I don't know if this is true but I would like it to be: Suppose $(\Omega,\mathbb{B},\mu)$ is a probalility space, where $\mathbb{B}$ is atomless, and $\mathbb{A}$ is non tirvial $\sigma$-algebra contained in $\mathbb{B}$. If $f,g$ are two $\mathbb{A}$-measurable functions and for all $A\in\mathbb{A}$ we got that $\int_A fd\mu=\int_A gd\mu$ then $f=g$ a.s ?  If this isn't true I could say something about $f$ and $g$ ?
How to show $f=g$ almost everywhere  Let two real-valued measurable functions $f$ and $g$ be such that for any measurable set $E$ the integrals $\int_E f\,d\mu$  and  $\int_E g\,d\mu$ coincide.  Could you please help.  It seems obvious. we could define the integrals as charges. can we do sth from here. but I do not know how to approach.
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! I had a problem with disappearing marker as the data sum up to more than 100 while xmax was set to 100. See .  As written in the answer to the cited question I tried to avoid the problem by normalizing the value to 100. But this lead to (much smaller) rounding errors in previously correct rows and now the marker disappear there. See the following example. Before the normalisation the column F ist affected, after the column B.   How can I in a reliable way avoid this rounding problem? Or at least get a clear error message. I have quite a number of this plots and I don't want to have to check them all the time for missing markers.    \documentclass{scrreprt} \usepackage{pgfplots} \pgfplotsset{compat=1.13} \usepackage{pgfplotstable} \pgfplotsset{  my stackbar plot/.style={              xbar stacked,              xmin=0,xmax=100,              symbolic y coords={A,B,C,D,E,F,G},              ytick=data,              nodes near coords={xxx},}}  \begin{document}  \pgfplotstableread[col sep=space]{  text   --  -    +       ++  A      0.0 1.7 13.8    84.5  B      0.0 0.6 20.1    79.3  C      0.0 1.9 13.2    84.9  D      0.0 1.6 27.9    70.5  E      1.3 3.9 19.5    75.3  F      0.0 1.4 15.0    83.7  G      0.3 1.7 24.8    73.2  }\data   \pgfplotstablecreatecol[create col/expr={\thisrow{--}+\thisrow{-}+\thisrow{+}+\thisrow{++}}]{sum}\data %  \pgfplotstablecreatecol[create col/copy=--]{--o}\data  \pgfplotstablecreatecol[create col/copy=-]{-o}\data  \pgfplotstablecreatecol[create col/copy=+]{+o}\data  \pgfplotstablecreatecol[create col/copy=++]{++o}\data  \pgfplotstablecreatecol[create col/expr={100/\thisrow{sum}*\thisrow{--o}}]{--}\data  \pgfplotstablecreatecol[create col/expr={100/\thisrow{sum}*\thisrow{-o}}]{-}\data  \pgfplotstablecreatecol[create col/expr={100/\thisrow{sum}*\thisrow{+o}}]{+}\data  \pgfplotstablecreatecol[create col/expr={100/\thisrow{sum}*\thisrow{++o}}]{++}\data  \pgfplotstablecreatecol[create col/expr={\thisrow{--}+\thisrow{-}+\thisrow{+}+\thisrow{++}}]{sumnew}\data  \pgfplotstabletypeset[columns={text,sum}   ,precision=10,columns/text/.style={string type}]\data \quad \pgfplotstabletypeset[columns={text,sumnew},precision=10,columns/text/.style={string type}]\data   %\pgfplotstablesave{\data}{pgfplotstempout.dat} \begin{tikzpicture} \begin{axis}[my stackbar plot]  \addplot table [x expr = \thisrow{--o},y=text] {\data};  \addplot table [x expr = \thisrow{-o}, y=text] {\data};  \addplot table [x expr = \thisrow{+o}, y=text] {\data};  \addplot table [x expr = \thisrow{++o}, y=text] {\data}; \end{axis} \end{tikzpicture} \quad \begin{tikzpicture} \begin{axis}[my stackbar plot]  \addplot table [x expr = \thisrow{--},y=text] {\data};  \addplot table [x expr = \thisrow{-}, y=text] {\data};  \addplot table [x expr = \thisrow{+}, y=text] {\data};  \addplot table [x expr = \thisrow{++}, y=text] {\data}; \end{axis} \end{tikzpicture}  \end{document} Is there any measurable time dilation when Earth reaches perihelion? Can we measure such a phenomena relative to the motion of the outer planets? Sequences or 'chains' of adjoint functors Suppose we have (some categories and some functors such that) $F_1$ is left adjoint to $G_1$, $G_1$ left adjoint to $F_2$, $F_2$ left adjoint to $G_2$. Will $F_1$ then be equal to $F_2$ (and $G_1$ to $G_2$)?  Or can there exist arbitrarily long 'chains' of adjoint functors that do not trivialize? Voldemort couldn't kill Harry as long as he called the Dursleys home. Why didn't he just murder the Dursleys? This would prevent Harry from being able to live there. Is the Formula Logically Valid? This is a very basic query I can't figure out....  Let's say I have a two column table like this:  userid  |  roleid --------|--------    1    |    1    1    |    2    1    |    3    2    |    1   I want to get all distinct userids that have roleids 1, 2 AND 3.  Using the above example, the only result I want returned is userid 1.  