8
$\begingroup$

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.

$\endgroup$
0
4
$\begingroup$

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

$\endgroup$
12
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.