 4:17 AM
@user193319 Define $h(x)=f(x)-g(x)$. Then $\{x \mid f(x) < g(x) \} =\{x \mid f(x)-g(x) < 0 \}=\{x \mid h(x) < 0 \}=h^{-1}(-\infty,0)$ which is measurable since h is measurable
In this question
1  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 fun...

in the answer actually, why for $t=0$ we have $\phi^{-1}([-\infty, 0]) = \phi^{-1}([-\infty,0)) \cup\{0\} = (-\infty, 0]$ ? why to split the preimage and {0}? and why $\frac{1}{x}<0\cup{0}=(-infty,0]$? Could you explain please @MartinSleziak ? 5:02 AM
@user193319 I think your seems ok. Of course, one has to be a bit careful with $f-g$ with functions that can have infinite values. But if we define $\infty-\infty=0$ and $-\infty-(-\infty)=0$, everything should work ok.
@Isa Well, dealing with $0$ separately seems natural here, since the function is defined differently for zero.
$\phi^{-1}(-\infty,0]$ contains only zero and negative numbers - just look at the graph. Or notice that for $x>0$ you have $\frac1x>0$.