Let $\mu$ be a finite measure on $\mathbb{R}$. What is the definition of the operator $d$ in the expression: $d\mu$. For example, I have an exercise where at one point:

\begin{equation} d\mu(x) = \frac{d x}{1+x^2} \end{equation}

I would said this a "differential" of $\mu$, but I can not find any definition of this kind on the Internet.


This is a notation related to Radon-Nykodym theorem. In this context, this means that for each non-negative measurable function, $$\int_{\mathbb R}f(x)d\mu(x)=\int_{\mathbb R}\frac{f(x)}{1+x^2}dx.$$

  • $\begingroup$ Or, written another way, $\frac{d\mu}{dx}=\frac{1}{1+x^2}$. $\endgroup$
    – icurays1
    Oct 11 '13 at 20:17
  • $\begingroup$ IMO this does not quite answer the question. Expressions like $dx$, $df$ are just differentials (particular cases of differential forms). What it $d\mu$? $\endgroup$
    – Alexey
    Apr 4 '15 at 9:51
  • $\begingroup$ My intuition suggests me that $d\mu$ must be something like "density" of $\mu$, but i think that in mathematics measure density is a relative notion: only density of one measure with respect to another can be defined. $\endgroup$
    – Alexey
    Apr 4 '15 at 9:58
  • $\begingroup$ Moreover, one should ask not only for the meaning of $d\mu(x)$, but for the meaning of $f(x)d\mu(x)$. I would hope for a concise and precise answer, like in the case of $\int_a^b f(x)dx$ (here $f(x)dx$ is a differential form, to be integrated over the oriented interval $[a,b]$). $\endgroup$
    – Alexey
    Apr 4 '15 at 10:10
  • 1
    $\begingroup$ @Alexey $\int_{\mathbb R} g(x)\,dx$ probably means integration of $g$ with respect to the Lebesgue measure (or a different implicitly given measure) here. The measure $\mu$ is then defined by its density with respect to Lebesgue. Nothing in the question nor the answer are meant to be differential forms. $\endgroup$
    – epimorphic
    Apr 8 '15 at 1:34

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