$f(X)$ measurable, but $f$ not measurable

Let $(\Omega,F,P)$ be a probability space. Suppose $f\colon \mathbb R\to\mathbb R$ is some function such that $f(X)$ is measurable for every real valued random variable $X$. I am curious if $f(X)$ is necessarily $\sigma(X)$-measurable. I tried to conclude with Doob-Dynkin lemma but to do we would need $f$ to be $\mathcal{B}(\mathbb R)$ measurable. Does someone has an idea or is this is false in general?

It is possible that $f(X)$ is measurable for every random variable $X$ but $f$ is not Borel. For example, if $\mathcal{F}=\mathcal{P}(\Omega)$, then $f(X)$ is always measurable regardless of $f$ and $X$. Obviously, we cannot infer that $f$ is a Borel function.

To complement the previous answer. Let $\left(\Omega,\mathcal{F},\mathbb{P}\right)=\left(\mathbb{R},2^{\mathbb{R}},\delta_{0}\right)$. Let $f\,:\,\mathbb{R}\longrightarrow \mathbb{R}$ be the indicator of some non-Borel set $S$ and note that $f(X)$ is measurable for any $X$ (as mentioned in Danny's post). Let $X$ be the identity from $\mathbb{R}$ to $\mathbb{R}$, the latter endowed with $\mathcal{B}\left(\mathbb{R}\right)$. Then, $X^{-1}\left(f^{-1}\left(\left\{1\right\}\right)\right)=S\notin \sigma(X)=\mathcal{B}(\mathbb{R})$.