Let \(f\) be measurable and \(g(x)=f(x)^2\). Prove that \(g\) is measurable, stating which set you check for \(a<0\) and for \(a\ge 0\). What is the correct conclusion?
Choose one option.
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C. For \(a<0\) the set \(\{g>a\}\) is all of X; for \(a\ge0\) it is \(\{f>\sqrt a\}\cup\{f<-\sqrt a\}\), which is measurable; hence g is measurable
For a<0 one has {g>a}=X, and for a≥0 the condition g>a is equivalent to |f|>√a, that is f>√a or f<-√a, both measurable. The option using only {f>a} ignores the negative values of f.
A function f: X → ℝ is measurable with respect to a σ-algebra \(\mathcal{A}\) on X if the preimage \(f^{-1}(B)\) belongs to \(\mathcal{A}\) for every Borel set B of ℝ.