Exercise · ZW-MEASURABLE-FUNCTION-007 · ★★☆ Intermediate

Let \(f_n(x)=x^n\) on \([0,1]\). What is the pointwise limit \(f\) and why is it measurable?

Choose one option.

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  • B. \(f=\chi_{\{1\}}\), because it is a pointwise limit of measurable functions

Explanation

For x<1 the limit is 0 and at x=1 it is 1, so the limit is the indicator of {1}. It is measurable because pointwise limits of measurable functions are measurable.

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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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