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I am trying to get an explanation in words, or math, of what the $d\mu$ means in an integration statement. Such as: $$\int f \ d\mu$$ How does the measure change our old "calculus" notion of integration? What is going on here that is different?

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2 Answers 2

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I actually struggled with this concept in grad school since I was studying applied math and was sort of thrust into higher level theory without building it up rigorously like I assume would be done in a pure math program.

If we are integrating over a space $X$, I sometimes prefer the notation $\int_X f(x)\mu(dx)$. I like to think of it as splitting the space we are integrating over into infinitesimal pieces, but we have to take the measure of those infinitesimal pieces as they may not all be identical under $\mu$. I'm used to working with probability measures, and at least in that case, you can often think of it similar to the way Lebesgue and Riemann integration are developed.

Create a disjoint partition $X=\cup_{k=1}^N A_k$, and define the sum which will approximate the integral using appropriately chosen sample points $x_k\in A_k$.

$$\int_X f(x)\mu(dx) \approx \sum_{k=1}^N f(x_k) \mu(A_k).$$

Ideally, taking the limit as $N\rightarrow\infty$ (carefully refining the partition and choosing appropriate sample points as $N$ increases) will make the sum converge to the integral. What we are doing here is effectively approximating the function $f$ with a simple function which is constant on a finite collection of measurable sets.

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    $\begingroup$ can I think it as $\int_X f(x) \mu(dx)$ = $\int_X f(x) \mu(x) dx$ as well? Where $\mu(x)$ is a function just giving different weights at different location. $\endgroup$ Aug 18, 2021 at 3:26
  • $\begingroup$ @ArtificiallyIntelligence I've seen it written $\int f \mu(dx)$ and $\int f d\mu$ most frequently. $\int f(x)\mu(x)dx$ to me looks like a standard integral of the product of functions $f(x)\mu(x)$ w.r.t. $x$ like from calculus. Generally, $dx$ simply implies you are doing standard integration, i.e. Riemann or Lebesgue (if $x$ is the independent variable). You need to signify that the measure being used is something else. $\endgroup$
    – jdods
    Aug 18, 2021 at 15:04
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    $\begingroup$ Thanks. I figured it out. I should use Radon derivative, it should be $\mu^{\prime}$. $\endgroup$ Aug 18, 2021 at 21:35
  • $\begingroup$ @jdods Can you explain this with respect to the standard integrals? I still have trouble understanding $\int_X f(x) \mu(dx)$. If i understand correctly, $dx$ here is not the integration operator but something else similar to a subset of $X$, why can't this be written as something more understandable such as $\int_X f(x)\mu(\text{subset that x is in})dx$, where $\mu(\text{subset that x is in})$ is a measure/weight on that particular subset instead? $\endgroup$
    – smaillis
    Mar 1 at 3:24
  • $\begingroup$ @smaillis maybe best to eliminate "d" altogether because it is different from the idea of derivative in $\mathbb R$ in one sense. One could instead write $\int_{(X,\mu)}f$ without a $d$. The integral is defined in terms of simple functions (with the integral of a simple function just equal to a finite sum). Try to understand it in the limiting sense I put in my answer where we have a partition $(A_k)$ and we are refining this partition so that $\mu(A_k)$ is sufficiently small to make the simple function $\sum_k f(x_k)\mathbf 1_{A_k}$ a "good approximation" to $f$. Does that help? $\endgroup$
    – jdods
    Mar 1 at 17:58
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Think of it physically: each measure assigns different weights to given sets: consider for example the particular case $d\mu=df(x)=f'(x)dx$ for a well behaved $f(x)$. Here you can really see the difference between the "ordinary" measure $dx$, which does not care about the location of the set, and $f'(x)dx$, which indeed does! In formulas: \begin{equation} \int_{[0,1]}dx =1= \int_{[1,2]}dx \end{equation} But in general \begin{equation} \int_{[0,1]}df(x) = f(1)-f(0) \neq f(2)-f(1) = \int_{[1,2]}df(x) \end{equation} This is just an example, but the idea applies with general measures $\mu$; it is in this sense that this perspective conveys the concept - in my view - of a weighted measure. It allows far more general and powerful structures than the old Riemann integration idea with shrinking rectangles.

One can then go further and (try to) relate different measures, for example via the Radon-Nikodym theory. There's a new world out there. :-)

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  • $\begingroup$ if you think this way with differentiation, then you will find the formula in this page very confusing: en.wikipedia.org/wiki/Dirac_measure $\endgroup$ Aug 18, 2021 at 3:28
  • $\begingroup$ This is a neat and intuitive explanation of integration w.r.t. a measure. Thank you for the answer. $\endgroup$ Feb 26 at 19:22

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