Exercise · ZW-MEASURABLE-FUNCTION-008 · ★★☆ Intermediate

If f and g are measurable, why is \(f+g\) measurable?

Choose one option.

Show answer
  • A. Because \(\{f+g>a\}=\bigcup_{q\in\mathbb{Q}}\big(\{f>q\}\cap\{g>a-q\}\big)\), a countable union of measurable sets

Explanation

If \(f(x)+g(x)>a\), there is a rational \(q\) with \(f(x)>q\) and \(g(x)>a-q\), so \(\{f+g>a\}\) is a countable union of intersections of measurable sets. Neither the union of \(\{f>a\}\) and \(\{g>a\}\) nor continuity or boundedness justifies measurability.

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Review the concept: Measurable function

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.

\[f^{-1}(B)\in\mathcal{A}\ \forall B\in\mathcal{B}(\mathbb{R})\]

More practice problems: measurable function

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