What does $dx$ mean in a Lebesgue integral?

Question

This is an introduction for Lebesgue integral of simple function in Carothers' Real Analysis.

We say that a simple function $\phi$ is Lebesgue integrable if the set {$\phi$ $\ne$ 0} has finite measure. In this case, we may write the standard representation for $\phi$ as $\phi = \sum_{i=0}^{n} a_i \chi_{A_i}$, where $a_0 = 0, a_1, .., a_n$ are distinct real numbers, where $A_0 = \{\phi = 0\}, A_1, ..., A_n$ are pairwise disjoint and measurable, and where only $A_0$ has infinite measure, Once $\phi$ is so written, there is an obvious definition for $\int \phi$, namely, $$\int \phi = \int_{\mathbb R} \phi = \int_{-\infty}^{+\infty} \phi(x) dx = \sum_{i=1}^{n} a_i m(A_i)$$.

I've noticed that wikipedia's definition of Lebesgue integral(see here https://en.wikipedia.org/wiki/Lebesgue_integration) uses $d\mu$. So What does $dx$ or $d\mu$ mean in Lebesgue integral?

Update:

I don't think it is a exactly duplicate one coz I didn't mean using $d\mu$ instead of $dx$. Before my typing this question, I have read Rodyen's Real Analysis, 3rd and he also uses $dx$ in Lebesgue integral as well. $d\mu$ is just from wikipedia. I have this question in this May when I was reading Caorthers' book and during that time, I treated it as a whole of symbols and being equal to a fixed formula -- $\sum_{i=1}^{n}a_i m(A_i)$. And then when I was trying to solve some problems with this symbol in Lebesgue integral, I felt weird for quite a while, recalling the Riemann's definition and then realized "ohhh, man, it is not Riemann integral".

possible duplicate of Can someone show me why mathematicians use $d\mu$ instead of $dx$ for Lebesgue Integral over $u(x)$ — user251257, Aug 01 '15 at 00:34
You would wrote $d\mu$ for integration with respect to measure $\mu$. Carothers does not use $\mu$ for Lebesgue measure. So he does not write Lebesgue integrals using $d\mu$, — GEdgar, Aug 01 '15 at 02:50
@GEdgar: Yes. I wrote $dx$ which from carothers and $d\mu$ is from wikipedia. Carothers didn't tell any story of what $dx$ means in Lebesgue integral. — Bear and bunny, Aug 01 '15 at 02:52
@GEdgar: Before my post of this question, I've also checked definition of Royden's Real Analysis, 3rd. He used $dx$ as well. So I think $d\mu$ is not seriously enough when you write a Lebesgue integral. — Bear and bunny, Aug 01 '15 at 02:57

score 8 · Accepted Answer · answered Aug 01 '15 at 02:26

The meaning of the $dx$ in a Lebesgue integral, or any other integral, depends on the extended framework in which the theory is situated. If you are working in a framework in which the measure is the main focus, where there is a wide class of possible measures, for example on a differentiable manifold or in a topological vector spaces, or something more exotic, then the $dx$ would have to be replaced by $d\mu$ to indicate which measure you are talking about. If you're just talking about the standard measure on a Cartesian space $\mathbb{R}^n$, then $dx$ is adequate. The $x$ then serves as a dummy variable (or "bound variable" in mathematical logic) so that you can use inline functions as integrands. For example $\int x^2\,dx$. If the measure is variable, you might like something like $\int x^2\,d\mu(x)$, which indicates the choice of measure and also the dummy variable.

In the framework of differential forms, the $dx$ has a different meaning. But that's a different kettle of fish.

In reality, integration theory has diverged in so many ways, the notations of the 17th and 18th centuries are no longer adequate. The "generalization thrusts" of integration theory include the following.

Generalization from Riemann integrable to very general kinds of functions.
Generalization of the class of measurable sets to a maximal class which extends the Borel measurable sets.
Generalization of the measure from the standard invariant measure on $\mathbb{R}^n$ to axiomatically defined very general measures.
Generalization from real-valued functions to distributions such as Radon measures, Schwartz distributions, Stieltjes integrals, etc.
Generalization from volume measures to surface measures, involving maybe differential forms, for example, as in the Stokes theorem.
Generalization to fractional dimensional measures, like Carathéodory measures, Hausdorff measures etc.
Generalization to extremely discontinuous domain sets, as in geometric measure theory.
Generalization to infinite-dimensional spaces, and spaces which are not vector spaces.

In these generalized frameworks, the traditional notations for integrals look more and more and more inadequate to describe what the integral is. So what I'm saying is that the problem is much bigger than you describe. The integral notation is just a shorthand, and you have to fill in the details by reading the context. I.e all of the notations are wrong or inaccurate or incomplete. And the extent of the inaccuracy or ambiguity is greater in the more generalized frameworks.

Wow, awesome! It is all clear. Benefit a lot from your answer. — Bear and bunny, Aug 01 '15 at 02:47
Thank you. I like your word choice "dummy variable". I have been thinking about that for a well. — aysljc, Feb 03 '21 at 06:23

score 5 · Answer 2 · edited Apr 13 '17 at 12:21

5

I answered a related question here.

At the very least, $dx$ is a variable-binder. The integral $\displaystyle\int_{\mathbb R} f(x,y)\, d\mu(x)$ binds the variable $x$ and leaves $y$ free, so the that value of the expression depends on the value of $y$ but not on any value of anything called $x$.

edited Apr 13 '17 at 12:21

Community

1

answered Aug 01 '15 at 00:40

Michael Hardy

1

Your previous answer is good, but it only talks about $dx$. It would be nice to hear how the $d\mu(x)$ notation differs from the $dx$ notation. For example, aside from what you said about binding the variable, is the only other purpose for the $d\mu(x)$ notation that tells us which measure we are using to integrate? Or is there somehow more to $d\mu(x)$ just like there is more to $dx$ than to tell us which variable to integrate with respect to ($dx$ can also intuitively and rigorously be thought of as an infinitesimal change in $x$, for example). – layman Aug 01 '15 at 01:05
@user46944 : Among the things to bear in mind here is that the dimensions, or units if you like, of $\mu$ get multiplied by those of $f$. $\qquad$ – Michael Hardy May 01 '17 at 02:08

score 1 · Answer 3 · answered Aug 01 '15 at 00:34

1

The notation $dx$ used for integration on the line is an abbreviation for $d\mu(x)$ where $\mu$ here denotes Lebesgue measure. It is a universally-observed convention for $\mathbb{R}$ or $\mathbb{R^d}$.

answered Aug 01 '15 at 00:34

ncmathsadist

50,127

1

If I understand it correctly, $d\mu(x)$ is just notation for the integral with respect to the measure $\mu$, while in calculus (i.e., the Riemann integral) $dx$ can be thought of as an infinitesimal change in $x$. In other words, the function that the $d\mu(x)$ notation serves is simply to tell us which measure we are using. – layman Aug 01 '15 at 00:36
Every time when I see a $dx$ in Lebesgue integral, it will recall my habit of solving integral by Riemann's definition. – Bear and bunny Aug 01 '15 at 00:39

What does $dx$ mean in a Lebesgue integral?

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