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 Does a positive-measure subset of the unit interval almost surely intersect a random translation of some countable subgroup of $\mathbb{R}$?. I'm asking it to help me solve my question Do the Birkhoff averages of a measurable stationary homogeneous Markov process in continuous time "converge to the right limit"?. ]


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.
 A: Yes your $F$ works. This is a result of B. Jessen, "On the Approximation of Lebesgue Integrals by Riemann Sums", Annals of Mathematics Second Series, Vol. 35, No. 2 (Apr., 1934), pp. 248-251.
I found this by googling "Strong Sweeping Out" and "Riemann sums". If you do the Riemann sum along multiples of $1/n$, however, things don't work at all as you take the limit as $n\to\infty$. You might look at the notes of Wierdl and Rosenblatt in the Cambridge University Press book for more info.
EDIT: So here are some more details of why your $F$ works. 
Define 
$$
\Lambda(f)(x)=\limsup_{n\to\infty}\frac{1}{2^n}\sum_{i=0}^{2^n-1}f(x+i/2^n).
$$
By Jessen's theorem, if $f$ is 1-periodic, then $\Lambda(f)(x)=\int_0^1 f$ for Lebesgue-a.e. $x$. 
For $q\in\mathbb Q$, define a map $\Pi_q$ sending a measurable function $f$ to the measurable 1-periodic function agreeing with $f$ on $[q,q+1)$ and let $\Lambda_q=\Lambda\circ\Pi_q$. By Jessen's theorem, $\Lambda_q(f)(x)=\int_q^{q+1}f$ for Lebesgue a.e. $x$. 
Let $A_q=\{x\colon \Lambda_q(f)(x)=\int_q^{q+1}f\}$. 
If $f$ is bounded, it's straightforward to see that $\Lambda_q(f)(x)\to\Lambda(f)(x)$ as $q\to x$ and also that $\int_q^{q+1}f\to\int_x^{x+1}f$ for all $x$. 
Now if $x\in\bigcap_q A_q$ (a set of full Lebesgue measure), we have $\Lambda(f)(x)=\lim_{q\to x}\Lambda_q(f)(x)
=\lim_{q\to x}\int_q^{q+1}f=\int_x^{x+1}f$. 
By the way, I believe that Jessen's theorem follows quickly from the backwards martingale convergence theorem if you take the $\sigma$-algebras to be $\mathcal F_n=\{B\in\mathcal B\colon B+2^{-n}=B\pmod 1\}$. 
