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.

Let

  • $n\in\mathbb{N}$,
  • $I_j\subseteq\mathbb{R}$, $1\le j\le n$, intervals,
  • $I:= I_1\times\dots\times I_n\subseteq\mathbb{R}^n$,
  • $f_j:I_j\to\mathbb{R}$, $1\le j\le n$, Lebesgue measurable,
  • $g:I\to\mathbb{R}$ continuous,
  • $h:I\ni(x_1,\dots,x_n)\mapsto g(f_1(x_1),\dots,f_n(x_n))\in\mathbb{R}$.

Then $h$ is Lebesgue measurable.

Do you know a textbook reference of this statement?

share|improve this question
    
So, it is certainly proved in any respectable measure theory book that $(x_1,\ldots,x_n) \to (f_1(x_1),\ldots,f_n(x_n))$ is Lebesgue measurable. Now you compose this from the left with a Borel measurable function. Where's the problem? –  t.b. Jan 7 '12 at 2:23
    
@t.b. The issue may be that many books discuss measurability of real or complex-valued functions only, allowing one to state this question, but not to draw on results about measurability of $\mathbb{R}^n$-valued functions like $(x_1, x_2, \dots, x_n) \mapsto (f_1(x_1), \dots, f_n(x_n))$. At least, this is the difficulty I ran into when looking for a quick reference that could be used as a black box. (Many of the books I have that discuss measure theory are not "respectable.") –  leslie townes Jan 7 '12 at 2:40
    
Thanks to both of you. My problem is that it's a long time since I dealt with some basic measure theory. My original intent was to find a proof for the formula: $f^{(-n)}(x) = \int_a^x \int_a^{\sigma_1} \cdots \int_a^{\sigma_{n-1}} f(\sigma_{n}) \, d\sigma_{n} \cdots \, d\sigma_2 \, d\sigma_1$ where $f$ is a Lebesgue integrable real-valued function. I will write down my results and ask another question about correctness of the argument. –  precarious Jan 7 '12 at 10:12
    
For reference: math.stackexchange.com/questions/97182/… –  precarious Jan 7 '12 at 14:37

1 Answer 1

The $n=2$ case is essentially covered by Theorems 1.7 and 1.8 in Rudin's Real and Complex Analysis.

It's not exactly what you want--- he assumes the domains of the $f_i$ are all equal to the same measurable space--- but this isn't essential to the proof, and once you see the proof you will understand how to modify it accordingly.

And of course the idea for general $n$ is the same (and exactly that which was outlined in t.b.'s comment; what he alludes to about left composition and measurability is a generalization of Rudin's Theorem 1.7).

share|improve this answer
    
Thank you. The case $n=2$ is good enough for my application. As far as I can see it, Rudin assumes $f_j:I\to\mathbb{R}$. So it seems to me that I have to change the argument involving a rectangles' preimage under $f$: Let $R\subset\mathbb{R}^2$ be an open interval. Then $R$ is the cartesian product of two open intervals $A,B\subset\mathbb{R}$. It holds $$ f^{-1}(R)=f_1^{-1}(A)\times f_2^{-1}(B). $$ The latter set is measurable in $\mathbb{R}^2$. Is this the adaption you had in mind? –  precarious Jan 7 '12 at 14:42
    
@precarious: Yes. –  leslie townes Jan 8 '12 at 6:52

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.