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If $\{f_i, 1 \le i \le m\}$ is a set of real valued Borel functions on $\mathbb R$, how to show a vector of functions, $(f_1, f_2,..., f_m)$ is a measurable mapping from $(\mathbb R^m, \mathcal{B}(\mathbb R^m))$ to $(\mathbb R^m, \mathcal{B}(\mathbb R^m))$ itself?

It seems to be an obvious result. However, I don't know how to prove it.

I think it is equivalent to show i-th entry of projection of a open set in $\mathbb R^n$ is still open in $\mathbb R$.

Eugene Zhang
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Bear and bunny
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  • If V is an open subset of $R^n$ then V is a union of a family of sets ,each of which is equal to $\prod _{i=1...n} A_i$ where each $A_i$ is an open real interval. The projection of such a set onto the $i$-th co-ordinate is $A_i$. – DanielWainfleet Sep 03 '15 at 07:57
  • @user254665: I have proved projection of open set is still open coz every point in open set is an interior and can be covered by a open ball in $\mathbb R^n$ inside and then projection of the ball will be open interval that contains projection of the point. – Bear and bunny Sep 03 '15 at 13:37
  • You should probably clarify how you define $F = (f_1, \dots, f_m)$. I assume you mean $F(x_1, \dots, x_m) = (f_1(x_1), \dots, f_m(x_m))$? Then you should show two separate claims: 1) If $g_1, \dots, g_k$ are real valued Borel measurable functions on $\Bbb{R}^n$, show that $x \mapsto (g_1(x), \dots, g_k (x))$ is measurable. 2) If $f$ is a real valued Borel measurable function on $\Bbb{R}$, show that $(x_1, \dots, x_n) \mapsto f(x_i)$ is measurable. Then combine 1) and 2). – PhoemueX Sep 03 '15 at 16:41
  • @PhoemueX: Your comment reminds me of confusing definition of F = $(f_1, \dots, f_m)$. Actually the exercise doesn't say the definition and initially, I treat it as $(f_1(x), \dots, f_m(x))$, the same $x$ working for m kinds of functions. And I think I need to show i-th projection of preimage of a open set in $\mathbb R^m$ and then a product of finite Borel sets in $\mathbb R$ will be a Borel set in $\mathbb R^m$. If $F(x_1, \dots, x_m) = (f_1(x_1), \dots, f_m(x_m))$, the problem will be sadly complicated. – Bear and bunny Sep 03 '15 at 17:09

2 Answers2

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Since $\{f_i, 1 \le i \le m\}$ is a set of Borel measurable functions $$ \{x:f_i(x)<c_i\}\subset \mathcal{B}(\mathbb R) $$ for all possible $c_i\in\Bbb{R}$. So $$ \{x:f_1(x)<c_1\}\times \{x:f_2(x)<c_2\}\times\cdots\times\{x:f_m(x)<c_m\}\subset \mathcal{B}(\mathbb R)^m\subset \mathcal{B}(\mathbb R^m) $$ for all possible $c_1\cdots c_m\in\Bbb{R}$. This means $(f_1, f_2,..., f_m)$ is Borel measurable.

Edit: $\quad$ We prove $\mathcal{B}(\mathbb R)^m\subset \mathcal{B}(\mathbb R^m)$. The idea is inspired by PhoemueX's comment.

Let $\overrightarrow{A_i}=\mathbb {R^i}×A_i×\mathbb {R^{m−i−1}}$, where $A_i\:(1\leqslant i \leqslant m)$ are Borel set. Then clearly $\overrightarrow{A_i}\subset \mathcal{B}(\mathbb R^m)$.

Now for any set $B\in\mathcal{B}(\mathbb R)^m$ $$ B=A_1\times\cdots \times A_m=\bigcap_{i=1}^m \overrightarrow{A_i}\in\mathcal{B}(\mathbb R^m) $$ for $\overrightarrow{A_i}\subset \mathcal{B}(\mathbb R^m)$ and Borel $\sigma$-algebra is closed under countable intersection. So we have $$ \mathcal{B}(\mathbb R)^m\subset \mathcal{B}(\mathbb R^m) $$

Eugene Zhang
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  • $\mathcal{B}(\mathbb R)^m=\mathcal{B}(\mathbb R^m)$, hmmm, why? – Bear and bunny Sep 03 '15 at 13:34
  • Open projection set in $\mathbb R$ is equivalent to open set in $\mathbb R^n$ and then $\mathcal{B}(\mathbb R)^m=\mathcal{B}(\mathbb R^m)$ coz both are produced by open sets? – Bear and bunny Sep 03 '15 at 13:57
  • Thanks. I agree with that. Unfortunately, "projection of a open set in $R^m$ is still open in R" will fail to work. Please see Phoemue's comments below. I'm not sure whether your answer will help the proof or the proof itself is actually incorrect. I'm working on it. Hope to hear from your idea. – Bear and bunny Sep 03 '15 at 22:55
  • Hmmmmmmm, make sense! Must vote. – Bear and bunny Sep 04 '15 at 03:41
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You're right. you have that the image of an open set under projection is open, then the preimage of that is measurable in $\mathbb{R}^1$. Then, the product of measurable sets is measurable, and you are done.

JHalliday
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  • How to show a product of a finite Borel sets is still Borel? – Bear and bunny Sep 03 '15 at 05:33
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    Hmm I thought you just wanted this function to be measurable. It is easy if you just want that. see here http://math.stackexchange.com/questions/360876/is-the-product-of-two-measurable-subsets-of-rd-measurable-in-r2d – JHalliday Sep 03 '15 at 05:37
  • Still appreciate, hope someone can help me prove a product of finite Borel sets in $\mathbb R$ belongs to $\mathcal{B} (\mathbb R^m)$ – Bear and bunny Sep 03 '15 at 05:44
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    This isn't hard for an $F_\sigma$ or $G_\delta$ set, just take the product of the open\closed sets each time in the sequence. Probably some transfinite induction stuff lets you get this for any Borel set. I'm not sure beyond that. – JHalliday Sep 03 '15 at 08:10
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    @Bearandbunny: It suffices to show that if $A$ is Borel, then so is $\Bbb{R}^k \times A \times \Bbb{R}^{n-k-1}$ for every $0 \leq k < n$ (if you have that, take suitable intersections). For simplicity, let us assume $k=0$. Then set $G := {A \mid A \times \Bbb{R}^{n-1} \text{ Borel}}$. Show that this is a sigma algebra and that it contains all open sets. Note furthermore, that knowing that projections of open sets are open does not help you at all for this exercise. Further note that projections of Borel sets are not Borel in general! – PhoemueX Sep 03 '15 at 16:38
  • @PhoemueX: Scare me. "projections of Borel sets are not Borel in general!" is based on your understanding of $F(x_1, \dots, x_m) = (f_1(x_1), \dots, f_m(x_m))$? – Bear and bunny Sep 03 '15 at 17:12
  • @Bearandbunny: I don't really understand what you are asking right now, but maybe the following could be helpful: http://math.stackexchange.com/questions/910147/projection-of-a-set-g-delta. – PhoemueX Sep 03 '15 at 17:48
  • @PhoemueX: If that's true, both JHalliday's answer and my guess will fail to work. – Bear and bunny Sep 03 '15 at 21:31