# Arzelà–Ascoli for equi-Lebesgue continuous functions

Given a measurable subset $$A$$ of $$[0, 1]$$, a sequence of functions $$f_n: [0, 1] \to \mathbb R$$ is said to be equi-Lebesgue continuous on $$A$$ if for every $$x \in A$$, and $$\varepsilon > 0$$, there exists some $$\delta > 0$$ such that for all $$0 < r < \delta$$, we have

$$\frac{1}{2r} \int_{B_r (x)} \lvert f_n (x) - f_n (y)\rvert \, dy < \varepsilon$$

for all $$n \in \mathbb N$$.

Let $$f_n: [0, 1] \to \mathbb R$$ be a sequence of functions equibounded in $$L^\infty$$, that is, $$\sup_{n \in \mathbb N} \lVert f_n \rVert_{L^\infty} < \infty$$. Suppose further that there exists a subset $$E$$ of $$[0, 1]$$ of measure $$1$$ such that $$f_n$$ are equi-Lebesgue continuous on $$E$$.

Question: Does there exist a subsequence $$f_{n_k}$$ of $$f$$ converging a.e.?

• TeX note: \vert\vert should usually be \Vert. Compare, for example, $\lvert\lvert f\rvert\rvert$ \lvert\lvert f\rvert\rvert to $\lVert f\rVert$ \lVert f\rVert. I have edited accordingly. Oct 23, 2022 at 18:01
• Ah, thank you for the info. Oct 23, 2022 at 18:02

$$\newcommand\ep\varepsilon\newcommand\ze\zeta\newcommand{\al}{\alpha}\newcommand{\be}{\beta}\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$$The answer is yes.

Indeed, take any real $$\be>0$$. Let $$\begin{equation*} \al:=\be/2,\quad\ep:=\be^2/48,\quad\ze:=\eta:=\be/4. \end{equation*}$$ Write $$B_x(r):=[0,1]\cup(x-r,x+r)$$ instead of $$B_r(x)$$.

Without loss of generality (wlog), $$|f_n|\le M$$ on $$E$$ for some real $$M>0$$ and all $$n$$.

By the regularity of the Lebesgue measure, there is a compact subset $$K_\al$$ of $$E$$ such that $$\begin{equation*} |E\setminus K_\al|=|[0,1]\setminus K_\al|\le\al, \tag{0}\label{0} \end{equation*}$$ where $$|A|$$ denotes the Lebesgue measure of a subset $$A$$ of $$\R$$.

By the main condition in the OP, $$\begin{equation*} \forall x\in E\ \exists \de_{x,\ep}\in(0,\infty)\ \forall r\in[0,3\de_{x,\ep}]\ \forall n\ \end{equation*}$$ $$\begin{equation*} \int_{B_x(r)}|f_n(y)-f_n(x)|\,dy\le\ep|B_x(r)|. \tag{1}\label{1} \end{equation*}$$

Since $$K_\al$$ is compact, there is a finite set $$G_{\al,\ep}\subset K_\al$$ such that $$\begin{equation*} K_\al\subseteq\bigcup_{x\in G_{\al,\ep}}B_x(\de_{x,\ep}). \end{equation*}$$ Moreover, by the Vitali covering lemma, there is a finite set $$F_{\al,\ep}\subseteq G_{\al,\ep}$$ such that the balls $$B_x(\de_{x,\ep})$$ for $$x\in F_{\al,\ep}$$ are pairwise disjoint and $$\begin{equation*} K_\al\subseteq\bigcup_{x\in F_{\al,\ep}}B_x(3\de_{x,\ep}). \tag{1.5}\label{1.5} \end{equation*}$$

By \eqref{1} and Markov's inequality, $$\begin{equation*} |A_{n,r,x,\eta}|\le\frac\ep\eta\,|B_x(r)| \end{equation*}$$ for all natural $$n$$, all $$x\in F_{\al,\ep}$$, and all $$r\in[0,3\de_{x,\ep}]$$, where $$\begin{equation*} A_{n,r,x,\eta}:=\{y\in B_x(r)\colon|f_n(y)-f_n(x)|\ge\eta\}. \end{equation*}$$ So, recalling that the balls $$B_x(\de_{x,\ep})$$ for $$x\in F_{\al,\ep}$$ are pairwise disjoint, for $$\begin{equation*} A_{n,\ep,\eta}:=\bigcup_{x\in F_{\al,\ep}}A_{n,3\de_{x,\ep},x,\eta} \end{equation*}$$ we have $$\begin{equation*} |A_{n,\ep,\eta}|\le\sum_{x\in F_{\al,\ep}}\frac\ep\eta\,|B_x(3\de_{x,\ep})| \le3\frac\ep\eta\,\sum_{x\in F_{\al,\ep}}|B_x(\de_{x,\ep})|\le3\frac\ep\eta. \tag{2}\label{2} \end{equation*}$$

