Showing a sequence of functions converges uniformly Let $\{f_n\}$ be a decreasing sequence of continuous functions with $f_n:[0,1]\rightarrow\mathbb{R}$, with the property that there exists an $M\in(0,1)$ such that $|f_n|\leq M$. Furthermore, $f_n\rightarrow f$ pointwise, where $f$ is also continuous. Prove that the convergence is actually uniform. 
This is what I tried: Let $\epsilon>0$ be given. Then as $f$ is continuous on $[0,1]$ it is uniformly continuous. Let $\delta$ be such that $\delta$-close points map to $\frac{\epsilon}{3}$-close points. Partition $[0,1]$ into intervals, all whose length are less than $\delta$. Let $I_1,...,I_k$ be such intervals. Let $y_i$ be the midpoint of $I_i$. Let $N_i$ be such that $|f_n(y_i)-f(y_i)|<\epsilon/3$. Let $N$ be the max of all $N_i$. 
Let $x\in [0,1]$ be arbitrary. Then $x\in I_k$ for some $k$. Hence, if $n\geq N$, then: $$|f_n(x)-f(x)|\leq |f_n(x)-f_n(y_k)|+|f_n(y_k)+f(y_k)|+|f(y_k)+f(x)|$$By choice of $N$, we have that the middle term is less that $\epsilon/3$, by continuity of $f$, the last summand is also less than $\epsilon/3$, so I was wondering how can I finish it?
Thanks
 A: As mentioned by Salech Alhasov, this is Dini's Theorem. 
Here is a proof closer in spirit to the OP's approach:
The result will follow if we can show that the sequence with terms $g_n=f_n-f$ converges uniformly to 0. Note that  $(g_n)$ is a decreasing  sequence of continuous functions converging pointwise to 0
Fix $\epsilon>0$.
For each $t\in I=[0,1]$, choose 
$g_{m_t}$ so that $0\le g_{m_t}(t)<{\epsilon\over2}$. Since $g_{m_t}$ is continuous at $t$, there is a $\delta(t)>0$ so that 
$$0\le g_{m_t}(x)<\textstyle{\epsilon }\quad\text{ for all}\quad |x-t|\le \delta(t).$$ 
Now, since $[0,1]$ is compact, there is a finite set $\{t_1,\ldots, t_k\}\subset I$ such that $\bigl\{ B_{t_1}(\delta(t_1)),\ldots,  B_{t_k}(\delta(t_k))\bigr\}$ covers $[0,1]$. 
Let $m>\max\limits_{1\le i\le k}\{m_{t_i}\}$ and $x\in I$. Then $x\in
 B_{t_i}(\delta(t_i))$ for some $i$ and thus $\phantom{\epsilon\over2}$
$$
0\le g_m(x)\le g_{m_{t_i}}(x)<\epsilon.
$$
Thus $(g_n)$ converges uniformly to $0$ on $[0,1]$.
A: Actually, the assumption that $|f_n|\leq M$ is redundant, since it is given that each $f_n$ is continuous (hence bounded on $[0, 1]$), and the sequence is decreasing (i.e. I presume you mean that for each $m > n$ and each $x\in [0, 1]$, $f_m(x)\leq f_n(x)$. The assumption that the upper bound (that is, $M$) is in $(0, 1)$ is also not needed. We'll only use the fact that each $f_n$ is continuous, the interval $[0,1]$ is compact, the sequence $\{f_n\}$ is decreasing, and $f_n\rightarrow f$ pointwise on $[0, 1]$.
Observe that since $\{f_n\}$ is decreasing in the sense assumed above, for each $m > n$ and $x\in[0,1]$, we have $f(x)\leq f_m(x)\leq f_n(x)$. Given $\epsilon > 0$ and $n\in\mathbb{N}$, set
$$
O_n := \{x\in[0,1]: f_n(x) - f(x) < \epsilon\}.
$$
By continuity of $f_n$ and $f$ (which is assumed in the original problem), for each $n$, $O_n$ is open. Let $C_n = O_n^c$. Then $C_n$ is compact and, moreover, by pointwise convergence of $\{f_n\}$ to $f$ and the fact that $\{f_n\}$ is decreasing, the sequence $\{C_n\}_{n\in\mathbb{N}}$ forms a nested sequence of compact sets; that is, $C_{n+1}\subset C_n$ for each $n\geq 1$. Now, if for each $n$, $C_n\neq \emptyset$, then
$$
C:= \bigcap_{n\in\mathbb{N}}C_n
$$
is nonempty, and for each $x\in C$, $f_n(x)$ does not converge to $f$. This is contradiction of pointwise convergence, hence for some $n_0$, $C_{n_0}$ must be empty. In other words, for some $n_0\in\mathbb{N}$, 
$$
f_{n_0}(x) - f(x)\leq \epsilon
$$
for each $x\in[0,1]$. By monotonicity of the sequence $\{f_n\}$ (in the sense presumed above), for each $m\geq n_0$, we have also 
$$
f_m(x) - f(x) \leq \epsilon
$$
for each $x\in[0,1]$. Uniform convergence is thus proved.
A: It is Dini's theorem.
