Convergence of local means implies converge ae? Let $f,f_n \in L^1(\mathbb{R},\mathbb{R}_+)$ with $\int_{\mathbb{R}} f = \int_{\mathbb{R}} f_n = 1$, $(\sqrt{f_n})'$ bounded in $L^2$, $\nabla \sqrt{f}\in L^2$ and such that $$\int_{ p+[0,1/n]} f_n = \int_{ p+[0,1/n]} f$$ for any $p \in \mathbb{Z}/n$ and any $n \in \mathbb{N}$. How to prove that $f_n$ converges to $f$ ae ?
 A: $\newcommand{\R}{\mathbb{R}}
\newcommand{\N}{\mathbb N}
\newcommand{\Z}{\mathbb{Z}}$
We should, and will, assume that 
\begin{equation}
\int_p^{p+1/n}f_n=\int_p^{p+1/n}f \tag{0}
\end{equation}
for all $n\in\N$ and all $p\in\Z/n$. (If (0) is assumed only for $p\in\N/n$, then the conclusion will obviously be false in general.) 
The key here is 

Lemma 1: If a function $g$ is such that $\int_\R g^2=1$ and $\int_\R g'^2=C^2$ for some real constant $C\ge0$, then $|g|\le2\sqrt C$. 

Lemma 1 will be proved at the end of this answer. 
Let $f_0:=f$, $g_0:=g:=\sqrt f$, and $g_n:=\sqrt f_n$, so that for $n=0,1,\dots$
\begin{equation*}
 \int_\R g_n^2=1 \quad \text{and}\quad\int_\R {g'_n}^2\le C
\end{equation*}
for some real $C>0$, whence, by Lemma 1,
\begin{equation*}
 |g_n|\le2\sqrt C. \tag{1}
\end{equation*}
Next, for all $n=0,1,\dots$  and all real $x$ and $y$ such that $x\le y$, 
\begin{equation*}
 |g_n(y)-g_n(x)|\le\int_x^y|g'_n(u)|\,du\le\sqrt{y-x}\,\sqrt{\int_\R g'_n(u)^2\,du}
 \le C\sqrt{y-x},
\end{equation*}
whence, in view of (1), 
\begin{equation*}
 |f_n(y)-f_n(x)|=|g_n(y)-g_n(x)||g_n(y)+g_n(x)|\le 4C\sqrt C\sqrt{y-x}. \tag{2}
\end{equation*}
Take now any $x\in\R$. Then for each $n\in\N$ and some $p=p_{n,x}\in\Z/n$, we have $x\in[p,p+1/n]$ and hence, by (2), 
\begin{equation*}
 |f_n(x)-\bar f_{n,p}|\le 4C\sqrt C\sqrt{1/n}  
\end{equation*} 
and 
\begin{equation*}
 |f(x)-\bar f_{n,p}|=|f(x)-\bar f_{n,0,p}|\le 4C\sqrt C\sqrt{1/n},   
\end{equation*} 
where 
\begin{equation}
 \bar f_{n,p}:=\frac1{1/n}\int_p^{p+1/n}f_n=\bar f_{n,0,p}:=\frac1{1/n}\int_p^{p+1/n}f, 
\end{equation}
which yields 
\begin{equation*}
 |f_n(x)-f(x)|\le 8C\sqrt C\sqrt{1/n},   
\end{equation*}
so that $f_n\to f$ uniformly. 

It remains to present
Proof of Lemma 1: Without loss of generality, $g(0)\ge0$ and $C\in(0,\infty)$. For any $x\in[0,x_*]$, where $x_*:=\frac14\,g(0)^2/C^2$, we have 
\begin{equation*}
 g(x)\ge g(0)-\int_0^{x_*}|g'(u)|\,du\ge g(0)-\sqrt{\int_0^{x_*}\,du}\;\sqrt{\int_\R g'(u)^2\,du}
= g(0)/2, 
\end{equation*}
whence 
\begin{equation*}
 1=\int_\R g(x)^2\,dx\ge\int_0^{x_*}g(x)^2\,dx\ge \tfrac14\,g(0)^2 x_*=\tfrac1{16}\,g(0)^4/C^2, 
\end{equation*}
whence $|g(0)|\le2\sqrt C$. Similarly, $|g(t)|\le2\sqrt C$ for all real $t$. $\Box$
