Proof: $f: \Bbb R \to \Bbb R^2 , x \mapsto (x,0)$ is $ \mathcal B^1 - \mathcal B^2$ measurable but not $\mathcal L^1 - \mathcal L^2$ measurable

Question

I have to prove the above. I know that:

$f$ is $\mathcal B - \mathcal B^2$ measurable $\Leftrightarrow$ $f^{-1} (B) \in \mathcal B^1 \: \:\forall B \in \mathcal B^2$

(for $\mathcal L^1 - \mathcal L^2$ measurable just swap the $\mathcal B$ with $\mathcal L$). I don't really see why this should be violated for $\mathcal L$.

Anny tipps or ideas? Thanks in advance!

Answer 1

I guess $\mathcal{L}$ means the Lebesgue sigma algebra so take $A\not\in \mathcal{L}^1$ then $A = f^{-1}((A,0))$ but $(A,0) \in \mathcal{L}^2$ because it's a nullset.