The OP’s main question has already been resolved in the comments. The following is a long comment addressing OP’s follow-up question in comments, namely, we show that under AC, we can construct a function $f: \mathbb{R} \to \mathbb{R}$ s.t. $f(A) = \mathbb{R}$ for any $A \subset \mathbb{R}$ with positive measure.
Let $\mathfrak{c}$ be the cardinality of $\mathbb{R}$. Let $(r_\lambda)_{\lambda < \mathfrak{c}}$ be an enumeration of $\mathbb{R}$ in which every element appears $\mathfrak{c}$ many times. (This can be arranged, for example, by fixing a bijection $\iota: \mathfrak{c} \to \mathbb{R} \times \mathbb{R}$ and let $r_\lambda$ be the first coordinate of $\iota(\lambda)$.) Recall that the collection of all Borel sets has cardinality $\mathfrak{c}$, so the same holds for the collection of all Borel sets with positive measure. Let $(A_\lambda)_{\lambda < \mathfrak{c}}$ be an enumeration of such sets.
We now construct, through transfinite induction, a function $f_\lambda$ for each $\lambda < \mathfrak{c}$ s.t., for each $\lambda$, $f_\lambda$ is a function defined on a subset of $\mathbb{R}$ of cardinality smaller than or equal to $\max\{\aleph_0, |\lambda|\}$, and has codomain $\mathbb{R}$. Furthermore, $f_\lambda \subset f_\kappa$ whenever $\lambda \leq \kappa < \mathfrak{c}$.
Assume for some $\lambda < \mathfrak{c}$, all $f_\kappa$ for $\kappa < \lambda$ have been constructed. We observe that $\cup_{\kappa < \lambda} f_\kappa$ is a function defined on a subset of $\mathbb{R}$ with cardinality smaller than or equal to $\max\{\aleph_0, |\lambda|\} < \mathfrak{c}$. Recall that any Borel set of positive measure must have cardinality $\mathfrak{c}$, whence we may choose (by another transfinite induction), $x_{\lambda, \eta} \in A_\eta$ for each $\eta \leq \lambda$, s.t. $x_{\lambda, \eta_1} \neq x_{\lambda, \eta_2}$ for any $\eta_1 \neq \eta_2$ and $x_{\lambda, \eta} \notin \text{dom}(\cup_{\kappa < \lambda} f_\kappa)$. Let,
$$f_\lambda = (\bigcup_{\kappa < \lambda} f_\kappa) \cup \{(x_{\lambda, \eta}, r_\lambda): \eta \leq \lambda\}$$
Now, having finished the inductive construction, we note that $\cup_{\lambda < \mathfrak{c}} f_\lambda$ is a partially defined function from $\mathbb{R}$ to $\mathbb{R}$. Extend it arbitrarily to a fully defined function $f: \mathbb{R} \to \mathbb{R}$. We claim that $f$ satisfies the requisite property. Indeed, for any $A \subset \mathbb{R}$ of positive measure, it must contain a Borel set of positive measure as a subset, say, $A_\eta \subset A$ for some $\eta < \mathfrak{c}$. Let $r \in \mathbb{R}$. Since each real number appears in the enumeration $(r_\lambda)_{\lambda < \mathfrak{c}}$ $\mathfrak{c}$ many times, there exists $\kappa \geq \eta$ s.t. $r_\kappa = r$. But then by definition of $f_\kappa$, we see that $x_{\kappa, \eta} \in A_\eta$ and,
$$f(x_{\kappa, \eta}) = f_\kappa(x_{\kappa, \eta}) = r_\kappa = r$$
So $r \in f(A_\eta) \subset f(A)$. Since $r \in \mathbb{R}$ is arbitrary, this shows $f(A) = \mathbb{R}$.
Remark 1: The construction of such an $f$ necessarily depends on some AC. Indeed, any such $f$ satisfies the following strong non-measurability property: for any $K \subset \mathbb{R}$ that is neither empty nor the entire $\mathbb{R}$, $f^{-1}(K)$ cannot be measurable, regardless of whether $K$ is measurable or not. Indeed, if $f^{-1}(K)$ were to be measurable, then so is $\mathbb{R} \setminus f^{-1}(K) = f^{—1}(\mathbb{R} \setminus K)$. Thus, either $f^{-1}(K)$ or $f^{-1}(\mathbb{R} \setminus K)$ has to have positive measure, contradicting the assumption on $f$. However, if I’m not mistaken, the above proof only used AC to the extent that $\mathbb{R}$ is well-orderable. That is definitely far weaker than the full AC.
Remark 2: The above proof shows that, given any collection $\Sigma$ of at most $\mathfrak{c}$ subsets of $\mathbb{R}$, each of cardinality $\mathfrak{c}$, one may construct an $f: \mathbb{R} \to \mathbb{R}$ s.t. $f(A) = \mathbb{R}$ for all $A \in \Sigma$. Note that this stands in contrast to the case where we require $f(A) = \mathbb{R}$ for all $A \subset \mathbb{R}$ of cardinality $\mathfrak{c}$, which is impossible to arrange, as has been demonstrated in this answer to another question of the OP’s.
