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Continuous functions on Euclidean spaces are Borel measurable

Statement

Assume the Axiom of Countable Choice. Let $n, m \geq 1$ . Every continuous map $f : R^{n} \to R^{m}$ is Borel measurable in the sense of Borel measurable and Lebesgue measurable functions on $R^{n}$ .

Statement references

Borel measurable and Lebesgue measurable functions on $R^{n}$

Facts & Assumptions

Given: The Axiom of Countable Choice, natural numbers $n, m \geq 1$ , and a continuous function $f : R^{n} \to R^{m}$ .

[L1]

A continuous map has Borel preimages of Borel sets. (A continuous map has Borel preimages of Borel sets)

Proof

technique · direct

1.1L1

Let $B \subseteq R^{m}$ be Borel. By [L1], the preimage $f^{- 1} (B)$ is a Borel subset of $R^{n}$ .

2.1step 1.1∎

By the definition of Borel measurability on Euclidean spaces, step 1.1 says [step 1.1] exactly that $f$ is Borel measurable.

Depends on

Dependency tree · 2 levels

Sources

Sheldon Axler, Measure, Integration and Real Analysis, Section 2B