The graph of Borel measurable function whose range is a separable metrisable space

Question

If $Y$ is separable and $f : X \to Y$ is Borel measurable, then the graph of $f$ is Borel.

On page 14, Lemma 2.3, (iii) of this online note, given, $\{U_n\}_{n \in \Bbb N}$, a basis for the topology of a metrisable space $Y$, the graph of $f$ is:

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I can't see why this formula represents the graph of $f$. It seems to me it's an empty set.

Is $Y$ assumed to be metrisable? Otherwise, why would it even have a countable basis? — tomasz, May 29 '13 at 23:15
I spelled this out for the graph of a function with values in $\mathbb{R}^n$ here maybe it helps understanding where this comes from. — Martin, May 30 '13 at 07:22

score 2 · Accepted Answer · answered May 29 '13 at 23:39

2

You're right; it's empty, as $U_n$ cover $Y$.

It's a typo, most likely meant to be $$\bigcap_{n=0}^\infty \left( \{(x,y)\mid y\notin U_n\} \cup \{(x,y) \mid x\in f^{-1}[U_n]\} \right)$$

answered May 29 '13 at 23:39

tomasz

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The graph of Borel measurable function whose range is a separable metrisable space

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