Let $\sigma(f)$ be the smallest $\sigma$-algebra on $[0,1]$ which makes $f$ measurable. I claim that $\overline{P_f} = L^2(I, \sigma(f), m)$, the space of all square-integrable $\sigma(f)$-measurable functions. In particular, $P_f$ is dense iff the completion of $\sigma(f)$ under $m$ contains all the Borel sets (equivalently, all the Lebesgue measurable sets).

One direction is easy: it is clear that $L^2(I, \sigma(f), m)$ is a closed subspace of $L^2(I, \mathcal{B}, m)$ (pass to an a.e.-convergent subsequence). And $P_f$ is clearly contained in the former, since any Borel function of $f$ remains $\sigma(f)$-measurable.

For the other direction, we use the multiplicative system theorem. This can be seen as a functional version of the monotone class theorem, or a measurable version of the Stone–Weierstrass theorem.

**Theorem.** Suppose $H$ is a vector space of real-valued bounded measurable functions on some measurable space $X$, which contains the constants and is closed under bounded pointwise convergence of sequences (i.e. if $f_n \in H$, $f_n \to f$ pointwise, and $|f_n| \le C$ for all $n$, then $f \in H$). Suppose $M \subset H$ is closed under pointwise multiplication, and let $\mathcal{G}$ be the $\sigma$-algebra generated by $M$ (i.e. the smallest $\sigma$-algebra on $X$ that makes all functions from $M$ measurable). Then $H$ contains all bounded $\mathcal{G}$-measurable functions.

You can find this (in a slightly stronger form) in Dellacherie, *Probabilities and Potential*, page 14, or in Janson, *Gaussian Hilbert Spaces*, appendix A. The proof is elementary and makes use of the classical Weierstrass approximation theorem. Dellacherie doesn't give any background, but I've heard the result credited to Dynkin.

Let $M = P_f$, which is obviously closed under multiplication.

Let $H = \overline{P_f} \cap L^\infty$ consist of all bounded measurable functions which are in the $L^2$-closure of $P_f$. Since $P_f$, its closure, and $L^\infty$ are all vector spaces, $H$ is a vector space. $H$ contains constants because $P_f$ did. And if $f_n \in H$ and $f_n \to f$ boundedly, then we also have $f_n \to f$ in $L^2$ by the dominated convergence theorem (using $C$ as the dominating function, since we are in a finite measure space). So $H$ satisfies all hypotheses.

Clearly $\mathcal{G} = \sigma(M) \supset \sigma(f)$, so the theorem asserts that $H$ contains all bounded $\sigma(f)$-measurable functions. Now for any $g \in L^2(I, \sigma(f), m)$, let $g_n = \max(-n, \min(g, n))$ be a truncation of $g$, which is bounded and still $\sigma(f)$-measurable. Then $g_n \in H \subset \overline{P_f}$ and $g_n \to g$ in $L^2$, so $g \in \overline{P_f}$.