Why is this true? suppose $T$ is a binary variable and $x$ is a continuous scalar, and $g(x)=E[T|x]$ is the conditional expectation of $T$. Suppose I estimate $g(x)$ using kernel regression $\widehat{g}(x)=\frac{\sum_{i=1}^{n}T_i K_h(X_i-x)}{\sum_{i=1}^{n} K_h(X_i-x)}$ using data $\{T_i,X_i\}_{i=1}^{n}$, where $K_h(\cdot)$ denotes the kernel function. Let $0<\epsilon<t_1<t_2<1-\epsilon<1$ for a small constant $\epsilon>0$. Define set $L(t_1)=\{x:g(x)\geq t_1\}$ and $L(t_2)=\{x:g(x)\geq t_2\}$. Define an estimated set  $L_{n}(t_2)=\{x:\widehat{g}(x)\geq t_2\}$. In a paper I read about the following unproven fact
$ Pr(L_{n}(t_2)\subset L(t_1))\rightarrow 1$ as $n \rightarrow \infty$.
I'm wondering why this is true.
Intuitively this seems true as $L(t_2)\subset L(t_1)$ and $L_n(t_2)$ should be "consistent" in some sense for $L(t_2)$. But how to rigorously prove it? Any comments, insights or reference are welcome. Thanks!
 A: It's not going to be true without a bit more in assumptions.  Suppose $X$ were sampled from $U[0,1]$. There's no way $\hat g(x)$ would be consistent for $g(x)$ at, say, $x=42$.
Even if $X\sim N(0,1)$, the range of $n$ observations from $X$ would be $O(\sqrt{\log n})$, so if $g(x)=(1+\sin x)/2$ there would be lots of $L_n(t_2)$ with basically no information.
So, let's assume that $X$ are iid from some distribution and we are talking about a compact set $K$ where the density of $X$ is bounded away from zero, and where $g$ is continuous at each point and $\hat g_n$ is equicontinuous at each point (which follows from a bound on $g'(x)$ plus consistency of $\hat g_n'(x)$).
Let's try for a contradiction by supposing there is a sequence $x_n$ of points with $\hat g_n(x_n)\geq t_2$ (ie, in $L_n(t_2)$) but $g(x_n)<t_1$ (ie, not in $L(t_1)$.  By compactness, this sequence has a convergent subsequence $x_{n_k}\to x_\infty\in K$.  By continuity of $g$, $g(x_\infty)=\lim_{k\to\infty} g(x_{n_k})<t_1$. And by equicontinuity of $\hat g_n$ at $x_\infty$, $\hat g_n(x_{n_k})\to g(x_\infty)>t_2$. Which is a contradiction, so there is no such sequence $x_n$ and $L_n(t_2)\subset L(t_1)$