How do I do this? Vitali-type set with given outer measure Difference between Hashing a Password and Encrypting it How do I set up a "secure" open resolver? This is a  about securing public DNS resolvers   Open DNS servers seem pretty neat and convenient, as they provide IP addresses that we can use consistently across our company regardless of where they are located. Google and OpenDNS provide this functionality, but I'm not sure that I want these companies to have access to our DNS queries.  I want to set up something like this for use by our company, but I hear a lot about this being dangerous practice (particularly in regards to ) and I want to make sure that we do this right. What things do I need to keep in mind when building this type of environment? Proving any number of the form $2^n$ can't be written as the sum of k consecutive numbers. I think i have proved that any number of the form $2^n$ can't be written as the sum of k consecutive positive integers, but i would be grateful if i could have some clarification as to whether this is a correct and valid proof or not.  Let $g=x+(x+1)+(x+2)+(x+3)+...+(x+(k-1))$, hence g is the sum of k consecutive positive integers.  $g=kx+\sum_{d=1}^{k-1}d$  Hence, $g=kx+\frac{k(k-1)}{2}$  Rearranging for x we find, $x=\frac{g}{k}-\frac{(k-1)}{2}$  So, for certain values of $k$, we can be sure that a positive integer value of $x$ can be found, and hence $g$ can be written as the sum of $k$ consecutive positive integers.There are two cases for $k$ to considor:  Case 1)  $k$ is odd. If $x$ exists, it must be the case $k$ | $g$, as for odd values of $k$, $\frac{(k-1)}{2}$ will alwyas be an integer, and hence $x$ will.  Case 2)  $k$ is even. $k$ does not divide $g$, but $\frac{g}{k}$ must be a multiple of $\frac{1}{2}$, and hence equals $\frac{h}{2}$, for some odd $h$.  Letting $j=2^n$, assume that we can write $j$ as the sum of $k$ consecutive positive integers.  If $k$ is odd, then if a $k$ exists, it must divide $j$, by Case 1. But, $j$ has no odd factors as it is a power of two, hence $k$ could not be odd.  If $k$ is even, by Case 2 it mut be the case $k$ does not divide $j$. Also, $\frac{j}{k}=\frac{h}{2}$, for some odd $h$.  But, this implies $2j$ has an odd factor, $h$. but, as $2j$ is a power of $2$, it can't have odd factors. So, $k$ can't be odd.  Hence, $k$ is neither odd or even, and this is a contradiction to our assumption. Hence no $x$ exists, and hence and number of the form $2^n$ can't be written as the sum of $k$ consecutive positive integers.  Any feedback on this proof would be appreciated. In a closed system, I know the rider wouldn't be able to change the combined center of gravity of the system by moving his weight around. If he pushes the bike to lean to the right, he would go to the left by the amount that would keep the combined cog the same.  But a bike with tires and a road under it is not a closed system, the tires wouldn't let the lower part of the bike left or right when tryingo to initiate lean. Does this mean it is possible to push with your bodies inertia to the bike, and the combined cog will be changed? I thing if you do so above the cog of the bike, you could, but I am not sure.  Edit: My question is not about if you can move the center of gravity of the bike at all. My question is about if can you change the combine cog of the system by moving around on the bike?  English is not my mother tongue, sorry for the confusion I want to run a script on all files with two specific extensions on a disk with a hierarchy of multiple nested folders.  My script works fine when I choose a folder further down the line with fewer subfolders but I'd like to run it on an entire archive with over a thousand folders.  The AppleScript takes forever to get all the files and then exits without returning anything and without executing the operations after getting the files.  No timeout message, no error message.  And the Finder becomes unresponsive (and remains stuck after the script exits).   