Theorem. If there exists a free ultrafilter $\mathcal U$ on $\omega$, then there exists a non-measurable susbet of the real line.

Proof. First observe that the free ultrafilter $\mathcal U$ is a non-measurable subset of the Cantor cube $\{0,1\}^\omega$ with respect to the standard product measure on $\{0,1\}^\omega$ (here we identify subsets of $\omega$ with their characteristic functions, which are elements of the Cantor cube $\{0,1\}^\omega$).

Next, consider the standard Cantor ladder map $$f:\{0,1\}^\omega\to[0,1]\subset\mathbb R,\;\;f:(x_n)_{n\in\omega}\mapsto\sum_{n\in\omega}\frac{x_n}{2^{n+1}}$$ and observe that $f$ is measure-preserving in the sense that for any measurable subset $B\subset [0,1]$ the set $f^{-1}(B)$ is measurable in $\{0,1\}^\omega$ and has product measure equal to the Lebesgue measure of $B$.

Since the symmetric difference $\mathcal U\triangle f^{-1}(f(\mathcal U))$ is at most countable, the set $f^{-1}(f(\mathcal U))$ is not measurable in $\{0,1\}^\omega$ and hence its image $f(\mathcal U)$ is not measurable in $[0,1]$.

Theorem. If there exists a free ultrafilter $\mathcal U$ on $\omega$, then there exists a non-measurable subset of the real line.

Proof. First observe that the free ultrafilter $\mathcal U$ is a non-measurable subset of the Cantor cube $\{0,1\}^\omega$ with respect to the standard product measure on $\{0,1\}^\omega$ (here we identify subsets of $\omega$ with their characteristic functions, which are elements of the Cantor cube $\{0,1\}^\omega$).

Next, consider the standard Cantor ladder map $$f:\{0,1\}^\omega\to[0,1]\subset\mathbb R,\;\;f:(x_n)_{n\in\omega}\mapsto\sum_{n\in\omega}\frac{x_n}{2^{n+1}}$$ and observe that $f$ is measure-preserving in the sense that for any measurable subset $B\subset [0,1]$ the set $f^{-1}(B)$ is measurable in $\{0,1\}^\omega$ and has product measure equal to the Lebesgue measure of $B$.

Since the symmetric difference $\mathcal U\triangle f^{-1}(f(\mathcal U))$ is at most countable, the set $f^{-1}(f(\mathcal U))$ is not measurable in $\{0,1\}^\omega$ and hence its image $f(\mathcal U)$ is not measurable in $[0,1]$.