$\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}\newcommand\ga{\gamma}$For natural $n$, let $E_n$ be the set of all points in $\R^2$ with "polar coordinates" $(r,t)$ in the set $$F_n:=\bigcup_{i=1}^\infty\bigcup_{j=1}^n F_{n;i,j}\subset(0,\infty)\times(0,2\pi]\subset\R^2,$$ where \begin{equation*} F_{n;i,j}:=\Big(\frac{i-1}n,\frac in\Big]\times(t^*_{n;i,j},t^{**}_{n;i,j}], \end{equation*} \begin{equation*} t^*_{n;i,j}:=\frac\pi n\Big(2j-2+\frac{1+(-1)^i}2\Big), \quad t^{**}_{n;i,j}:=\frac\pi n\Big(2j-1+\frac{1+(-1)^i}2\Big). \end{equation*} Slightly more precisely, \begin{equation*} E_n:=\{(r\cos t,r\sin t)\colon(r,t)\in F_n\}. \tag{0}\label{0} \end{equation*}
Here is the intersection (colored light blue) of the set $E_{20}$ with the unit disk:
So, we have a kind of chessboard in polar coordinates over the entire plane.
Let $\ga^{\otimes2}$ denote the standard normal distribution over $\R^2$. Let $$\nu_n(B):=\ga^{\otimes2}(E_n\cap B)$$ for all Borel subsets $B$ of $\R^2$. For any unit vector $u\in\R^2$, let $\nu_{n,u}$ be the pushforward of the measure $\nu_n$ under the projection map $\R^2\ni x\mapsto p_u(x):=u\cdot x\in\R$, where $\cdot$ denotes the dot product.
Is it then true that $\nu_{n,u}$ converges (as $n\to\infty$) in total variation to $\frac12\,\ga$, where $\ga$ is the standard normal distribution over $\R$?
Discussion:
The intuition behind this conjecture is as follows:
- recall that $\nu_{n,u}$ is the pushforward of the measure $\nu_n$ under the map $p_u$;
- note that $\ga$ is the pushforward of the measure $\ga^{\otimes2}$ under the map $p_u$;
- note that the total $\nu_n$-mass of $\R^2$ is $\frac12$ of the total $\ga^{\otimes2}$-mass of $\R^2$ and this fraction, $\frac12$, seems to stay almost uniform over $\R^2$ for large $n$;
- so, it seems plausible that $\nu_{n,u}$ will be close to $\frac12$ of $\ga$ -- even in the total variation sense, because the projection $p_u$ should smooth out the density of the measure $\nu_n$.
Note that, to prove that $\nu_{n,u}$ converges in total variation to $\frac12\,\ga$, it is enough (in view of dominated convergence) to show that the density of $\nu_{n,u}$ (w.r.t. the Lebesgue measure over $\R$) converges to the density of $\frac12\,\ga$ almost everywhere. That is, (in view of the rotational symmetry of the measure $\ga^{\otimes2}$) it is enough to show that for each unit vector $u\in\R^2$ and almost all $s\in\R$ we have \begin{equation*} \int_\R dt\,g(t)\,1(sv+tu\in E_n)\to\frac12, \tag{10}\label{10} \end{equation*} where $g$ is the standard normal density and $v$ is either one of the two unit vectors in $\R^2$ orthogonal to $u$. So, the conjectured effect of the indicator $t\mapsto1(sv+tu\in E_n)$ in \eqref{10} is to approximately halve the total standard normal probability mass of $\R$, for almost all $s\in\R$.
The latter indicator is rather complicated, since the set $E_n$ is rather complicated. So, it makes sense to spell out this indicator. First here is the trivial remark that, by symmetry, without loss of generality $s>0$ in \eqref{10}. Next, the equation of any straight line in $\R^2$ in polar coordinates $r,t$ is of the form $r\cos(t-t_0)=s$ for some real $s>0$ and some real $t_0$. So, in view of \eqref{0}, we can rewrite \eqref{10} as \begin{equation*} H_{t_0,n}(s):=\sum_{i=1}^\infty\sum_{j=1}^n \ga(T_{n;i,j}(t_0,s))\to\frac12 \tag{10a}\label{10a} \end{equation*} for each $t_0\in[0,\pi)$ and almost all real $s>0$, where \begin{equation*} \begin{aligned} T_{n;i,j}(t_0,s)&:=\Big\{s \tan(t-t_0)\colon t\in(t^*_{n;i,j},t^{**}_{n;i,j}],\ \cos(t-t_0)\in\Big[\frac{ns}{i},\frac{ns}{i-1}\Big)\Big\} \\ & =\Big\{s \tan u\colon u\in(t^*_{n;i,j}-t_0,t^{**}_{n;i,j}-t_0],\ \cos u\in\Big[\frac{ns}{i},\frac{ns}{i-1}\Big)\Big\}, \end{aligned} \end{equation*} with $\frac{ns}{i-1}:=\infty$ for $i=1$. (Of course, here we can express $\tan u$ in terms of $\cos u$, depending on which of the four quarters of the interval $[0,2\pi)$ the value of $u\mod2\pi$ is in. Also, because $\cos\le1$, we can replace $\sum_{i=1}^\infty$ in \eqref{10a} by $\sum_{i=\lfloor ns\rfloor+1}^\infty$. Even for $i\ge\lfloor ns\rfloor+1$, the sets $T_{n;i,j}(t_0,s)$ will be empty for most of the $j$'s.)
Update: Below are the graphs $\{(s,H_{1,n}(s)-\frac12)\colon0<s<4\}$ for $n=5$ (black), $n=10$ (blue), and $n=20$ (green). If I was not mistaken, we seem to have some improvement going from $n=5$ to $n=10$, but (to my surprise) no improvement going from $n=10$ to $n=20$.
Perhaps, $n=20$ is still a small number in this context, even though the calculations for $n=20$ are already rather heavy.
Or perhaps the conjecture is false. Perhaps this "chessboard" is too "regular" and has to be modified somehow to get the desired effect.
