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 $\mathcal F_n=\{B\in\mathcal B\colon B+2^{-n}=B\pmod 1\}$.