Recalling that $$|f_n|\le M$$ on $$E$$ for all $$n$$ and $$F_{\al,\ep}\subset E$$, and passing to a subsequence if needed, wlog we have $$f_n(x)\to g(x)\ \forall x\in F_{\al,\ep}$$ (as $$n\to\infty$$), where $$g$$ is some real-valued function on $$F_{\al,\ep}$$, so that for some natural $$n_{\al,\ep,\ze}$$ we have $$\begin{equation*} n\ge n_{\al,\ep,\ze}\implies\forall x\in F_{\al,\ep}\ |f_n(x)-g(x)|\le\ze. \end{equation*}$$ So, if $$m,n\ge n_{\al,\ep,\ze}$$ and $$y\in B_x(3\de_{x,\ep})\setminus A_{m,\ep,\eta}\setminus A_{n,\ep,\eta}$$ for some $$x\in F_{\al,\ep}$$, then $$\begin{equation*} |f_m(y)-f_n(y)|\le|f_m(y)-f_m(x)|+|f_m(x)-g(x)|+|g(x)-f_n(x)|+|f_n(x)-f_n(y)| \le\eta+\ze+\ze+\eta, \end{equation*}$$ whence, in view of \eqref{1.5}, $$\begin{equation*} |f_m(y)-f_n(y)|\le2\eta+2\ze=\be \end{equation*}$$ if $$m,n\ge n_{\al,\ep,\ze}$$ and $$y\in K_\al\setminus A_{m,\ep,\eta}\setminus A_{n,\ep,\eta}$$.

So,
$$\begin{equation*} |\{x\in[0,1]\colon |f_m(y)-f_n(y)|>\be\}|\le|[0,1]\setminus K_\al| +|A_{m,\ep,\eta}|+|A_{n,\ep,\eta}| \le\al+2\times3\frac\ep\eta=\be \end{equation*}$$ if $$m,n\ge N_\be:=n_{\al,\ep,\ze}=n_{\be/2,\be^2/48,\be/2}$$. So, the sequence $$(f_n)$$ is Cauchy convergent in measure, and hence convergent in measure. So, a subsequence of $$(f_n)$$ is convergent almost everywhere, as claimed.

An almost the same proof will work for the corresponding general statement for functions $$f_n$$ on $$[0,1]^d$$ for any natural $$d$$ and, even more generally, for any complete separable metric space $$S$$ with a finite doubling Borel measure $$\mu$$ over $$S$$, so that $$\mu(B_x(3r))\le C\mu(B_x(r))$$ for some real $$C>0$$, all $$x\in S$$, and all real $$r>0$$, where $$B_x(r)$$ is, of course, the ball in $$S$$ of radius $$r$$ centered at $$x$$.

Also, the main condition in the OP can be relaxed to the following:

$$\begin{equation} \forall x\in E\ \forall\ep>0\ \exists\de>0\ \forall n \end{equation}$$ $$\begin{equation} \int_{B_x(\de)} |f_n (x)-f_n (y)|\,dy<\ep|B_x(\de)|. \end{equation}$$

• This is very impressive. I will give this a good read by tomorrow. Oct 23, 2022 at 22:10
• Just got done reading this, very nicely done. Only one thing, when you write $3 \frac{\varepsilon}{\eta} \sum_{x \in F_{\alpha, \varepsilon}} |B_x (\delta_{x, \varepsilon})| \leq 9 \frac{\varepsilon}{\eta}$, could you have gotten away with $3 \frac{\varepsilon}{\eta}$ on the RHS instead? Since the sum in question is less than $1$ if I’m not mistaken. Oct 24, 2022 at 10:01
• @NateRiver : I have made this change, by also changing the definition of the ball. Oct 24, 2022 at 14:34
• Thank you! Just wanted to make sure I was not missing anything. Oct 24, 2022 at 14:42

The usual equi-continuity $$L^1$$ is $$\sup_n\|f_n-\tau_h f_n \|_1=o(1)\ \qquad \text{as } h\to0$$ (Here $$(\tau_hf)(x)=f(x+h)$$; the $$f_n$$ are to be zero-extended to $$\mathbb R$$ so that $$\tau_h f_n$$ is well defined). Note that in your case the a.e. convergence implies $$L^1$$ convergence by dominated convergence, thus you are actually looking for a compactness theorem in $$L^1$$.

