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Let $f\colon \mathbb R^{n} \to \mathbb R^{m}$ a Borel function. Prove that the graph $\{(x,f(x)) \mid x \in \mathbb{R}^n\}$ is Borel in $\mathbb R^{n+m}$.

Martin
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Malaq
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    You could and should include your attempts. I'm sure you won't receive downvotes anymore once you will have done that. – Davide Giraudo Jan 25 '13 at 12:52
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    I'm noob, sorry – Malaq Jan 25 '13 at 13:10
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    I don't understand why two of this person's questions are getting downvoted, while most other questions that show no effort (e.g.) are happily answered without any negative feedback. –  Jan 25 '13 at 13:57
  • @Rahul I can't know for sure. The first two questions (this and the other heavily downvoted one) were posted in quick succession, probably because they were already prepared. When I first saw this question it only had one downvote and I think it had three when I posted. The usual lottery in voting plus trigger-happy piling up, I guess. – Martin Jan 25 '13 at 14:13

1 Answers1

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Choose a countable base $\{B_k\}_{k \in \mathbb{N}}$ for the topology of $\mathbb{R}^m$.

By definition, $\operatorname{graph}f = \left\{(x,y) \in \mathbb{R}^{n+m} \mid y = f(x)\right\}.$ Now $$ \begin{align*} y \neq f(x) &\iff \exists k \in \mathbb{N}\colon (f(x) \in B_k) \land (y \notin B_k) \\ &\iff \exists k \in \mathbb{N}\colon (x,y) \in f^{-1}(B_k) \times B_{k}^c \end{align*} $$ and Borel measurability of the graph follows from $$\mathbb{R}^{n+m} \setminus \operatorname{graph}f = \bigcup_{k \in \mathbb{N}} f^{-1}(B_k) \times B_{k}^c.$$

Martin
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  • Thanks Martin. I have many exercices. Would you like to help me? – Malaq Jan 25 '13 at 13:49
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    I think there are many people out there who are more than willing to help you. Bear in mind: the more work and own effort you show the more will people be willing to help you and write good and detailed answers. You'll gain much more if you try to do as much of your homework on your own and ask for help here only if you are really stuck or confused. – Martin Jan 25 '13 at 14:08
  • Thanks for this. I just gave an answer regarding a converse statement here (if the graph is Borel then $f$ is Borel), and I linked that to your answer: https://math.stackexchange.com/questions/4484012/let-b-in-mathcal-b-y-and-c-in-mathcal-bx-times-y-is-x-in-x-mi?noredirect=1#comment9403539_4484012 – Michael Jul 01 '22 at 05:25
  • Your proof technique above seems quite useful in the more general case when the range is a general Polish space. When the range is $\mathbb{R}^m$ another proof is: [Suppose $f$ is Borel. Define the Borel measurable function $g(x,y) = f(x)-y$, then $g^{-1}({0}) = {(x,y) : f(x)=y} = graph f$ is Borel.] – Michael Jul 01 '22 at 05:34