Sigma algebra unequal to its universal completion

Question

Let $(\Omega, \mathcal{F}, P)$ be a probability space. The completion of $\mathcal{F}$ w.r.t. $P$ is the smallest sigma-algebra that contains $\mathcal{F}$ and all subsets of $P$-nullsets in $\mathcal{F}$. The universal completion is defined as the intersection of all universal completions of $\mathcal{F}$ w.r.t all probability measures on $(\Omega, \mathcal{F})$. What is a (as minimal as possible) working example of a sigma-algebra that differs from its universal completion.

Answer 1

Example: Let $\Omega$ be the space of continuous functions mapping $[0,1]$ to $\Bbb R$. When endowed with the uniform norm, $\Omega$ is a Banach space. The Borel $\sigma$-algebra on $\Omega$ coincides with the $\sigma$-algebra generated by the projection maps $\Omega\ni f\to f(t)\in\Bbb R$, $t\in [0,1]$. It is known ["The set of continuous nowhere differentiable functions," Pacific J. Math. 83 (1979) 199–205] that the subset $M\subset\Omega$ comprising the nowhere differentiable elements of $\Omega$ is not a Borel set but is co-analytic (that is, the complement of an analytic subset of $\Omega$). Thus, because analytic sets are universally measurable, $M$ is universally measurable.