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This is an textbook exercise: If $(\Omega, \mathcal{F}, \mu)$ is a complete metric space, then a function $f:\Omega \rightarrow \mathbb{R}$ is measurable iff sets of the form {$\omega: | f(\omega)\le t, t\in \mathbb{R}$} belong to $\mathcal{F}$. Here $\mathbb{R}$ is equipped with the sigma algebra of Lebesgue measurable sets, $\mathcal{M(\mathbb{R})}$.

My attempt: I want to show that $\forall M\in \mathcal{M(\mathbb{R})}$, $f^{-1}(M)\in \mathcal{F}$. I know that $M$ is the union of a Borel set $B$ and a null set $N$ in $\mathbb{R}$. So I just need to show $f^{-1}(N)\in \mathcal{F}$ for any null set $N$. But I don't know how to use the completeness of $\mathcal{F}$. If I can show there exists a Borel set $\tilde{B}$ such that $\tilde{B} \supset N$ meanwhile $f^{-1}(\tilde{B})$ is null in $(\Omega, \mathcal{F}, \mu)$ then it's done, but I'm not sure if this is possible.

I know that Lebesgue measurable function is almost a Borel measurable function but both definitions use preimage of Borel sets only, I'm not sure how to deal with preimage of a Lebesgue measurable set.

I have a feeling this should be pretty straightforward but I wasn't able to make progress. Any hint is appreciated.

TomG
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Indeed as commented by Sassatelli this claim should be false. There is a result that "The preimage of a Borel set under a continuous function is a Borel set. The preimage of Lebesgue measurable set under a continuous function may not be Lebesgue measurable. " (The source is here.)

So take a continuous function $f:(\mathbb{R},\mathcal{M}(\mathbb{R}),\lambda) \rightarrow (\mathbb{R},\mathcal{M}(\mathbb{R}),\lambda)$. The domain is a complete measure space, and since intervals are Borel the inverse image of intervals are Borel and hence in $\mathcal{M}(\mathbb{R})$, but the preimage of a general Lebesgue measurable set may not be in $\mathcal{M}(\mathbb{R})$.

TomG
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