# Helly's selection theorem

In mathematics, Helly's selection theorem (also called the Helly selection principle) states that a uniformly bounded sequence of monotone real functions admits a convergent subsequence. In other words, it is a sequential compactness theorem for the space of uniformly bounded monotone functions. It is named for the Austrian mathematician Eduard Helly. A more general version of the theorem asserts compactness of the space BV$_{loc}$ of functions locally of bounded total variation that are uniformly bounded at a point.
The theorem has applications throughout mathematical analysis. In probability theory, the result implies compactness of a tight family of measures.

## Statement of the theorem

Let (f$_n$)$_{n∈N}$ be a sequence of increasing functions mapping a real interval I into the real line R, and suppose that it is uniformly bounded: there are a,b ∈ R such that a ≤ f$_n$ ≤ b for every n ∈ N. Then the sequence (f$_n$)$_{n∈N}$ admits a pointwise convergent subsequence.

## Proof
### Step 1. An increasing function f on an interval I has at most countably many points of discontinuity.

Let $A=\{x\in I:f(y)\not \rightarrow f(x){\text{ as }}y\rightarrow x\}$, i.e. the set of discontinuities, then since f is increasing, any x in A satisfies $f(x^{-})\leq f(x)\leq f(x^{+})$, where $f(x^{-})=\lim \limits _{y\uparrow x}f(y)$,$f(x^{+})=\lim \limits _{y\downarrow x}f(y)$, hence by discontinuity, $f(x^{-})<f(x^{+})$. Since the set of rational numbers is dense in R, $\prod _{x\in A}[{\biglf(x^{-}),f(x^{+}){\bigr )}\cap \mathrm {Q} ]$ is non-empty. Thus the axiom of choice indicates that there is a mapping s from A to Q.
It is sufficient to show that s is injective, which implies that A has a non-larger cardinity than Q, which is countable. Suppose x$_1$,x$_2$∈A, x$_1$<x$_2$, then $f(x_{1}^{-})<f(x_{1}^{+})\leq f(x_{2}^{-})<f(x_{2}^{+})$, by the construction of s, we have s(x$_1$)<s(x$_2$). Thus s is injective.

### Step 2. Inductive Construction of a subsequence converging at discontinuities and rationals.

Let $A_{n}=\{x\in I;f_{n}(y)\not \rightarrow f_{n}(x){\text{ as }}y\to x\}$, i.e. the discontinuities of f$_{n,}$ $A=(\cup _{n\in \mathrm {N} }A_{n})\cup (\mathrm {I} \cap \mathrm {Q} )$, then A is countable, and it can be denoted as {a$_n$: n∈N}.
By the uniform boundedness of (f$_n$)$_{n∈N}$ and B-W theorem, there is a subsequence (f$^{(1)}$$_n$)$_{n∈N}$ such that (f$^{(1)}$$_n$(a$_1$))$_{n∈N}$ converges. Suppose (f$^{(k)}$$_n$)$_{n∈N}$ has been chosen such that (f$^{(k)}$$_n$(a$_i$))$_{n∈N}$ converges for i=1,...,k, then by uniform boundedness, there is a subsequence (f$^{(k+1)}$$_n$)$_{n∈N}$ of (f$^{(k)}$$_n$)$_{n∈N}$, such that (f$^{(k+1)}$$_n$(a$_{k+1}$))$_{n∈N}$ converges, thus (f$^{(k+1)}$$_n$)$_{n∈N}$ converges for i=1,...,k+1.
Let $g_{k}=f_{k}^{(k)}$, then g$_k$ is a subsequence of f$_n$ that converges pointwise in A.

