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  • $n\in\mathbb{N}$,
  • $I_j\subseteq\mathbb{R}$, $1\le j\le n$, 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?

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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:… – precarious Jan 7 '12 at 14:37

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).

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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

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