I'm trying to prove the following identity: If $f:X \rightarrow \mathbb{R}$ is a bounded, continuous function, and $\mu$ is a borel measure in $X$, then $\int f d\mu = \int_0^1 \mu(\{x \in X ; f(x)&gt;t\})dt$. (W/ loss of generality, one may consider $0\leq f\leq 1$. This equality is used to prove one of the portmanteau theorem affirmations, as in chapter 8 of Gordan Z.'s lecture notes:  As he says, it is a particular case of Problem 5.23, but i'm trying to obtain it by scratch. My attempt was: name the set $A_t=\{x \in X ; f(x)&gt;t\} $. Then $ \int_0^1 \mu(A_t)dt= \int_0^1 \int \mathcal{X}_{A_t} d\mu dt=\int \int_0^1 \mathcal{X}_{A_t} dt d\mu$. Then it remains to prove that $\int_0^1 \mathcal{X}_{A_t} dt $ is equal to $f$, but I got stuck at this point.
Let $f\colon X\to\overline{\mathbb{R}}_{\geq 0}$ be a measurable function. Show that     $$ \int f\, d\mu=\int\limits_0^{\infty}\mu(\left\{x\in X: f(x)&gt;t\right\})\, dt. $$     (The right integral is to be read as an Lebesgue-integral.)      Hello, my idea is to approximate $f$ by the functions $$ f_n:=2^{-n}\sum\limits_{k=1}^{\infty}\chi_{\left\{x\in X: f(x)&gt; k/2^n\right\}}. $$  With this approximation by simple functions I got $$ \int\limits_X f\, d\mu=\lim\limits_{n\to\infty}\int\limits_X f_n\, d\mu=\lim\limits_{n\to\infty}\sum\limits_{k=1}^{\infty}\frac{k}{2^n}\cdot\mu\left(\left\{x\in X: f_n(x)=k/2^n\right\}\right)\\=\lim\limits_{n\to\infty}\sum\limits_{k=1}^{\infty}\frac{k}{2^n}\cdot\mu\left(\left\{x\in X:f(x)&gt;k/2^n\right\}\right) $$  This is the point where I do not come along anymore... is that right to this point and if yes: How can I continue in order to get the desired right side?  With regards
How to switch language keyboard combination? I am writing a cover letter and CV using moderncv. Everything works fine so far, but the text in the cover letter is not justified. I think it is good that it is not hyphenated, but it would look much nicer if the text was justified.  Here is a minimal example:  \documentclass{moderncv} \moderncvstyle{classic}  \usepackage{lipsum} % just for dummy text  \firstname{John} \familyname{Doe}  \begin{document} \recipient{Prof.\ Dr.\ Foo Bar}{} \opening{Dear Professor Dr.\ Bar,} \closing{Yours sincerely,} \enclosure[Attached]{curriculum vit\ae}  \makelettertitle \lipsum[1] \makeletterclosing  \clearpage  \makecvtitle  \section{Education} \cventry{2009--present}{PhD Student}{here}{and}{there}{} \end{document} Why is human vision restricted to 400-700 nm? Across the electromagnetic spectrum, 400-700 nm is a narrow spectrum of frequencies and focused in the region of short wavelengths. For example, radio waves cover a large range of frequencies unexploited by the visual system. So what biological reason is there that evolved us to use such a small frequency bandwidth for vision? How do I build a native (command line) executable to run on Android? What are good out of the box alternative geocoding solutions to Esri's services? Esri's  will retire at the end of the year, what are good GIS services that can replace the North American and US Streets geocoders?  I have totally excluded Esri's replacement, the  as it is a subscription service. Fixing the sucktastic merge UI (for tags) When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$? Prove that if $n$ is not the square of a natural number, then $\sqrt{n}$ is irrational. Unable to Retrieve Metadata The cartesian product $\mathbb{N} \times \mathbb{N}$ is countable I'm examining a proof I have read that claims to show that the Cartesian product $\mathbb{N} \times \mathbb{N}$ is countable, and as part of this proof, I am looking to show that the given map is surjective (indeed bijective), but I'm afraid that I can't see why this is the case. I wonder whether you might be able to point me in the right direction?  Indeed, the proof begins like this:  "For each $n \in \mathbb{N}$, let $k_n, l_n$ be such that $n = 2^{k_n - 1} \left(2l_n - 1 \right)$; that is, $k_n - 1$ is the power of $2$ in the prime factorisation of $n$, and $2 l_n - 1$ is the (necessarily odd) number $\frac{n}{2^{k_n - 1}}$."  It then states that $n \mapsto \left(k_n , l_n \right)$ is a surjection from $\mathbb{N}$ to $\mathbb{N} \times \mathbb{N}$, and so ends the proof.  I can intuitively see why this should be a bijection, I think, but I'm not sure how to make these feelings rigorous?   