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
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.
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