Exercise 7.13

(analambanomenos, fixed after kyp44 found errors in the original solution)
(a) Following the steps of the hint: (i) Show that some subsequence $\{f_{n_i}\}$ converges at all rational points $r$, say, to $f(r)$.

Let $\{r_m\}$ be an enumeration of the rational points of $R$. Since $\{f_n(r_1)\}\subset [0,1]$, a compact set, there is a subsequence $\{f_{1,n_j}(r_1)\}$ which converges to $f(r_1)$, by Theorem 2.41. Similarly, the $\{f_{1,n_j}(r_2)\}$ has a subsequence $\{f_{2,n_j}(r_2)\}$ which converges to $f(r_2)$. Continuing in this fashion, we get a sequence of subsequences $\{f_{m,n_j}\}$ such that each $\{f_{m,n_j}\}$ is a subsequence of $\{f_{m-1,n_j}\}$ and $f_{m,n_j}(r_m)$ converges to $f(r_m)$. Letting $f_{n_i}=f_{i,n_i}$, we get a subsequence of $\{f_n\}$ which converges on all rational points $r$ of $R$ to $f(r)$.

(ii) Define $f(x)$, for any $x\in R$, to be $\sup <!--\; FUNCTION\; APPLICATION\; -->f(r)$, the sup being taken over all rational points $r\le x$.

First we need to show that this new definition of $f(r)$ agrees with the previous one; let $f^{\text{∗}}(r)$ be the new definition of $f(r)$. It is clear from the definition that $f^{\text{∗}}(r)\ge f(r)$. Suppose $f^{\text{∗}}(r)>f(r)$, that is, there is a rational point $q<r$ and a real number $a$ such that $f(q)>a>f(r)$. Since $f(q)=\lim <!--\; FUNCTION\; APPLICATION\; -->f_{n_i}(q)$, and $f(r)=\lim <!--\; FUNCTION\; APPLICATION\; -->f_{n_i}(r)$, there is an integer $N$ such that for all $i\ge N$ we have $f_{n_i}(q)>a>f_{n_i}(r)$, but this contradicts the fact that these functions are monotonically increasing.

It is clear from the definition that $f$ is monotonically increasing.

(iii) Show that $f_{n_i}(x)\to f(x)$ at every $x$ at which $f$ is continuous.

Suppose $f$ is continuous at $x$ and let $𝜀>0$. Then there are rational numbers $r_1<x<r_2$ such that

Since $f_{n_i}(r_1)\to f(r_1)$ and $f_{n_i}(r_2)\to f(r_2)$, there is an integer $N$ such that for all $i>N$ we have

And since $f_{n_i}$ is monotonically increasing, we have for all $i>N$

That is, $|f(x)-f_{n_i}(x)|<𝜀$, so $f_{n_i}(x)\to f(x)$ as $i\to \infty$.

(iv) Show that a subsequence of $\{f_{n_i}\}$ converges at every point of discontinuity of $f$ since there are at most countably many such points.

Since $f$ is monotonically increasing, $f$ has at most countably many simple points of discontinuity by Theorems 4.29 and 4.30. Let $\{x_m\}$ be an enumeration of the points of discontinuity of $f$. Repeating the argument of step (i), there is a subsequence $\{f_{1,n_j}(x_1)\}$ of $\{f_{n_i}(x_1)$ which converges to a value $a$ between 0 and 1. We can continue this way for all the $x_m$, getting a sequence of subsequences $\{f_{m,n_j}\}$ of $\{f_{n_i}\}$ such that each $\{f_{m,n_j}\}$ is a subsequence of $\{f_{m-1,n_j}\}$ for which $f_{m,n_j}(x_m)$ converges. Letting $f_{n_k}=f_{k,n_k}$, we get a subsequence of $\{f_n\}$ which converges at all the points $\{x_m\}$. We can redefine $f(x_i)$ at these points to be those values.

Since we have already shown that $\{f_{n_k}\}$ converges at the points where $f$ is continuous, we have $f_{n_k}(x)$ converging to $f(x)$ for all real values $x$.

(b) Let $K$ be a compact subset of $R$ and let $𝜀>0$. It was shown in step (iii) of part (a) that, since $f$ is everywhere continuous, $f(x)=\lim <!--\; FUNCTION\; APPLICATION\; -->f_{n_k}(x)$ for all points $x$. Also, we are using the initial definition of $f$, so that is it monotonically increasing. For each $x\in K$ there is a ${\delta }_x>0$ such that for all $y\in [x-{\delta }_x,x+{\delta }_x]$ we have