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Let $f:\mathbb R^n \to \mathbb R^n$ be a measurable mapping (assuming Lebesgue measure).

What we could say about an image of Borel set $B$. Is the set $f(B)$ measurable?

Nikita Evseev
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1 Answers1

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Let $C \subset \Bbb{R}$ denote the Cantor set, and $V \subset \Bbb{R}$ denote the Vitali set. It is well known that $C$ is Borel of measure zero, $V$ is not measurable, and that they both have the cardinality of continuum. Let $f:C \longrightarrow V$ be any bijection. Extend $f$ to a function $f': \Bbb{R} \longrightarrow \Bbb{R}$ setting $f'(x)=0$ if $x \notin C$.

Then $f'$ is measurable (check it), and maps the Borel set $C$ into the non-measurable set $V$.

Crostul
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    I feel like some clarification is in order. Let $\mathcal{B}$ be the Borel $\sigma$-algebra and $\bar{\mathcal{B}}$ the Lebesgue $\sigma$-algebra (i.e. the completion of $\mathcal{B}$ with respect to the Lebesgue measure). Is the question asking about $(\bar{\mathcal{B}},\bar{\mathcal{B}})$-measurable functions? This would be a bit strange, as even continuous functions don't have to be $(\bar{\mathcal{B}},\bar{\mathcal{B}})$-measurable, as shown in https://math.stackexchange.com/questions/1516137. – Julian Newman Oct 31 '19 at 16:55
  • What would make more sense is for the question to be asking about $(\bar{\mathcal{B}},\mathcal{B})$-measurable functions (or equivalently, functions that are Lebesgue-almost everywhere equal to a $(\mathcal{B},\mathcal{B})$-measurable function). In this case, as the answer given shows, the image of a $\mathcal{B}$-measurable set need not be $\bar{\mathcal{B}}$-measurable. But, if we are talking about $(\mathcal{B},\mathcal{B})$-measurable functions, then it is the case that the image of every $\mathcal{B}$-measurable set is $\bar{\mathcal{B}}$-measurable. – Julian Newman Oct 31 '19 at 16:59