\mathbb N$. Then $f_{n_k} \to f$ and $g_{n_k} \to g$ in measure, so by Proposition 3.1.3 there is a subsequence $\{n_{k_j}\}\subset \{n_k\}$ such that $f_{n_{k_j}} \to f$ and $g_{n_{k_j}} \to g$ almost everywhere. (To see this, take a subsequence of $\{n_k\}$ that works for $f$. Then that subsequence has a subsequence which works for both $f$ and $g$.) Therefore $f_{n_{k_j}} g_{n_{k_j}} \to fg$ almost everywhere. As $\mu$ is finite, Proposition 3.1.2 implies that $f_{n_{k_j}} g_{n_{k_j}} \to fg$ in measure. Now here is the magic part: Suppose that $f_n g_n$ does not converge to $fg$ in measure. Then there is some $\epsilon &gt; 0$ such that $\mu\{|f_n g_n - fg| &gt; \epsilon\}$ does not converge to zero. Thus there is some $\delta &gt; 0$ such that $\mu\{|f_n g_n - fg| &gt; \epsilon\} &gt; \delta$ for infinitely many $n$. In other words, there is a subsequence $\{n_k\} \subset \mathbb N$ such that $\mu\{|f_{n_k} g_{n_k} - fg| &gt; \epsilon\} &gt; \delta$ for all $n_k$. Clearly this subsequence cannot have any further subsequence $\{n_{k_j}\}$ for which $f_{n_{k_j}} g_{n_{k_j}} \to fg$ in measure, but this contradicts what we showed above. Consequently, $f_n g_n$ must converge to $fg$ in measure.  Is this argument legitimate? It almost seems too easy.
I'm trying to prove the following statements in Folland's book.  Let $(X,\mathcal{M},\mu)$ be a measure space. If $f_n\to f$ in measure and $g_n\to g$ in measure, then $f_n+g_n\to f+g$ in measure and $f_ng_n\to fg$ in measure (in this last case, assume $\mu(X)&lt;\infty$).  The first one is already done, but the second one is giving me trouble.   My attempt: We can take subsequences $f_{n_k}$ and $g_{n_k}$ such that $f_{n_k}\to f$ a.e. and $g_{n_k}\to g$ a.e. In this case, we know that $f_{n_k}g_{n_k}\to fg$ a.e. Note that $\mu(X)&lt;\infty$, so we can use Egoroff's theorem to conclude that $f_{n_k}g_{n_k}\to fg$ almost uniformly. Finally, almost uniform convergence implies convergence in measure, therefore $f_{n_k}g_{n_k}\to fg$ in measure.   The problem is that I just have convergence in measure for some subsequence of $f_ng_n$ and I need convergence in measure for the whole sequence. Also I don't know if using Egoroff's theorem is unnecessary, if this is the case, I would appreciate to see some more elementary prove.  Detail: We say that $f_n\to f$ almost uniformly if for every $\varepsilon &gt; 0$ there is some $E\subset X$ measurable, such that $\mu(E) &lt;\varepsilon$ and $f_n\to f$ uniformly on $E^c$.  Thank you!
Option + click in iTerm In Terminal, Option + click allows me to place my cursor at any point on the current line-- great for editing really long commands.  How can I do this in iTerm2?  Update: According to , the feature has been around since 0.7.0 (2003). I've also noticed that if I hold Command+Option, the cursor changes into the familiar cross like in Terminal...  but clicking still does not reposition my cursor. How do we know the universe is expanding, and not that its contents are shrinking? Is there a physical reason why not to think that instead of space expanding, all physical constants and parameters are shrinking (including of course the instruments we use to measure the constants) and space is static, or is it a case of Occam's razor? I am trying to under stand how to build a barrel shifter using multiplexers. I understand how the barrel shifter works but I don't get how you decide the number of multiplexers to use, and how the shifting is done with the multiplexer basically.  So far wikipedia doen't help much. And the best I have so far is   But it does not really answer my question. It's the first time I'm using Tikz on Latex. I have to draw this block diagram (this is a low resolution snip from a paper)   In Tikz I have done this:   That is very similar but I need some adjustments, like the u_i arrow and the node sum after the H(s) block.  