Past exam of the probability, statistics and options I course of the University of Oxford / 2010 / D / 1 (D1)

Use the countable identityIf $f(x)>b$, some $n$ has $1/n\le f(x)-b$, proving the nontrivial inclusion. Each preimage on the right is measurable by statistical hypothesis, so closure of a sigma-algebra under countable unions proves the strict-superlevel preimage is measurable.

For each real $a$,The final strict-superlevel set is measurable by part 1 and the given criterion for measurable functions. A complement relative to the measurable domain is measurable, so $-g$ is measurable.

If $f$ is measurable, then for $a\le 0$, $\{f^{+}\ge a\}=E$, and for $a>0$, $\{f^{+}\ge a\}=\{f\ge a\}$. Thus the given superlevel criterion proves $f^{+}$ measurable. Part 2 makes $-f$ measurable, and the same argument applied to $(-f{)}^{+}$ proves $f^{-}$ measurable.

Conversely $f=f^{+}-f^{-}$. If both parts are measurable, part 2 and the permitted closure under addition prove $f$ measurable. Therefore $f$ is measurable exactly when its positive and negative parts are measurable.

Since $|f|=f^{+}+f^{-}$, the previous part and closure under addition prove measurability of the absolute value.

The converse fails. Choose a nonmeasurable subset $N$ of $[0,1]$ and define $f=1$ on $N$, $f=-1$ on its complement. Then $|f|\equiv 1$ is measurable, whereas $f^{-1}([0,\infty ))=N$ is not. Thus measurability of $|f|$ does not imply measurability of $f$.