### Step 3. g$_k$ converges in I except possibly in an at most countable set.

Let $h_{k}(x)=\sup _{a\leq x,a\in A}g_{k}(a)$, then , h$_k$(a)=g$_k$(a) for a∈A, h$_k$ is increasing, let $h(x)=\limsup \limits _{k\rightarrow \infty }h_{k}(x)$, then h is increasing, since supremes and limits of increasing functions are increasing, and $h(a)=\lim \limits _{k\rightarrow \infty }g_{k}(a)$ for a∈ A by Step 2. By Step 1, h has at most countably many discontinuities.
We will show that g$_k$ converges at all continuities of h. Let x be a continuity of h, q,r∈ A, q<x<r, then $g_{k}(q)-h(r)\leq g_{k}(x)-h(x)\leq g_{k}(r)-h(q)$,hence
$\limsup \limits _{k\rightarrow \infty }{\biglg_{k}(x)-h(x){\bigr )}\leq \limsup \limits _{k\rightarrow \infty }{\biglg_{k}(r)-h(q){\bigr )}=h(r)-h(q)$
$h(q)-h(r)=\liminf \limits _{k\rightarrow \infty }{\biglg_{k}(q)-h(r){\bigr )}\leq \liminf \limits _{k\rightarrow \infty }{\biglg_{k}(x)-h(x){\bigr )}$
Thus,
$h(q)-h(r)\leq \liminf \limits _{k\rightarrow \infty }{\biglg_{k}(x)-h(x){\bigr )}\leq \limsup \limits _{k\rightarrow \infty }{\biglg_{k}(x)-h(x){\bigr )}\leq h(r)-h(q)$
Since h is continuous at x, by taking the limits $q\uparrow x,r\downarrow x$, we have $h(q),h(r)\rightarrow h(x)$, thus $\lim \limits _{k\rightarrow \infty }g_{k}(x)=h(x)$

### Step 4. Choosing a subsequence of g$_k$ that converges pointwise in I

This can be done with a diagonal process similar to Step 2.
With the above steps we have constructed a subsequence of (f$_n$)$_{n∈N}$ that converges pointwise in I.

## Generalisation to BV$_{loc}$

Let U be an open subset of the real line and let f$_n$ : U → R, n ∈ N, be a sequence of functions. Suppose that (f$_n$) has uniformly bounded total variation on any W that is compactly embedded in U. That is, for all sets W ⊆ U with compact closure W̄ ⊆ U,
       $\sup _{n\in \mathbf {N} }\left(\|f_{n}\|_{L^{1}(W)}+\|{\frac {\mathrm {d} f_{n}}{\mathrm {d} t}}\|_{L^{1}(W)}\right)<+\infty ,$
    where the derivative is taken in the sense of tempered distributions.
Then, there exists a subsequence f$_{nk}$, k ∈ N, of f$_n$ and a function f : U → R, locally of bounded variation, such that
* f$_{nk}$ converges to f pointwise almost everywhere;
* and f$_{nk}$ converges to f locally in L$^1$ (see locally integrable function), i.e., for all W compactly embedded in U,
       $\lim _{k\to \infty }\int _{W}{\big |}f_{n_{k}}(x)-f(x){\big |}\,\mathrm {d} x=0;$$^{:132}$
* and, for W compactly embedded in U,
       $\left\|{\frac {\mathrm {d} f}{\mathrm {d} t}}\right\|_{L^{1}(W)}\leq \liminf _{k\to \infty }\left\|{\frac {\mathrm {d} f_{n_{k}}}{\mathrm {d} t}}\right\|_{L^{1}(W)}.$$^{:122}$

## Further generalizations

There are many generalizations and refinements of Helly's theorem. The following theorem, for BV functions taking values in Banach spaces, is due to Barbu and Precupanu:
Let X be a reflexive, separable Hilbert space and let E be a closed, convex subset of X. Let Δ : X → [0, +∞) be positive-definite and homogeneous of degree one. Suppose that z$_n$ is a uniformly bounded sequence in BV([0, T]; X) with z$_n$(t) ∈ E for all n ∈ N and t ∈ [0, T]. Then there exists a subsequence z$_{nk}$ and functions δ, z ∈ BV([0, T]; X) such that
* for all t ∈ [0, T],
       $\int _{[0,t)}\Delta (\mathrm {d} z_{n_{k}})\to \delta (t);$
* and, for all t ∈ [0, T],
       $z_{n_{k}}(t)\rightharpoonup z(t)\in E;$
* and, for all 0 ≤ s < t ≤ T,
       $\int _{[s,t)}\Delta (\mathrm {d} z)\leq \delta (t)-\delta (s).$