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Given the following definition of a measurable function:
Let $(X, \mathfrak{M})$ and $(X, \mathfrak{N})$ be two measure spaces. A function $f:X \longrightarrow Y$ is $(\mathfrak{M} , \mathfrak{N)}$-measurable if $f^{-1}(E) \in \mathfrak{M} \; \;$ $\forall E \in \mathfrak{N}$.

Is it true that if $f: \mathbb{R} \longrightarrow \mathbb{R} $ is $ (\mathcal{B}(\mathbb{R}), \mathcal{B}(\mathbb{R}))$-measurable, then it's also $ (\mathcal{L}(\mathbb{R}), \mathcal{L}(\mathbb{R}))$-measurable (with $\mathcal{B}(\mathbb{R})$ being the Borel $\sigma$-algebra and $\mathcal{L}(\mathbb{R})$ being the Lebesgue $\sigma$-algebra)?

My professor seems to take for granted that it's true, however, since $\mathcal{B}(\mathbb{R}) \subsetneq \mathcal{L}(\mathbb{R})$, I don't really understand why it's impossible that there is a subset $E \in \mathcal{L} (\mathbb{R}) \setminus \mathcal{B}(\mathbb{R}) $ such that $f^{-1}(E) \notin \mathcal{L} (\mathbb{R})$. I tried looking for a counterexample but I couldn't find one.

Mulstato
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    Mind that the standard terminology in real analyisis is to call $f:\Bbb R^n\to\Bbb R$ Lebesgue measurable if it is $(\mathcal L(\Bbb R^n),\mathcal B(\Bbb R))$-measurable in the measure-theoretic sense (as a special case of calling $\mathcal E$-measurable a rela- valued function $f:(X,\mathcal E)\to\Bbb R$ if it is $(\mathcal E,\mathcal B(\Bbb R))$-measurable). Are you sure that your professor is referring to $(\mathcal L(\Bbb R),\mathcal L(\Bbb R))$-measurability? – Sassatelli Giulio Jun 03 '24 at 14:50
  • @SassatelliGiulio Thank you very much for providing the counterexample I was looking for. You are probably right that she wasn't referring to $(\mathcal{L}(\mathbb{R}),\mathcal{L}(\mathbb{R}))$-measurability. – Mulstato Jun 03 '24 at 15:16
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    It's a common confusion. For instance, if you take a course in real analysis and another in probability back to back, you may end up wondering why $\text{measurable}\circ \text{measurable}=\text{measurable}$ is false in some discussions and true in other discussions: that is why. – Sassatelli Giulio Jun 03 '24 at 15:19

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This is not true. Let $A \subset \mathbb{R}$ be a fat Cantor set. Then $A$ is homeomorphic to the usual Cantor set $B \subset \mathbb{R}$ and let $g: A \to B$ be such a homeomorphism. Define,

$$f: \mathbb{R} \to \mathbb{R}, \; f(x) = \begin{cases} g(x), & \text{ if }x \in A\\ -1, & \text{ otherwise} \end{cases}$$

Since $g$ is continuous, it is easy to verify that $f$ is Borel-to-Borel measurable. However, it is not Lebesgue-to-Lebesgue measurable. Indeed, as $A$ has positive measure, there exists a non-Lebesgue-measurable $C \subset A$. But as $B$ is null, $g(C)$, being a subset of $B$, a null set, is Lebesgue-measurable. Hence, as $C = f^{-1}(g(C))$ is not Lebesgue-measurable, we have $f$ is not Lebesgue-to-Lebesgue measurable.

David Gao
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  • In fact, using Tietze extension theorem, one may choose $f$ to even be continuous but still not Lebesgue-to-Lebesgue measurable. – David Gao Jun 03 '24 at 14:55
  • As Sassatelli Giulio mentioned in comments though, it’s likely that your professor was referring to Lebesgue-to-Borel measurability when they said Lebesgue-measurable, as is standard in measure theory. This is also why I used the terminology “Lebesgue-to-Lebesgue measurable” in my answer, as to be clear what I was referring to is not the standard Lebesgue measurability. – David Gao Jun 03 '24 at 14:56