I suppose I'd say that the map is surjective since given any $\left(k_n , l_n \right) \in \mathbb{N} \times \mathbb{N}$ we can simply take $n$ indeed to be equal to $2^{k_n - 1} \left(2l_n - 1 \right)$ and note that $k_n - 1 \geq 0$ and thus $2^{k_n - 1}$  is both greater or equal to one so is a natural number (making the obvious inductive argument, noting that multiplication on $\mathbb{N}$ is closed), and similarly that $2 l_n - 1 \geq 2\cdot 1 - 1 = 1$ and is also a natural number, and thus the product of these two, $n$ must also be a natural number. Is it just as simple as this?  I suppose my gut feeling in the proving that the map is injective would be to assume that $2^{k_n - 1} \left(2 l_n - 1 \right) = 2^{k_m - 1} \left(2 l_m - 1 \right)$ and then use the Fundamental Theorem of Arithmetic to conclude that $n = m$. Is this going along the right lines? The 'implicit' definition of the mapping has me a little confused about the approach.    On a related, but separate note, I am indeed aware that if $K$ and $L$ are any countable sets, then so is $K \times L$, so trivially, taking the identity mapping we see trivially that this map is bijective and therefore that $\mathbb{N}$ is certainly countable (!), and thus so is $\mathbb{N} \times \mathbb{N}$. Hence, it's not really the statement that I'm interested in, but rather the exciting excursion into number theory that the above alternative proof provides. What is the difference between CentOS "DVD" vs "Everything" ISOs Names of some months don't make sense I'm not a native English speaker but I'm always trying to do my best. Unfortunately I have a real problem with dates for some odd reason, I couldn't learn when was my birthday until I was 12 years old.  Anyway, as I was learning names of the months on English, and since I know some Latin from high school I noticed that English months don't match up with their Latin numeral counterparts. For example:  September - 9. month October - 10. month November - 11. month December - 12. month  While on latin septem, octo, novem and decem are words for 7, 8, 9 and 10.  So, does anyone know why is that. At first I thought it had something to do with switch from Julian to Gregorian calendar but I figured it's too small of a difference.  Anyone? :) Is sum of two orthogonal matrices singular? I am trying to solve following problem.      Let $A, B \in \mathbb{R}^{n\times n}$ be an orthogonal matrices and $\det(A) = -\det(B)$. How can it be proven that $A+B$ is singular?   I could start with implication:  $\det(A)=-\det(B) \Rightarrow B$ is created from $A$ by swapping two lines or columns.   But I am not sure if this implication is correct. Is every sufficiently large positive integer of the form $ab + ac + bc + 1$? Is every sufficiently large positive integer $A$ of the form $ab + ac + bc + 1$   where $a,b,c$ are some positive integers larger than some given positive integer $d$ ?  How large is sufficiently large , in other words how do we know our number $A$ is sufficiently large ( as a function of $d$ ) ?    Update :  I read the following about positive integers that are not of the form $ ab + ac + bc $:    And I quote about that sequence above : " Note that for n in this sequence, n+1 is either a prime, twice a prime, the square of a prime, 8 or 16 "  This is why I considered the weird +1 in $ab + ac + bc + 1$  So composites larger than 16 not a prime , twice a prime or the square of a prime are of the form $ab + ac + bc + 1$.  And I wonder why !  I guess that relates strongly to the Original posted problem. Maybe I should have mentioned that before , sorry.  Will Jagy's answer gave the same OEIS link which I already knew. This also explains some of the comments.  Im no expert in genus theory or Galois theory but I assume this has a simple answer. What is the analytic continuation of the Riemann Zeta Function When should I not kill -9 a process? diophantine equation: $x^2 +y^2 = z^n$ Prove that $x^2+y^2=z^n$ has a solution $(x, y, z)$ in $\mathbb{N}$ for all $n\in\mathbb{N}$  I tried to prove this by induction, but couldn't. ( This was probably because the solution for some $n$ isn't necessarily related to the solution for $n+1$)  I can't seem to see any other way other than induction for proving the statement. Any help/hints on solving this problem, or any alternative approach will be highly appreciated :) Inequality between $\ell^p$-norms Simulating the Windows key on an old keyboard What is the point of isomorphism concept?