Helly's selection theorem Can someone guide me to a reference (preferably open access online) stating and proving Helly's selection theorem for sequences monotone  uniformly bounded  functions on $[0,1]$. Something that can actually be taught without introducing the ideas of bounded total variation or probability theory.
Just one straight proof and coherent statement for students in an introductory real analysis course. 
 A: Here is the proof from my lecture notes; I expect it is Helly's original proof. Today the theorem would perhaps be seen as an instance of weak$^*$ compactness.
Christer Bennewitz
Lemma(Helly).
Suppose $\{\rho_j\}_1^\infty$ is a uniformly bounded sequence of increasing functions on an interval $I$. Then there is a subsequence converging pointwise to an increasing function.
Proof. Let $r_1,r_2,...$ be a dense sequence in $I$, for example an
enumeration of the rational numbers in $I$. By Bolzano-Weierstrass’
theorem we may choose a subsequence $\{\rho_{1j}\}_{j=1}^\infty$ of 
$\{\rho_j\}_1^\infty$ so that $\rho_{1j}(r_1)$ converges. 
Similarly, we may choose a subsequence $\{\rho_{2j}\}_{j=1}^\infty$ of 
$\{\rho_{1j}\}_{j=1}^\infty$ such that $\rho_{2j}(r_2)$ converges; as a subsequence of $\{\rho_{1j}(r_1)\}_{j=1}^\infty$ also 
$\{\rho_{2j}(r_1)\}_{j=1}^\infty$ converges. 
Continuing in this fashion, we obtain a sequence of sequences $\{\rho_{kj}\}_{j=1}^\infty$, $k=1,2,\dots$ such that each sequence is a subsequence of those preceding it, and such that 
$\rho(r_n)=\lim_{j\to\infty}\rho_{kj}(r_n)$ exists for $n\le k$. Thus 
$\rho_{jj}(r_n)\to\rho(r_n)$ as $j\to\infty$ for every $n$, since 
$\{\rho_{jj}(r_n)\}_1^\infty$ is a subsequence of 
$\{\rho_{nj}(r_n)\}_{j=1}^\infty$ from $j=n$ on. Clearly $\rho$ is increasing, so if $x\in I$ but $\ne r_n$ for all $n$, we may choose an increasing subsequence $\{r_{j_k}\}_1^\infty$ of $\{r_j\}_1^\infty$
converging to $x$, and define $\rho(x)=\lim_{k\to\infty}\rho(r_{j_k})$.
Suppose $x$ is a point of continuity of $\rho$. If $r_k<x<r_n$ we get
$\rho_{jj}(r_k)−\rho(r_n)\le\rho_{jj}(x)−\rho(x)\le\rho_{jj}(r_n)−\rho(r_k)$. Given $\epsilon>0$ we may choose $k$ and $n$ such that 
$\rho(r_n)−\rho(r_k)<\epsilon$. We then obtain
$$
−\epsilon\le\liminf_{j\to\infty}(\rho_{jj}(x)−\rho(x))\le\limsup_{j\to\infty}
(ρ_{jj}(x)−\rho(x))\le\epsilon.
$$
Hence $\{\rho_{jj}\}_1^\infty$ converges pointwise to $\rho$, except possibly in points of discontinuity of $\rho$. But there are at most countably many such discontinuities, $\rho$ being increasing. Hence repeating the trick of extracting subsequences, and then using the ‘diagonal’ sequence, we get a subsequence of the original sequence which converges everywhere in $I$.