You want the Fréchet-Kolmogorov theorem, the $$L^p$$ version of Ascoli-Arzelà, which indeed follows from it, by regularising via convolution. A nice reference is Functional Analysis by H. Brezis.

• Wow, very nice! Oct 23, 2022 at 18:55
• Actually your condition is not was it is usually called Lebesgue or $L^1$ equicontinuity, the usual hypothesis of th F.K. theorem (my apologies). So I will add some further comment which is in order Oct 23, 2022 at 20:05
• @PietroMajer : It is unclear to me how the $L^1$ equicontinuity follows from the conditions in the OP, where $\delta$ may depend on $x$. Also, you need the equi-tightness for the compactness. How do you get it? Oct 23, 2022 at 22:05
• Hm you’re right.. the convergence is uniform in $n$ but not necessarily $x$. Oct 23, 2022 at 22:10
• Does it follow from the fact that the $f_n$ are equibounded, and a dominated convergence argument? I don’t immediately see how to work it out though. Oct 23, 2022 at 22:21

I am giving another answer which does use the Vitali-covering lemma.

Your assumption can be seen as a way to ensure a uniform $$\mathrm{L}^1$$ approximation by a specific regularizing kernel ; by the way, you implicitly assume some kind of extension of the functions $$f_n$$ outside $$[0,1]$$ (most probably by $$0$$). In the sequel I work on the flat torus $$\mathbf{R}/\mathbf{Z}$$, the arguments being completely similar if you extend by $$0$$ outside. I use the notations $$\dot{\in}$$ for "bounded in" and $$\ddot{\in}$$ for "relatively compact in".

Fix an integrable kernel $$\rho\in\mathrm{L}^1(\mathbf{T})$$. Then, for any $$(f_n)_n\dot{\in}\mathrm{L}^\infty(\mathbf{T})$$, one has $$(f_n\star \rho)_n\ddot{\in}\mathrm{L}^1(\mathbf{T})$$ (that is : compactness of the map $$f\mapsto f\star\rho$$). You can check this by using RFK theorem (equi-continuity is given by $$\tau_h(f\star \rho) = f\star \tau_h\rho$$) or use Banach-Alaoglu theorem to extract weak$$-\star$$ convergence of $$(f_n)_n$$ in $$\mathrm{L}^\infty(\mathbf{T})$$ and check that the corresponding subsequence $$(f_{n_m})_m$$ satisfies $$(f_{n_m}\star\rho_k)_m\ddot{\in}\mathrm{L}^1(\mathbf{T})$$, by duality.

Now, if $$\rho:=\frac12\mathbf{1}_{[-1,1]}$$ and $$\rho_k(x):=k\rho(kx)$$, your assumptions implies

$$\begin{equation} \tag{1}\sup_{n\in\mathbf{N}}\|f_n-f_n\star \rho_k\|_1 \operatorname*{\longrightarrow}_{k\rightarrow +\infty} 0. \end{equation}$$

For a fixed $$k\in\mathbf{N}$$, what we have seen above gives you $$(f_n \star \rho_k)_n\ddot{\in}\,\mathrm{L}^1(\mathbf{T})$$ and by diagonal extraction you can assume (omitting the extraction) that for all $$k\in\mathbf{N}$$, $$(f_n\star\rho_k)_n$$ is converging in $$\mathrm{L}^1(\mathbf{T})$$. Writting, for $$n,p\in\mathbf{N}$$ \begin{align*} f_n-f_p = \stackrel{A_{n,k}}{\overbrace{f_n-f_n\star\rho_k}} + \stackrel{B_{n,p,k}}{\overbrace{f_n\star\rho_k-f_p\star\rho_k}}+\stackrel{C_{p,k}}{\overbrace{f_p\star\rho_k-f_p}}, \end{align*} you first use (1) to pick $$k$$ large enough so that $$\sup_{n,p}\|A_{n,k}\|_1+\|B_{p,k}\|_1$$ is very small. Then, for this fixed $$k$$, you know that $$(f_n\star\rho_k)_n$$ is converging in $$\mathrm{L}^1(\mathbf{T})$$: it satisfies the Cauchy criterion which is thus transferred to $$(f_n)_n$$ by a usual $$\varepsilon$$-argument. Eventually, you conclude by completeness of $$\mathrm{L}^1(\mathbf{T})$$ and another extraction to get a.e. convergence.