Sequence of monotone functions converging to a continuous limit, is the convergence uniform?

Question

I'm reading some extreme value theory and in particular regular variation in Resnick's 1987 book Extreme Values, Regular Variation, and Point Processes, and several times he has claimed uniform convergence of a sequence of functions because "monotone functions are converging pointwise to a continuous limit". I am finding this reasoning a little dubious.

EDIT: Thanks to some comments below I realized I was confused, I really want each $f_n$ to be a monotone function on $[a,b]$, so it isn't quite like Dini's theorem. Formally I am wondering:

Let $f_n:[a,b]\rightarrow\mathbb{R}$ be a sequence of non-decreasing functions converging pointwise to the continuous function $f$. Is the convergence uniform?

My thoughts:

I have been thinking that since the set of discontinuities of each function $f_n$ is at most countable, the set of discontinuities of the entire sequence is at most countable...something something? Any thoughts would be greatly appreciated!

Answer 1

Let $f:=\lim f_n$, and let $\varepsilon >0$ be given. We have to show that if $n$ is large enough, then $$\vert f_n(t)-f(t)\vert\leq\varepsilon\quad \hbox{for all $t\in[a,b]$}\, .$$

Since $f$ is assumed to be continuous, it is uniformly continuous on the compact interval $[a,b]$. So we may find a subdivision $a=t_0<t_1\dots <t_K=b$ of $[a,b]$ such that the oscillation of $f$ on each interval $[t_i,t_{i+1}]$ is less than $\varepsilon/2$.

Since $f_n(t_i)\to f(t_i)$ as $n\to\infty$ for $i=0,\dots ,K$, one can find $N$ such that if $n\geq N$, then $$\vert f_n(t_i)-f(t_i)\vert\leq\varepsilon/2\quad\hbox{for $i=0,\dots K$}\, .$$

Let us check that $\vert f_n(t)-f(t)\vert\leq \varepsilon$ for every $n\geq N$ and all $t\in [a,b]$.

Fix $n\geq N$, and take any $t\in [a,b]$. One can choose $i$ such that $t\in [t_i,t_{i+1}]$. Since the functions $f_n$ are non-decreasing we have $$f_n(t_i)\leq f_n(t)\leq f_n(t_{i+1})\, . $$ Since $\vert f_n(t_i)-f(t_i)\vert$ and $\vert f_n(t_{i+1})-f(t_{i+1})\vert$ are not greater than $\varepsilon/2$, it follows that $$f(t_i)-\varepsilon/2\leq f_n(t)\leq f(t_{i+1})+\varepsilon/2\, . $$ Moreover, since the oscillation of $f$ on $[t_i,t_{i+1}]$ is less than $\varepsilon/2$ and since $t\in [t_i,t_{i+1}]$, we also have $f(t_i)\geq f(t)-\varepsilon/2$ and $f(t_{i+1})\leq f(t)+\varepsilon/2$. Altogether, this gives $$f(t)-2\varepsilon/2\leq f_n(t)\leq f(t)+2\varepsilon/2\, , $$ which concludes the proof.

Answer 2

Definition. Let $X$ be a set and let $Y$ be a metric space. A sequence of mappings $u_n:X \to Y$ converges uniformly to $u:X \to Y$ if $$\lim_{n \to \infty} \sup_{x \in X} d(u_n(x),u(x)) = 0.$$

Proof. Assume $(u_n)$ does not converge uniformly to $u$. There exists $\varepsilon > 0$ and a sequence $(x_n) \subset [a,b]$ such that for all $n$, $$\varepsilon \le |u_n(x_n) - u(x_n)|.$$ Since $[a,b]$ is compact, $(x_n)$ contains a convergent subsequence. Assume, without loss of generality, that $(x_n)$ converges to $x_0 \in [a,b]$. By the Heine-Cantor theorem, $u:[a,b] \to \mathbb{R}$ is uniformly continuous. Thus, there exists $\delta > 0$ such that \begin{equation} x,y \in (x_0-\delta,x_0+\delta) \implies |u(x) - u(y)| < \frac{\varepsilon}{2}.\tag{1}\label{1} \end{equation} Since $(x_n)$ converges to $x_0 \in (x_0 - \delta, x_0 + \delta)$, there exists $N$ such that if $n \ge N$, then $x_n \in (x_0-\delta,x_0+\delta)$. Since $(u_n)$ converges to $u$ pointwise, there exists $K \ge N$ such that if $k \ge K$, then \begin{equation} |u_k(x_0+\delta) - u(x_0+\delta)| < \frac{\varepsilon}{2} \tag{2}\label{2} \end{equation} and \begin{equation} |u_k(x_0-\delta) - u(x_0-\delta)| < \frac{\varepsilon}{2}. \tag{3}\label{3} \end{equation} Fix $k \ge K$. We can deduce from $\eqref{2}$ and $\eqref{3}$ that $$u_k(x_0+\delta) < u(x_0 + \delta) + \frac{\varepsilon}{2}$$ and $$u(x_0-\delta) - \frac{\varepsilon}{2} < u_k(x_0-\delta).$$ Since $u_k$ is nondecreasing and $x_0 - \delta < x_k < x_0 + \delta$, $$u(x_0-\delta) - \frac{\varepsilon}{2} < u_k(x_0-\delta) \le u_k(x_k) \le u_k(x_0+\delta) < u(x_0 + \delta) + \frac{\varepsilon}{2}.$$ Subtract by $u(x_k)$ to get \begin{equation} u(x_0-\delta) - u(x_k) - \frac{\varepsilon}{2} < u_k(x_k) - u(x_k) < u(x_0 + \delta) - u(x_k) + \frac{\varepsilon}{2}.\tag{4}\label{4} \end{equation} By $\eqref{1}$, we know that $$-\frac{\varepsilon}{2} \le u(x_0-\delta) - u(x_k)$$ and $$u(x_0 + \delta) - u(x_k) \le \frac{\varepsilon}{2}.$$ Combining these deductions with $\eqref{4}$, we obtain $$-\varepsilon=-\frac{\varepsilon}{2} -\frac{\varepsilon}{2} < u_k(x_k) - u(x_k) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$ whence $$|u_k(x_k) - u(x_k)| < \varepsilon.$$ By assumption, $\varepsilon \le |u_k(x_k) - u(x_k)|$, so we get the contradiction $\varepsilon < \varepsilon$.