set myFolder to choose folder with prompt "Choose a folder:"     tell application "Finder"         try             set eafFiles to (every file in entire contents of myFolder whose name ends with ".eaf") as alias list         on error             try                 set eafFiles to ((every file in entire contents of myFolder whose name ends with ".eaf") as alias) as list             on error                 set eafFiles to {}             end try         end try         try             set pfsxFiles to (every file in entire contents of myFolder whose name ends with ".pfsx") as alias list         on error             try                 set pfsxFiles to ((every file in entire contents of myFolder whose name ends with ".pfsx") as alias) as list             on error                 set pfsxFiles to {}             end try         end try         set myFiles to eafFiles &amp; pfsxFiles     end tell      repeat with CurrentFile in myFiles         set CurrentFile to CurrentFile as string          do shell script "perl /path/to/perl/script.pl " &amp; quoted form of CurrentFile     end repeat Suppose that sequences of real numbers satisfy:  \begin{align*}   a_n &amp;= \sum\limits_{r=0}^n \binom{n}{r} b_r \\ \end{align*}  Prove that:  \begin{align*}   (-1)^n b_n &amp;= \sum\limits_{s=0}^n \binom{n}{s} (-1)^s a_s \\ \end{align*}  My work:  The $n=0$ case:  \begin{align*}   a_0 &amp;= b_0 \\   b_0 &amp;= a_0 \\ \end{align*}  Then $n=1$ case:  \begin{align*}   a_1 &amp;= \binom{1}{0} b_0 + \binom{1}{1} b_1 = b_0 + b_1 \\   -b_1 &amp;= \binom{1}{0} a_0 - \binom{1}{1} a_1 = a_0 - a_1 \\   b_1 &amp;= a_1 - a_0 = b_0 + b_1 - a_0 = b_1 \\ \end{align*}  The inductive step:  \begin{align*}   (-1)^{n+1} b_{n+1} &amp;= \sum\limits_{s=0}^{n+1} \binom{n+1}{s} (-1)^s a_s \\ \end{align*}  I'm not sure where to go from here. I tried substituting $a_s$ into that last equation and it didn't help. Any ideas? Limit of $\cos(x)/x$ as $x$ approaches $0$ As the title says, I want to show that the limit of  $$\lim_{x\to 0} \frac{\cos(x)}{x}$$  doesn't exist.  Now for that I'd like to show in a formally correct way that   $$\lim_{x\to 0^+} \frac{\cos(x)}{x} = +\infty$$  I'm sure this is right since $\displaystyle\lim_{x\to 0^+} \cos(x) = 1$ and $\displaystyle\lim_{x\to 0^+} x = 0$, but since $\displaystyle\lim_{x\to 0^+} x = 0$ I can't just say:  $$\lim_{x\to 0^+} \frac{\cos(x)}{x} = \frac{1}{\displaystyle\lim_{x\to 0^+} x}$$   Things I tried: expand the fraction and use L'Hospital's rule (This didn't seem to yield results even though I tried quite a few things, should it? If it does, should it always work? It worked for similar problems) and going by the definition of the limit which didn't work for me either.  I'm grateful for any hints for doing this problem and similar problems especially. I cannot see WiFi option in my Ubuntu 15.04. What should I do? Revert asap  lspci -nnk | grep 0280 -A2 output:-  03:00.0 Network controller [0280]: Qualcomm Atheros Device [168c:0042] (rev 30) Subsystem: Foxconn International, Inc. Device [105b:e09a]   rfkill list output:-  0: hci0: Bluetooth Soft blocked: no Hard blocked: no 1: acer-wireless: Wireless LAN Soft blocked: no Hard blocked: no      dmesg | grep ath [    9.545199] ath10k_pci 0000:03:00.0: pci irq msi-x interrupts 8 irq_mode 0 reset_mode 0 [   10.029684] ath10k_pci 0000:03:00.0: Direct firmware load for ath10k/cal-pci-0000:03:00.0.bin failed with error -2 [   10.029701] ath10k_pci 0000:03:00.0: Direct firmware load for ath10k/QCA9377/hw1.0/firmware-5.bin failed with error -2 [   10.029704] ath10k_pci 0000:03:00.0: could not fetch firmware file 'ath10k/QCA9377/hw1.0/firmware-5.bin': -2 [   10.029711] ath10k_pci 0000:03:00.0: Direct firmware load for ath10k/QCA9377/hw1.0/firmware-4.bin failed with error -2 [   10.029713] ath10k_pci 0000:03:00.0: could not fetch firmware file 'ath10k/QCA9377/hw1.0/firmware-4.bin': -2 [   10.029719] ath10k_pci 0000:03:00.0: Direct firmware load for ath10k/QCA9377/hw1.0/firmware-3.bin failed with error -2 [   10.029721] ath10k_pci 0000:03:00.0: could not fetch firmware file 'ath10k/QCA9377/hw1.0/firmware-3.bin': -2 [   10.029726] ath10k_pci 0000:03:00.0: Direct firmware load for ath10k/QCA9377/hw1.0/firmware-2.bin failed with error -2 [   10.029728] ath10k_pci 0000:03:00.0: could not fetch firmware file 'ath10k/QCA9377/hw1.0/firmware-2.bin': -2 [   10.029733] ath10k_pci 0000:03:00.0: Direct firmware load for ath10k/QCA9377/hw1.0/firmware.bin failed with error -2 [   10.029735] ath10k_pci 0000:03:00.0: could not fetch firmware (-2) [   10.029737] ath10k_pci 0000:03:00.0: could not fetch firmware files (-2) [   10.029739] ath10k_pci 0000:03:00.0: could not probe fw (-2) How do I transcode a Javascript string to ISO-8859-1? Every open set in $\mathbb{R}$ is a countable union of closed sets When you are performing time consuming operations in bash like installing new software, is it possible to write text that will appear at the prompt when the operation has finished.  E.g   Imagine running:  apt-get install eclipse-platform   Then it will use quite a while to finish while you see the installation log, meanwhile I want to create new folders (workspaces) which Eclipse later will use.  Is this possible without opening a new terminal (or tab)? Producing .bbl file with Lyx and JabRef How do I produce the .bbl file required for arXiv using JabRef for bibtex and Lyx?