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To be specific, $\phi(t)={1\over t} $ (if $t \neq 0$) or $0$ (if $t=0$).

Please check the process I prove is correct:

I used the definition of Borel measurable or Borel function which is

$\phi^{-1}([-\infty,t])\in B:$ Borel set

Since $\phi^{-1}([-\infty,t])=(0,1/t]\in B$, $\phi$ is Borel function.

Am I right?

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You are right in $\phi$ being a Borel function, but you made come mistakes computing the inverse images:

For $t < 0$, we have \begin{align*} x \in \phi^{-1}([-\infty, t]) &\iff \phi(x) \le t \\ &\iff \frac 1x \le t\\ &\iff x \ge \frac 1t \land x < 0 \end{align*} Hence $\phi^{-1}([-\infty, t]) = [\frac 1t, 0)$, which is a Borel set. For $t = 0$, we have $$ \phi^{-1}([-\infty, 0]) = \phi^{-1}([-\infty,0)) \cup\{0\} = (-\infty, 0] $$ and for $t > 0$, we have $$ \phi^{-1}([-\infty, t]) = \phi^{-1}([-\infty,0]) \cup \phi^{-1}((0,t]) = (-\infty, 0] \cup \left[\frac 1t, \infty\right) $$

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  • $\begingroup$ It really helped. You are awesome!! Thanks. $\endgroup$ – SW PARK Apr 2 '15 at 13:46

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