Oct 6, 2016

# Student Puzzle 15: solution

Editor Anirban DasGupta writes:

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

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 $x$, $\lim_{h \to 0}\,\frac{1}{2h}\,\int_{x-h}^{x+h}\, |f(t)-f(x)|dt \to 0$ 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.

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