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Can the integral of a "generic" bounded measurable function be determined by its values on the rationals?

[This question is an extension of my question http://mathoverflow.net/questions/210316/does-a-positive-measure-subset-of-the-unit-interval-almost-surely-intersect-a-ra. I'm asking it to help me solve my question http://mathoverflow.net/questions/199540/do-the-birkhoff-averages-of-a-measurable-stationary-homogeneous-markov-process-i. ]

Does there exist a measurable function $F\colon [0,1]^{\mathbb{Q}\cap [0,1]} \to [0,1]$ with the property that for every measurable $f\colon [0,\infty) \to [0,1]$, for Lebesgue-almost all $\tau \geq 0$ we have $$ \int_\tau^{\tau+1} f(t) \, dt \ = \ F\left( \, (f(\tau+q))_{q \in \mathbb{Q}\cap [0,1]} \, \right) \ ? $$

Given all the results to the effect that "measurable objects are approximately topological objects", it seems highly intuitive that the answer should be yes. In fact, it even seems intuitive to me that the function $$ F\left( \, (r_q)_{q \in \mathbb{Q}\cap [0,1]} \, \right) \ := \ \limsup_{n \to \infty} \, \frac{1}{2^n} \sum_{k=0}^{2^n-1} r_{\!\frac{k}{2^n}} $$ should work, but I have not managed to prove it.