An extended real value function is measurable (proof verification)

Question

For the second proof, it says "$\{x\in X: f_1(x) > \alpha\} = \{x \in X : f(x) >\alpha \} \cup B$". $f_1(x)$ cannot be equal to $-\infty$. Then, why is it union of $B$?

Answer 1

For e.g. $\alpha=-1$ we find that the following statements are equivalent:

• $f_1(x)>-1$
• $f_1(x)\in(-1,\infty)$
• $f_1(x)=0\vee f_1(x)\in(-1,0)\cup(0,\infty)$
• $f(x)\in\{-\infty,0,\infty\}\vee f(x)\in(-1,0)\cup(0,\infty)$
• $f(x)\in(-1,\infty]\vee f(x)=-\infty$
• $f(x)>-1\vee x\in B$

Answer 2

If $f(x) = -\infty$, then $f_1(x) = 0$.

Answer 3

If $\alpha < 0$ and $f_1(x) > \alpha$, then either $f(x)$ is finite and $f(x) = f_1(x) > \alpha$ or $f(x) = \infty$ or $f(x) = -\infty$. Thus, $$\{x: f_1(x) > \alpha\} \subset \{x: f(x) > \alpha\} \cup A \cup B.$$ But $A \subset \{x: f(x) > \alpha\}$, so this reduces to $$\{x: f_1(x) > \alpha\} \subset \{x: f(x) > \alpha\} \cup B.$$ Conversely, if $\alpha<0$ and either $f(x)> \alpha$ or $f(x) = -\infty$, then, in the former case $f(x)$ is either finite or = $\infty$, whence $f_1(x) = f(x)$ or $0$, respectively, whence $f_1(x)> \alpha$; and in the latter case $f_1(x)=0 > \alpha$. So, $$\{x: f_1(x) > \alpha\} \supset \{x: f(x) > \alpha\} \cup B.$$