Is there anybody that can help me with this?  my code:      \tikzstyle{block} = [draw, fill=white, rectangle,  minimum height=2em, minimum width=3em] \tikzstyle{sum} = [draw, fill=white, circle, radius=1mm, node distance=1.5cm] \tikzstyle{input} = [coordinate] \tikzstyle{output} = [coordinate] \tikzstyle{pinstyle} = [pin edge={to-,thin,black}]  \begin{tikzpicture}[auto, node distance=2cm,&gt;=latex'] \node [input, name=input] {}; \node [input, above of=input](input1) {}; \node [sum, right of=input] (sum) {}; \node [block, right of=sum] (controller) {$K(s)$}; \node [sum, right of=controller] (sum1) {}; \node [block, right of=sum1] (filterinv) {$H^{-1}(s)$}; \node [block, right of=filterinv] (system) {$G(s)$}; \node [output, right of=system] (output) {}; \node [output, above of=output] (output1) {}; \node [block, below of=filterinv] (filter) {$H(s)$}; \node [block, above of=controller] (delay) {$D(s)$}; \node [sum, below of=sum1] (sum2) {};  \draw [-&gt;] (filterinv) -- node[name=u] {$u_{i}$} (system); \draw [draw,-&gt;] (input) -- node {$s_{i-1}$} (sum); \draw [-&gt;] (sum) -- node {$e_{i}$} (controller); \draw [-&gt;] (system) -- node [name=q] {$q_{i}$}(output); \draw [-&gt;] (q) |- (filter); \draw [-&gt;] (filter) -- node {} (sum2); \draw [-&gt;] (controller) -- node {} (sum1); \draw [-&gt;] (sum1) -- node[name=xi] {$\xi_{i}$} (filterinv); \node [input, below of=sum1](input2) {}; \draw [draw,-&gt;] (input2) -- node {$r_{i}$} (sum2); \draw [-&gt;] (sum2) -| node[pos=0.99] {$-$}      node [near end] {$\tilde{s}_i$} (sum); \draw [-&gt;] (delay) -| node[] {}      node [near end] {} (sum1); \draw [draw,-&gt;] (input1) -- node[pos=0.15] {$u_{i-1}$} (delay); \draw [-&gt;] (u) |- node[pos=0.99] {$u_i$}      node [near end] {} (output1);  \end{tikzpicture} Proving that the Intersection of any Collection of Compact Sets is Compact I was trying to do this problem this way:  Let $\mathcal{B}=\{B_i\}$ be a collection of compact sets. By Heine-Borel, each of the $B_i$'s are closed and bounded. We already know that the intersection of a collection of closed sets is once again closed, so $\cap \mathcal{B}$ is closed.  If I can prove $\cap\mathcal{B}$ is bounded as well, I can use Heine-Borel to say it is compact. So I was not sure how to show it was bounded, but a friend showed me his way of doing it. He did it the following way:  Let $b,a\in\mathbb{R}$ be an upper and lower bound for $B_1$, respectively. Since $\cap\mathcal{B}\subset B_1, a$ is a lower bound for $\cap \mathcal{B}$ and $b$ is an upper bound for $\cap\mathcal{B}$. Therefore, $\cap\mathcal{B}$ is bounded and then by Heine-Borel, it is compact.  My question is on the paragraph right above. I don't understand why what he did means that $a$ and $b$ are lower and upper bounds for $\cap\mathcal{B}.$ Can anyone explain why this is? Thanks. Minecraft crashes on launch with EXCEPTION_ACCESS_VIOLATION, Problematic frame: ig4dev32.dll or ig4dev64.dll or ig4icd32.dll or ig4icd64.dll I have recently started reading and watching Naruto. I was wondering in what aspect does the clones made by different techniques differ? They all have the same potential and capabilities. Does $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ imply that $a_n$ convergent? Does the condition $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ imply that $a_n$ convergent?  When I say convergent I mean to $L\in \mathbb{R}$ or to $\infty$ or to $-\infty$  I have a gut feeling this statement is false, just can't find a series which contradicts it.  Thanks a lot How to I stop a launchd script from running on wake I have a simple launchd script that plays a sound every 30 minutes and it runs out of the user library ~/Library . It doesn't run if my machine is asleep, but does run the last event missed when the machine wakes. I think this is the expected behavior. My question is, how do I stop this happening. If I missed the half-hour event, I don't want it to be run at 20 minutes past the hour.  Here's the core of my script. It calls a Python script. I tried to put a block in the python script to check that I am within a few minutes of each 30 minutes, but that doesn't seem to work on wake. (Is it missing the system clock somehow?)  &lt;key&gt;ProgramArguments&lt;/key&gt; &lt;array&gt;     &lt;string&gt;/usr/local/bin/python3&lt;/string&gt;     &lt;string&gt;/Users/pheon/Documents/playsound.py&lt;/string&gt; &lt;/array&gt;  &lt;key&gt;StartCalendarInterval&lt;/key&gt; &lt;array&gt;     &lt;dict&gt;         &lt;key&gt;Minute&lt;/key&gt;         &lt;integer&gt;0&lt;/integer&gt;     &lt;dict&gt;         &lt;key&gt;Minute&lt;/key&gt;         &lt;integer&gt;30&lt;/integer&gt;     &lt;/dict&gt; &lt;/array&gt;   Here is a snippet from the python code which checks the time before playing the sound.  time0 = datetime.datetime.now() if (time0.minute % 30) &lt; 2:     subprocess.run(["/Users/pheon/bin/afplay-vol.sh", "1", bell],check=True) As we know, any proposition can translate to Godel numbers, such as  $$( \exists x)\space (x=sy)$$Its Godel numbers is $2^8 × 3 ^4 × 5^ {13} × 7^ 9 × 11^8 × 13^{13} × 17^ 5 × 19^ 7 × 23^ {17} × 29^ 9=a \space big \space number $  We also could get the character sequence "$( \exists x)\space (x=sy)$" through prime factorization of this big number.  