Proving this piecewise function is measurable.
Solution 1:
I see no disastrous problems with this proof, it looks good. I advocate one change, however. I would handle the case $\alpha=0$ separately:
Suppose, $\alpha = 0$. Then
$$\{x: f(x) < 0\} = \{x: f(x) < 0,\, g(x)\not=0\} = \{x: \frac{1}{g(x)} < 0\}= \{x: g(x) < 0\}.$$
The way you have it now, we get a division by zero:
Suppose, $\alpha \le 0$. Then,
$$\{x \in \mathbb{R}: f(x) < 0\} = \{x: \frac{1}{g(x)} < 0\} = \{x: g(x) > \frac{1}{0} \}.$$
In fact, even writing $1/0 = \infty$, I think this last step is no good.