Editor Anirban DasGupta writes:

Well done to Promit Ghosal (pictured below), at Columbia University, who sent a careful answer to this problem.

Promit Ghosal

The problem asked was the following. Take an integrable function $f$ on the unit interval, and for $x$ in the interval $[j/2^n,(j+1)/2^n)$, define $f_n(x)$ to be the average of $f$ over $[j/2^n,(j+1)/2^n)$.
Then, $f_n$ converges pointwise to $f$ for almost all $x$, and also converges to $f$ almost uniformly.

First, the specific partition $[j/2^n,(j+1)/2^n),j=0,1,\cdots ,2^n-1$ does not have much to do with the pointwise convergence; neither does the unit interval. We can conclude from real analysis that for any $f$ which is merely locally integrable, for almost all $x$, $\underset{h\to 0}{lim}\frac{1}{2h}{\int }_{x-h}^{x+h}|f(t)-f(x)|dt\to 0$ as $h\to 0$. A point $x$ satisfying this property is what analysts call a Lebesgue point of $f$. Almost all points $x$ are Lebesgue points for a locally integrable function. The result generalizes to higher dimensions, and to well shaped shrinking neighborhoods.

Now, how does one prove this probabilistically? Consider the family of sets ${\mathcal{A}}_n=\{[j/2^n,(j+1)/2^n),j=0,1,\cdots ,2^n-1\}$, and consider ${\mathcal{F}}_n$, the sigma-algebra generated by ${\mathcal{A}}_n$. Fix an $x$. Then in the probability space $([0,1],\mathcal{B},P)$, where $P$ is Lebesgue measure on $[0,1]$, the sequence $f_n(x)$ is a martingale for the filtration ${\mathcal{F}}_n$. This is a straight verification. It therefore follows from Doob’s martingale convergence theorem that $f_n(x)\stackrel{a.s.P}{\to }f(x)$. Moreover, the convergence is almost uniform by Egoroff’s theorem.