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I am wondering why Rudin used the following notation in his "Real and Complex Analysis". It is in Definition 8.7, as following.

If $(X, \mathscr{S}, \mu)$ and $(Y, \mathscr{T}, \lambda)$ are two $\sigma$-finite measure spaces, and $Q\in \mathscr{S}\times \mathscr{T}$, then define $(\mu\times \lambda)(Q)=\int_{X}{\lambda(Q_x) d\mu(x)}=\int_{Y}{\mu(Q^y) d\lambda(y)}$.

My question is why $\mu$ and $\lambda$ depend on $x$ and $y$ respectively? Aren't they fixed once the measure spaces $(X, \mathscr{S}, \mu)$ and $(Y, \mathscr{T}, \lambda)$ are given? What is the meaning of $\mu(x)$ and $\lambda(y)$ or the author wants to emphasize here?


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My understanding is it's by analogy with $dx$ for ordinary integration with respect to the Lebesgue measure. It describes the infinitesimal size of a small measurable set containing $x$. – Qiaochu Yuan May 20 '11 at 15:57
@Qiaochu: thanks. I thought initially there were some measures generated/induced by $x$ and $y$ by looking at the notation! But on the other hand, if you say $d\mu(x)$ is analogous to $dx$ as in the Riemannian case, isn't this in disagreement with the definition of Lebesgue's integral? – Qiang Li May 20 '11 at 16:06
By the way: this question reproduces (and note that Arturo's assertion in the comments on the main question is later on withdrawn in the comments on Byron's answer). – Did May 20 '11 at 16:48
up vote 2 down vote accepted

There is no meaning to $\mu(x)$ (and Rudin does not use this notation). On the other hand the integral of a function $f$ with respect to a measure $\mu$ is denoted in a variety of ways, and amongst these, $$ \int f \mathrm{d}\mu=\mu(f)=\int f(x) \mathrm{d}\mu(x)=\int f(x) \mu(\mathrm{d}x). $$

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thanks. Fair enough! Just thought $\mu(x)$ implies some function dependence as I am used to understand. This notation is confusing to me. – Qiang Li May 20 '11 at 16:10
Once again: you will not find $\mu(x)$ anywhere. What some people use is $\mathrm{d}\mu(x)$. – Did May 20 '11 at 16:37

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