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I am an engineer and I learned my Lebesgue integral from an engineering text which dumbed down a lot of stuff, most prominently all Lebesgue integrals were introduced as $\int_\Omega u(x) dx$ instead of $\int_\Omega u d\mu$.

I was basically told not to worry about it and just keep in mind if you are integrating over a point, then the measure of that point is zero hence the integral is zero. And I was assured that in most applications, Riemmanian Integration and Lebesgue Integration yields completely identical answers.

But now I am going through some stuff written by mathematicians and $d\mu$ is almost always used in place of $dx$ i.e. these notes. Is there any reason why I should care about this distinction?

Fraïssé
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  • Your integrating over a measure, hence $d \mu$ not $dx$. If you're integrating over a line use $dx$. – Zach466920 Jul 30 '15 at 19:44
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    In measure theory, integrals are usually written as $\int_X f, d\mu$ or $\int_X f(x), d\mu(x)$ (but more commonly the former, at least when it makes sense) to emphasize the dependence on the measure $\mu$. The difference between Riemann and Lebesgue integrals is certainly extremely important, but notation is just notation. I should also mention that the Lebesgue integral is defined with respect to an arbitrary measure, not necessarily Lebesgue measure on $\mathbb{R}$, and the 'completely identical answers' remark only applies (under certain circumstances) to Lebesgue measure. – anomaly Jul 30 '15 at 19:52
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    That notation can make important results inaccessible to everyone who has not taken measure theory. If you like, you can think of the "$d\mu$" as a generic "$d x_1 dx_2\cdots dx_n$," so it can be used for multi-dimensional problems. When dealing with probabilities and PDF functions, the "$d\mu$" can also replace a PDF "$f_X(x_1,\ldots, x_n)dx_1\cdots dx_n$." – Michael Jul 30 '15 at 20:50
  • The strange thing about those notes is that the author seems to stick to integration over the real numbers, so it does not seem like there is much advantage in using $d\mu$ in replace of $dx$. – Michael Jul 30 '15 at 20:53
  • Some alternative notes on Lebesgue integration (which I think also includes Fatou stuff) is in this link. Note that on page 72 the author acknowledges other notation, including $\int f(x) dx$ that coincides with notation you use from calculus. http://www.math.washington.edu/~burke/crs/555/555_notes/integration.pdf – Michael Jul 30 '15 at 21:56

2 Answers2

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It's a way of emphasizing that you're doing measure theory and using the Lebesgue integral, which is substantially more general than the Riemann integral. Among other things, it depends on a choice of measure (this is what $\mu$ refers to), and while choosing the Lebesgue measure reproduces the familiar answers you're used to from calculus, choosing other measures does other stuff. The $d \mu$ notation also continues to apply to multivariate integrals, whereas $dx$ really only applies to integrals over $\mathbb{R}$.

Qiaochu Yuan
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  • How does it emphasize in itself that it is the Lebesgue integral? The $\mu$ could be whatever measure. – quid Jul 30 '15 at 19:51
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    @quid: by "Lebesgue integral" I mean the measure-theoretic integral with respect to some measure. – Qiaochu Yuan Jul 30 '15 at 19:52
  • Thanks for the clarification. With this in mind it makes sense to me. – quid Jul 30 '15 at 19:57
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    The Lebesgue integral is not more general than the Riemann integral, as some functions, such as $x \mapsto \frac{\sin x}{x}$ an $\mathbb{R}$, are Riemann integrable, but not Lebesgue integrable. – Keba Jul 30 '15 at 20:24
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    @Keba: again, by "Lebesgue integral" I mean the measure-theoretic integral with respect to some measure, not necessarily the Lebesgue measure. – Qiaochu Yuan Jul 30 '15 at 20:32
  • @QiaochuYuan On page 6 the notes claim that $\int_0^{\infty} e^{-x}d\mu(x) = 1$, with no mention of what the measure $\mu$ is whatsoever! I think those notes are confusing. Would you agree the reader should likely replace the $d\mu$ with $dx$ everywhere it appears in those notes? (especially as it seems to stick to integration over the real number line)? – Michael Jul 30 '15 at 21:05
  • @Michael: On page 2 the notes clarify that $\mu$ is Lebesgue measure. I think you should not replace the $d\mu$ with $dx$ everywhere it appears; it's important to recognize the difference between the Lebesgue and Riemann integrals. Even if they ultimately give the same answer in familiar situations, they're built very differently, and the way you prove theorems about them is different. – Qiaochu Yuan Jul 30 '15 at 21:40
  • @QiaochuYuan Well then we can agree to disagree. It seems some folks do use notation $\int_{\mathbb{R}} f(x) dx$ even for the Lebesgue integral, which IMHO makes things more accessible to a wider audience. I agree that a reminder that the Lebesgue integral is different can be useful for people who have taken measure theory, though. – Michael Jul 30 '15 at 22:04
  • @Michael: look, these notes were written for a reason, and that reason was to teach some students about the dominated convergence theorem, which is a theorem about the Lebesgue integral. There is no intention on the part of the author for the notes to be accessible to a wider audience that doesn't know about the Lebesgue integral: the audience is students in a course involving the Lebesgue integral. If you don't want to learn about the Lebesgue integral, don't read these notes. – Qiaochu Yuan Jul 30 '15 at 22:44
  • Well, your way of looking at it is probably better. Now that I think about it, I suppose the $d\mu$ is a good reminder about chopping up the range of $f$, rather than the domain. – Michael Jul 31 '15 at 05:45
  • @Keba: "x $\mapsto \frac{\sin x}{x}$ an R, are Riemann integrable, but not Lebesgue integrable" why? Its Riemann integral on $[0, +\infty]$ is $\frac{\pi}{2}$. How about its Lebesgue integral? How to calculate? Can you explain more? Thanks. – Bear and bunny Aug 04 '15 at 20:01
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    @Bearandbunny: $f$ is Lebesgue integrable only if $|f|$ is Lebesgue integrable. (Both $f_+$ and $f_-$ need to be finite, but both are not finite in this case.) – Keba Aug 06 '15 at 14:02
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    @Keba: I think both of us have made a mistake here. $x \mapsto \frac{\sin x}{x}$ on $\mathbb R$ is not a Riemann Integral, it is an improper integral, isn't it? – Bear and bunny Aug 06 '15 at 22:34
  • @Bearandbunny: Yes, but an improper Riemann integral. – Keba Aug 07 '15 at 11:03
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In some cases I think that the answer is easier than we believe. As somebody said, the abstract integral of a function $f$ with respect to a measure space $(X,\Sigma,\mu)$ is denoted by $\int_X u \, d\mu$. This is the most economic piece of notation. For particular purposes, we may felle the need to introduce a dummy integration variable, and we write $\int_X u(x)\, d\mu(x)$.

Since the Lebesgue measure is particularly useful in mathematical analysis and in many other fields, mathematicians decide to save letters and ink, and they write $dx$ instead of $d\mathcal{L}(x)$. In higher dimension, the notation $d\mathcal{L}^N(x)$ is seldom seen, and everybody writes $dx_1\cdots dx_N$.

Of course all this comes from another little abuse of notation: we should not write $$ \int_{\mathbb{R}^2} e^{-x^2-y^2}\, dx\, dy $$ but instead

$\int_{\mathbb{R}^2} f \, d\mathcal{L}^2$, where $f(x,y)=e^{-x^2-y^2}$.

We all understand why we prefer the abuse over the more correct version.

Siminore
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