# Uniform convergence of monotone bounded functions

Let $(\mathbb{R}, \mathcal{B}, \mu)$ be a measurable space. Let $f: \mathbb{R} \to \mathbb{R}$ be monotone non-constant measurable function and $\exists a \exists b\forall x:a < f(x) < b$. Let $f_n \rightarrow f$ almost everywhere. Is it true that $f_n$ converges uniformly to $f$?

It seems that if $f_n$ are not monotone then it is not true since we can take $f_n = f + wave_n$. Also there is a Glivenko-Cantelli Theorem that states that empirical distributions converges uniformly to the real distribution. But empirical distributions have pretty specific form so maybe there is an example when it is not true?

• What it means $f$ from $X$ to $\mathbb{R}$ to be monotone ? – Leandro May 11 '16 at 7:37
• @Leandro oh I see, let's say $f:\mathbb{R}\to \mathbb{R}$. – Jihad May 11 '16 at 7:44

The conclusion does not hold, a counter-example is $f_n(x) = \arctan(n x)$. The limit function $$f(x) = \begin{cases} \frac \pi 2 & (x > 0) \\ 0 & (x = 0) \\ -\frac \pi 2 & (x < 0) \\ \end{cases}$$ is not continuous, so the convergence cannot be uniform.
• How does it reconciled with Glivenko-Cantelli Theorem? It seems that such $f(x)$ with little manipulations can be converted to a distribution function, isn't it? – Jihad May 11 '16 at 9:28