Such as$$243,000,000=2^6×3^5×5^6 $$ $$6\space\space\space\space\space 5\space\space\space\space\space 6 $$ $$\downarrow\space\space\downarrow\space\space\downarrow$$ $$0\space\space\space=\space\space\space 0 $$ So $243000000$ is the Godel numbers of "$0=0$"   There are four axioms in Godel's PM system, here are two of them: $$(p\lor p)\subset p$$ $$p\subset (p\lor q)$$ through these two axioms, we can prove "$(p\lor p)\subset(p\lor q)$" is a true theorem.   And these two axioms and a theorem all have a corresponding Godel numbers, can we prove this theorem is true through Godel numbers? If it works, how? I am on Ubuntu 12.04 64-bits  I accidentally deleted a NTFS 255 GB partition using Gparted, and yes it contains LOTS of important data.  Upon realizing the deletion I haven't yet created any file-system on it. So it still shows Unallocated.    I tried Testdisk , but it shows entire partitions created from the beginning.  This sdb is my secondary drive with Xubuntu and Kubuntu on it. And that 255 GB NTFS  partition was cross platform storage drive.  I am currently in sda with Ubuntu and Win7 in dual boot, which enables me to use it as Safe recovery option rather then Live USB drive.   First ,  should i allocate it to ? file-System for better recovery .  Second , i really need to recover data from this drive so please give me possible  suggestions and proven methods of data recovery.  Third ,does Testdisk requires deleted partition to be allocated for its recovery .  I am currently trying the answer seen here  with Testdisk scanning for partition second time, hoping to recover it.   It looks a duplicate but i can't confirm if the OP got it right. Since my first doubt is to allocate it or not.  If any clarification required please do comment ,i will add it.  Edit: TestDisk deeper search results.      The following partitions can't be recovered:          Partition               Start        End    Size in sectors     &gt;  HPFS - NTFS          22241 107 11 55592   3 27  535777280        FAT32 LBA            274460 158 19 460576 228 16 2989957948  NTFS, 274 GB / 255 GiB &amp; for FAT32 LBA 1530 GB / 1425 GiB How best to sum up lots of floating point numbers? %AX = (%AH + %AL)  So why not %EAX = (%SOME_REGISTER + %AX) for some register %SOME_REGISTER? Let $X$ and $Y$ be independent $N(0; 1)$ random variables. Let $Z = \sqrt{X^2+Y^2}$.  (a) Derive the marginal pdf of $Z$ and then using the marginal pdf to compute ${\rm E}[Z^2]$  (b) Can you propose a different way other than that in (a) to compute ${\rm E}[Z^2]$  (c) Compute ${\rm E}[Z]$.  This is the whole question that I was asked. I can do (c) but I don't know how to find the marginal pdf of $Z$ with the given information and can't seem to find any formulas. Any help would be appreciated. Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $ Can I use I2C over 2m cables? Keeping tables/figures close to where they are mentioned This is my last homework problem and I've been looking at it for a while. I cannot nail down what is wrong with this proof even though its obvious it is wrong based on its conclusion. Here it is:     Find the flaw in the following bogus proof by strong induction that   for all $n \in \Bbb N$, $7n = 0$.      Let $P(n)$ denote the statement that $7n = 0$.      Base case: Show $P(0)$ holds.      Since $7 \cdot 0 = 0$, $P(0)$ holds.      Inductive step: Assume $7·j = 0$ for all natural numbers $j$ where $0 \le j \le k$ (induction hypothesis). Show $P(k + 1)$: $7(k + 1) = 0$.      Write $k + 1 = i + j$, where $i$ and $j$ are natural numbers less than $k + 1$.  Then, using the induction hypothesis, we get $7(k + 1) = 7(i + j) = 7i + 7j = 0 + 0 = 0$. So $P(k + 1)$ holds.      Therefore by strong induction, $P(n)$ holds for all $n \in \Bbb N$.   So the base case is true and I would be surprised if that's where the issue is.  The inductive step is likely where the flaw is. I don't see anything wrong with the strong induction declaration and hypothesis though and the math adds up! I feel like its so obvious that I'm just jumping over it in my head. Switching a DIV background image with jQuery Integration of secant $$\begin{align} \int \sec x \, dx  &amp;= \int \cos x \left(  \frac{1}{\cos^2x} \right)  \, dx \\ &amp;= \int \cos x \left(  \frac{1}{1-\sin^2x} \right) \, dx \\ &amp; = \int\cos x\cdot\frac{1}{1-\frac{1-\cos2x}{2}} \, dx \\ &amp;= \int \cos x \cdot\frac{2}{1+\cos2x} \, dx  \end{align}$$  I am stuck in here. Any help to integrate secant?