The answer is "no". 

As it was noted by Martin Sleziak in order to preserve measurability, your function has to satify Luzin N property. Let me show that this is not the case. That any continuous, nowhere differentiable function $f$ maps a set of zero measure maps to a set of positive measure.

Note that for fixed $L<\infty$ and almost any $y\in f(\mathbb{I})$ there is an interval $[p,q]\subset \mathbb{I}$ such that $y\in f([p,q])$ and $$\lambda(f([p,q]))>L\cdot\lambda([p,q]),$$ so you are in the position to apply Vitali covering theorem.
 
Fix $\varepsilon>0$.
Applying Vitali covering theorem, you can pass to a closed subset $S\subset\mathbb{I}$ formed by a finite collection of closed intervals such that 
$$\lambda(f(S))>(1-\varepsilon)\cdot\lambda(f(\mathbb{I}))\quad\text{and}\quad \lambda(S)<\tfrac12\cdot \lambda(\mathbb{I}),$$
where $\lambda$ denotes Lebesgue measure.

It remains to iterate this construction for a sequence $\varepsilon_n\to 0$ such that 
$$\prod_n(1-\varepsilon_n)>0.$$