Spring 2014 - Problem 3

Solution.

1. For $n \geq 1$, let $\set{t_k^{\p{n}}}_{1 \leq k \leq n}$ be a partition of $\br{a, b}$. Then set $I_k^{\p{n}} = \pco{t_{k-1}^{\p{n}}, t_k^{\p{n}}}$ and

   $$\begin{aligned} L_n &= \sum_{k=1}^n \chi_{I_k^{\p{n}}} \inf_{t \in I_k^{\p{n}}} f\p{t} \\ U_n &= \sum_{k=1}^n \chi_{I_k^{\p{n}}} \sup_{t \in I_k^{\p{n}}} f\p{t}. \end{aligned}$$

   These are Borel functions since each $\chi_{I_k^{\p{n}}}$ is Borel. Observe that if $x$ is a point of continuity for $f$, then there exists (a unique) $k \geq 1$ such that $x \in I_k^{\p{n}}$, so

   $$U_n\p{x} - L_n\p{x} = \sup_{t \in I_k^{\p{n}}} f\p{t} - \inf_{t \in I_k^{\p{n}}} f\p{t} \xrightarrow{n\to\infty} 0.$$

   Conversely, if $U_n\p{x} - L_n\p{x} \to 0$ for some $x \in \br{a, b}$, then for all $\epsilon > 0$, there exists $N \in \N$ such that whenever $n \geq N$, we have

   $$\sup_{t \in I_{k_n}^{\p{n}}} f\p{t} - \inf_{t \in I_{k_n}^{\p{n}}} f\p{t} \leq \epsilon,$$

   where $k_n$ is such that $x \in I_{k_n}^{\p{n}}$ for all $n \geq 1$. Hence, if $\delta = d\p{x, \p{I_{k_N}^{\p{N}}}^\comp}$ and $\abs{x - y} \leq \frac{\delta}{2}$, we see $y \in I_{k_N}^{\p{N}}$ as well. Consequently,

   $$\abs{f\p{x} - f\p{y}} \leq \sup_{t \in I_{k_N}^{\p{N}}} f\p{t} - \inf_{t \in I_{k_N}^{\p{N}}} f\p{t} \leq \epsilon,$$

   so $x$ is a point of continuity if and only if $F\p{x} \coloneqq \lim_{n\to\infty} \p{U_n\p{x} - L_n\p{x}} = 0$. Since each $U_n - L_n$ is Borel, $F$ is Borel as a limit of these and so

   $$\tag{$*$} \set{x \in \br{a, b} \mid f \text{ continuous at } x} = F^{-1}\p{\set{0}}$$

   is Borel.

2. If $f$ is Riemann integrable, let $R\p{f}$ denote the Riemann integral of $f$ over $\br{a, b}$. Then by definition, the Darboux sums must converge to $R\p{f}$, so if $m$ denotes Lebesgue measure,

   $$R\p{f} = \lim_{n\to\infty} \int_{\br{a,b}} L_n \,\diff{m} = \lim_{n\to\infty} \int_{\br{a,b}} U_n \,\diff{m}.$$

   Since $f$ is bounded by some $M > 0$, $\abs{U_n - L_n} \leq M \in L^1\p{\br{a,b}}$, so by dominated convergence,

   $$\int_{\br{a,b}} F \,\diff{m} = \lim_{n\to\infty} \int_{\br{a,b}} U_n - L_n \,\diff{m} = 0.$$

   But $U_n - L_n \geq 0$, so $F \geq 0 \implies F = 0$ almost everywhere. From ($*$), we immediately see that $f$ is continuous almost everywhere.

   Conversely, suppose $f$ is continuous almost everywhere, so ($*$) implies that $F = 0$ almost everywhere. Thus,

   $$0 = \int_{\br{a,b}} F \,\diff{m} = \lim_{n\to\infty} \int_{\br{a,b}} U_n - L_n \,\diff{m} = \lim_{n\to\infty} \p{\int_{\br{a,b}} U_n \,\diff{m} - \int_{\br{a,b}} L_n \,\diff{m}}$$

   by dominated convergence again. Hence, we see that all Darboux sums converge to the same number, namely $R\p{f}$, so $f$ is Riemann integrable.