Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

Thanks.

share|improve this question
    
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 math.stackexchange.com/q/5230/6179 (and note that Arturo's assertion in the comments on the main question is later on withdrawn in the comments on Byron's answer).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
add comment

1 Answer

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, $$ \int f \mathrm{d}\mu=\mu(f)=\int f(x) \mathrm{d}\mu(x)=\int f(x) \mu(\mathrm{d}x). $$

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.