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.

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

• Thank you for drawing my attention to this. The paper that you cited seems to only consider the case that $f$ is 1-periodic (i.e. has a periodicity matching the width of the interval being integrated over). Is there any easy way to extend to the more general case? Or did you find a reference that gives the more general case? – Julian Newman Sep 15 '15 at 19:50
• My guess is that you could obtain the general case from this by fiddling around. – Anthony Quas Sep 15 '15 at 21:02
• @JulianNewman: Did you manage to obtain the result you wanted from the Jessen result? My guess is that you could create a finite set of periodic functions from your original $f$ and use them to show that for each $\epsilon$, you're within $\epsilon$ of the right thing on a set of full measure and then finish up by taking a countable intersection. – Anthony Quas Sep 16 '15 at 5:29
• I have to admit, I am really struggling to see how I can derive the general case as a corollary of the case where the periodicity coincides with (or is an integral factor of) the width of the range of integration. (Nonetheless, I might possibly be able to extend Jessen's method of proof to the more general situation; I have not gone through the paper yet.) – Julian Newman Sep 16 '15 at 16:01
• Thank you very much for this! (I had a vague feeling that continuity of $\Lambda_r(f)(x)$ in $r$ might be relevant, but couldn't see how to make a proof out of that.) In your argument for Jessen's theorem: taking $\mathbb{P}$ to be the Lebesgue measure on $[0,1]$, I can see that $\mathbb{E}[f|\mathcal{F}_n](x)$ is the $n$-th step in the Riemann-sum approximation of $\int_x^{x+1} f$; however, I cannot see why $\mathbb{E}[f|\mathcal{F}_\infty]$ is almost everywhere equal to the integral of $f$ (over a period). – Julian Newman Sep 17 '15 at 14:10