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context: i am currently learning about the Radon Nicodym Theorem.

I was wondering what the notation $du$ represent, where $u$ is a measure. I mean just $du$ by itself, not within an integral or anything.

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1  
Depending on your conventions, nothing. –  Qiaochu Yuan Mar 10 '11 at 23:51
    
$du$ can have different meaning depending on the context. –  Fabian Mar 10 '11 at 23:52
    
Maybe it helps if you look at this page (en.wikipedia.org/wiki/Differential_%28mathematics%29) and pages which are link from this page? –  Fabian Mar 10 '11 at 23:53
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The last name is usually romanized/spelled as "Nikodym". –  Arturo Magidin Mar 11 '11 at 0:17

2 Answers 2

up vote 0 down vote accepted

It represents the measure $u$. If (a function) $f$ is the Radon-Nikodym derivative (or density) of $u$ w.r.t. $v$, it is denoted by $du/dv$. Now, if $f=du/dv$, then $du=f dv$. For example, suppose that $u$ is an absolutely continuous probability distribution with density function $f$. Then, this can be denoted by $du(x)=f(x)dx$. That is, the right-hand side determines the measure $u$ by $u((a,b]) = \int_a^b {f(x)dx}$ (for all $-\infty < a < b < \infty$). Here, $dx$ stands for the Lebesgue measure on $\mathbb{R}$. Remark: an alternative notation for $du(x)$ is $u(dx)$.

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Excluding the example, this is essentially the same answer automorphism gave. Note also that $f$ is unique a.e. $v$. –  Shai Covo Mar 11 '11 at 1:05

Rudin for instance uses $d\mu$ in the following sense: $d\mu = fd\lambda$ just means the measure $\mu$ is defined by $\mu(E) = \int_E fd\lambda$ where for instance $f\in L^1(\mathbb{R})$. That's all it is: just a shorthand for this.

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