Hello I came across this problem and tried to prove it, I would just like to confirm that my proof is correct please as I think there is something wrong with it.   The problem says the following: Suppose $(f_n)$ is a sequence of integrable functions on a set A of finite measure. Show that if the sequence is uniformly convergent on A, then $\lim_{n\to \infty} \int_{A} f_n dm= \int_{A} \lim_{n\to \infty} f_n dm$.  My attenpt: Let $\epsilon&gt;0$ be fixed but arbitrary. Then by uniform convergence,there exists an N s.t. $|f_n-f|&lt;\epsilon$ for all n>N. Then $0\leq \int_A |f_n-f| \leq \epsilon m(A)$ for all n>N.  Now since $\epsilon&gt;0$ arbitrary and $m(A)&lt;\infty$ then $\int_A |f_n-f| \to 0$ which implies $0&lt;|\int_{A} (f_n-f)| \leq \int_{A} |f_n-f|$ and now taking the limit gives the desired result.  Could someone verify my proof? I am not sure when steps like since epsilon is arbitrary work.Also where did I use integrability of the $f_n$ ? Thanks.
Let $f_n, f \colon [a,b] \to \mathbb{R}.$ Show that, if $f_n \to f$ uniformly, then the Lebesgue integrals are equal, i.e. $\int f = \lim \int f_n$.  This is clearly true for continuous functions, but how do I handle the case of non-continuous functions?
Here $p$ and $q$ are co-prime integers. Is it necessarily an algebraic number? Visual Studio C# statement collapsing When editing really long code blocks (which should definitely be refactored anyway, but that's beyond the scope of this question), I often long for the ability to collapse statement blocks like one can collapse function blocks.  That is to say, it would be great if the minus icon appeared on the code outline for everything enclosed in braces.  It seems to appear for functions, classes, regions, namespaces, usings, but not for conditional or iterative blocks.  It would be fantastic if I could collapse things like ifs, switches, foreaches, that kind of thing!  Googling into that a bit, I discovered that apparently C++ outlining in VS allows this but C# outlining in VS does not.  I don't really get why.  Even notepad++ will so these collapses if I select the C# formatting, so I don't get why Visual Studio doesn't.  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Crash report: *** *** *** *** *** *** *** *** *** *** *** *** *** *** *** *** Build fingerprint: 'Xiaomi/polaris/polaris:10/QKQ1.190828.002/20.9.4:user/release-keys' Revision: '0' ABI: 'arm64' Timestamp: ... pid: ..., tid: ..., name: com.dashlane &gt;&gt;&gt; com.dashlane &lt;&lt;&lt; uid: ... signal 11 (SIGSEGV), code 1 (SEGV_MAPERR), fault addr 0x0 Cause: null pointer dereference          ...  backtrace:     #00 pc 00000000038621f0 /data/app/ com.google.android.trichromelibrary 438909034-YaxP9ttrIDcQBfleP5vm7Q==/base.apk! libmonochrome_64.so (offset 0x646000) (BuildId: ... )  (click  for the screenshot of the bug report) Semigroup analogue to the classification of finite simple groups? Is there a semigroup analogue to the classification of finite simple groups?  If so what are some of the major results?  --Edit, reference links--- Spoiler warning!  During the first escape attempt in Portal 2 (when GLaDOS is still in charge) she opens a door for you which you can choose to go through instead of following Wheatley (the blue-eyed ball). What happens if you go there? How do I evaluate $\int \frac{\mathrm{d}x}{e^x + 1} $? How do I solve evaluate $$\int \frac{\mathrm{d}x}{e^x + 1}\ ?$$  I know that I have to use $u$ substitution but I can't seem to find something to substitute with. Why is my variable unaltered after I modify it inside of a function? - Asynchronous code reference How to prove the function is a constant Let $f\in C[0,1]$ be such  that $\forall \phi\in C[0,1]$ with $\int_0^1\phi \, dx=0$,we have $\int_0^1f\phi \, dx=0$.  Then $f$ is constant.  May be someone can help me proof this problem. Thanks! $$T(n) = \sqrt{n} T \left(\sqrt n \right) + n$$  Master method does not apply here. Recursion tree goes a long way. Iteration method would be preferable.  The answer is $Θ (n \log \log n)$.  Can anyone arrive at the solution. Do Irrationals form a group? It's true that the irrationals do not form a group under addition or multiplication, but I want to find a binary operation "∗" such that ($\mathbb R \setminus \mathbb Q,∗$) is a group. It is possible by transpose of operation from $\mathbb R$, but I want a explicit example. Make 15 lines / ticks with Photoshop, each of them rotated of 20° This is what my GRUB 2 menu looks like so far:  I want to do a few things here:  I want to rename &quot;Windows Boot Manager (on /dev/sda1)&quot; to just &quot;Windows 8.1&quot;. I'd like to rename &quot;Ubuntu&quot; to &quot;Lubuntu 14.04.1&quot;. I'd also like to switch the order of the options to this:  Windows 8.1 Lubuntu 14.04.1 Advanced options for Lubuntu    (Windows will be the default OS) If it helps in any way, I'll explain the steps I took prior to this, since I am very new to Linux and don't know what information is required.  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Probably dumb question, but I would like to ask it anyway.  I was to solve this equation:  $z^4 = -16$  At first glance, the way to solve it would be (as any other equation):  $z^2 = \sqrt{-1} * 4 \lor z^2 = -\sqrt{-1} * 4$  $z^2 = 4i \lor z^2 = -4i$  $z = 2\sqrt{i} \lor z = -2\sqrt{i} \lor z = 2i\sqrt{i} \lor z = -2i\sqrt{i}$  However, in suggested answers, there are $z_1, z_2, z_3$ and $z_4$ found using trigonometric formulas. Is there something particulary wrong in my solution and the only proper one uses trigonometric form of complex number? I have a Spring MVC web app which uses Spring Security.  I want to know the username of the currently logged in user.  I'm using the code snippet given below .  Is this the accepted way?    I don't like having a call to a static method inside this controller - that defeats the whole purpose of Spring, IMHO.  Is there a way to configure the app to have the current SecurityContext, or current Authentication, injected instead?      @RequestMapping(method = RequestMethod.GET)   public ModelAndView showResults(final HttpServletRequest request...) {     final String currentUser = SecurityContextHolder.getContext().getAuthentication().getName();     ...   }