pointwise convergence and monotony implies uniformly convergence Let $(f_n)_n$ be a sequence of functions, $f_n:\mathbb{R}_+\rightarrow\mathbb{R}$
and suppose that for all $n$ the function $f_n$ is monotone decreasing (which means that for $u<v$ we have $f_n(u)\geq f_n(v)$).
Furthermore, assume that for all $x\geq 0$ we have $f_n(x)\rightarrow f(x)$ and that the limit function $f$ is monotone decreasing, too, and continuous.

Claim: $f_n\rightarrow f$ uniform on compact sets, i.e., for all $T>0$,
  $$\sup_{s\leq T}\vert f_n(s)-f(s)\vert\rightarrow 0.$$

How do I prove this?
 A: Let $T>0$ and $\epsilon>0$. Since the limit function $f$ is continuos in $\mathbb{R}_+$, it is uniformly continuous on the compact $[0,T]$, so there exists $\delta>0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$ for $0\leq x,y\leq T$. Consider a partition $0=x_0<x_1<\cdots<x_n=T$ with $|x_{i+1}-x_i|<\delta$. Now, $f_n\to f$ pointwise, so there exists $N$ such that $|f_n(x_i)-f(x_i)|<\epsilon$ for all $i$ whenever $n\geq N$. If $n\geq N$.
Now, let $n\geq N$ and $x\in [0,T]$ be arbitrary. Choose $i$ such that $x_i\leq x\leq x_{i+1}$. Then
$$f(x_i)-\epsilon<f_n(x_i)\leq f_n(x)\leq f_n(x_{i+1})<f(x_{i+1})+\epsilon<f(x_i)+2\epsilon$$
(the first and fourth inequalities follow from the choice of $N$, and the last from the choice of the partition $\left\{x_i\right\}$).
But also, $f(x_i)-\epsilon<f(x)<f(x_{i+1})+\epsilon<f(x_i)+2\epsilon$. Thus, both $f(x)$ and $f_n(x)$ belong to the interval $(f(x_i)-\epsilon,f(x_i)+2\epsilon)$ of length $3\epsilon$, so $|f_n(x)-f(x)|<3\epsilon$ for all $x\in[0,T]$ and all $n\geq N$. This shows that $f_n\to f$ uniformly on $[0,T]$, for all $T>0$.
