Fall 2020 - Problem 1

Solution.

1. Observe that by continuity,

   $$F\p{x} = \limsup_{n\to\infty} f\p{x, n} = \inf_N \sup_{q > N; q \in \Q} f\p{x, q},$$

   so because supremums and infimums of a countable family of measurable functions is measurable, it follows that $F$ is Borel measurable.

2. Let $\mathcal{F}$ be the set of Borel sets which satisfy this condition. Since $0 = \chi_\emptyset$ and $1 = \chi_{\br{0,1}}$, we see that $\emptyset, \br{0, 1} \in \mathcal{F}$. If $E$ is an open set, then $\chi_E$ is lower semi-continuous, so there exists a sequence $\set{f_n}_n$ of continuous functions which converge pointwise to $\chi_E$, and if for $n \leq y < n + 1$ we set

   $$f\p{x, y} = \p{y - n}f_{n+1}\p{x} + \p{\p{n+1} - y} f_n\p{x}$$

   then we see that $f$ is piecewise continuous and continuous at each endpoint, since each $f_n$ was. Thus, $\mathcal{F}$ contains open sets. If $E \in \mathcal{F}$, then let $f\p{x, y}$ be the associated continuous function. Observe that if $g\p{x, y} = 1 - f\p{x, y}$, then $g$ has range in $\br{0, 1}$, is continuous, and satisfies

   $$\limsup_{y\to\infty} g\p{x, y} = 1 - \limsup_{y\to\infty} f\p{x, y} = 1 - \chi_E = \chi_{E^\comp},$$

   so $\mathcal{F}$ is closed under complements. Finally, suppose $\set{E_n}_n \subseteq \mathcal{F}$ with corresponding continuous functions $\set{f_n}_n$. Let $E = \bigcup_{k=1}^\infty E_k$ and $\set{g_n}_n$ be the sequence $f_1, f_1, f_2, f_1, f_2, f_3, \ldots$, i.e., each $f_k$ appears infinitely often. For each $k \geq 1$, by regularity of the Lebesgue, there exists an open set $U_k \supseteq E$ such that $m\p{U_k \setminus E} < \frac{1}{k}$. Let $\phi_k$ be a piecewise linear approximation to $\chi_{U_k}$ such that $\phi_k$ and $\chi_{U_k}$ only differ on an interval of length $\frac{1}{k}$ such that $\phi_k = 1$ on $U_k$. As before, set for $n \leq y < n + 1$

   $$f\p{x, y} = \p{y - n}g_{n+1}\p{x, y}\phi_{n+1}\p{x} + \p{\p{n+1} - y} g_n\p{x, y}\phi_n\p{x}$$

   as before. Then by construction, if $x \in E_k$, then because $f_k$ appears infinitely often, it follows that $\limsup_{y\to\infty} g\p{x, y} = 1$. If $x \notin E_k$ for every $k$, then there exists $n$ large enough so that $\phi_n\p{x} = 0$, so by construction, $f\p{x, y} = 0$ for all $y \geq n$, and so $\limsup_{y\to\infty} g\p{x, y} = \chi_E$ almost everywhere, as required.