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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.?

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    $\begingroup$ 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. $\endgroup$
    – LSpice
    Oct 23, 2022 at 18:01
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    $\begingroup$ Ah, thank you for the info. $\endgroup$
    – Nate River
    Oct 23, 2022 at 18:02

3 Answers 3

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$\newcommand\ep\varepsilon\newcommand\ze\zeta\newcommand{\al}{\alpha}\newcommand{\be}{\beta}\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$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}

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    $\begingroup$ This is very impressive. I will give this a good read by tomorrow. $\endgroup$
    – Nate River
    Oct 23, 2022 at 22:10
  • $\begingroup$ 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. $\endgroup$
    – Nate River
    Oct 24, 2022 at 10:01
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    $\begingroup$ @NateRiver : I have made this change, by also changing the definition of the ball. $\endgroup$ Oct 24, 2022 at 14:34
  • $\begingroup$ Thank you! Just wanted to make sure I was not missing anything. $\endgroup$
    – Nate River
    Oct 24, 2022 at 14:42
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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.

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    $\begingroup$ Wow, very nice! $\endgroup$
    – Nate River
    Oct 23, 2022 at 18:55
  • $\begingroup$ 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 $\endgroup$ Oct 23, 2022 at 20:05
  • $\begingroup$ @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? $\endgroup$ Oct 23, 2022 at 22:05
  • $\begingroup$ Hm you’re right.. the convergence is uniform in $n$ but not necessarily $x$. $\endgroup$
    – Nate River
    Oct 23, 2022 at 22:10
  • $\begingroup$ 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. $\endgroup$
    – Nate River
    Oct 23, 2022 at 22:21
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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.

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