Suppose we are given a measure $d\mu$. I have seen in some physics textbooks that we often define a new measure in terms of this measure, let's say $$d\rho = f\,d\mu,$$ where $f$ is some integrable function.

My experience with measure theory is entirely self-taught, and so this may be incorrect, but I interpret this as saying, $$\rho(E) = \int_E f\,d\mu$$ for some $E$ in the $\sigma$-algebra (i.e. an application of the Radon-Nikodym Theorem).

Is this correct? If so, why is this often of interest in physics, I see it come up so often that I am beginning to wonder why this is. Is it simply out of notational convenience, or is it meant to show some underlying relation?

Or rather, should this be viewed in terms of infinitesimals instead of measure?


1 Answer 1


What you wrote is not an application of the Radon-Nikodym theorem. Given a measurable $f\geq 0$, the set function $\rho(E)=\int_Ef\,d\mu$ automatically defines a positive measure simply by monotone convergence; also we have absolute continuity $\rho\ll \mu$ (i.e for every measurable set $E$, $\mu(E)=0\implies \rho(E)=0$). The Radon-Nikodym theorem talks about the non-trivial converse!

You are right in that if $(X,\mathfrak{M})$ is the measurable space under consideration, then \begin{align} d\rho=f\,d\mu\,&:\iff \text{for every $E\in\mathfrak{M}$, $\rho(E)=\int_Ef\,d\mu$}. \end{align} So, the symbol on the left is shorthand notation for the right. The question of whether something should be viewed as a statement about measures/ differential forms etc boils down to context. A very common situation where this comes up is when computing center of masses, moments of inertia etc. One often starts out with a 'mass density' function $\rho$ on $\Bbb{R}^3$, and defines a measure $dm=\rho\,dV$, where $dV\equiv d\lambda_3$ denotes the 3-dimensional Lebesgue measure, and $dm$ now denotes the mass-measure for the body under consideration. Of course, not all types of masses can be described by such density functions with respect to Lebesgue measure (the prototypical example being point masses, so the Radon-Nikodym theorem doesn't apply to them). See this for a related discussion.

If you allow for $f$ to be not just non-negative, but if you allow $f\in L^1(\mu)$ to be signed or even complex-valued, then you get a corresponding signed/complex measure. Signed measures are also interesting, because we can use them to describe charge distributions.

  • $\begingroup$ Thanks! So this is really short-hand for when we describe a measure in terms of an integral and there's not much more to it? $\endgroup$
    – CBBAM
    Jul 16, 2022 at 19:23
  • 1
    $\begingroup$ @CBBAM right, it's just a quick way of writing down the definition of the new measure from old ones by prescribing a density, and encoding it in a familiar notation. $\endgroup$
    – peek-a-boo
    Jul 16, 2022 at 19:24

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