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Prove the function $f:\mathbb{R}\rightarrow \mathbb{R} $ defined by $$f(x)= \begin{cases} 1/p, &\text{if $x=p/q$ is rational}\\ 0, &\text{if x is irrational}\\ \end{cases}$$ is Borel measurable.

I began trying to show {$f\geq a$} is a Borel set. So I considered different values for a to find the x's, and unioned all the possible x values, if that makes any sense. But then I got all of $\mathbb{R}$.

I have been doing this for a long time and I am now very confused.

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If $a \le 0$ then $\{f \ge a\} = \mathbb{R}$ which is Borel. If $a > 0$ then $\{f \ge a\} \subset \{f > 0\} = \mathbb{Q}$. But every subset of $\mathbb{Q}$ is countable and hence Borel.

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Great, thank you! –  Ashley Sep 21 '12 at 2:18

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