=  [1/2,1]\times[0,1/2],$ $A_3 = [0,1/2]\times[1/2,1],$ $A_4 =  [1/2,1]\times[1/2,1],$ and $\mathbb{E}(Y|X) = 1$ if $X \in A_1\cup A_4$ and $0$ if $X\in A_2\cup A_3.$ Further assume $\alpha=1$. Then $c_1=0$.

\[condition 2\] The weighted probability measure $\mathbb{P}_\alpha$ has density $\rho_\alpha(X', X'')$ in $\Omega$. Moreover, $\rho_\alpha(X', X'') = \rho_{1,\alpha}(X') \rho_{2,\alpha}(X'') + \rho_{3,\alpha}(X', X''),$ where for all $(X', X'')\in \Omega$, $\rho_{3,\alpha}(X', X'') \le c_2 \rho_\alpha(X', X''),$ with $c_2\in [0,1]$ a constant.

Condition \[condition 2\] asserts that the joint density of $X', X''$ can be decomposed into an independent component plus a dependent component, where the dependent component is dominated by the independent component up to a constant. This condition controls the dependence between $X'$ and $X''$. Since given $X'$, $Y$ is independent of $X''$, this condition will also control the dependence between $Y$ and $X''$. The strength of the condition depends on the constant $c_2$. When $c_2=1$, the condition imposes no restrictions; when $c_2=0$, the condition asserts $X'$ and $X''$ are completely independent.

Using Conditions \[condition 1\]-\[condition 2\], we establish feature selection consistency in Theorem \[feature selection\].

\[feature selection\] If the optional steps in Algorithm 1 are enabled, condition \[condition 1\], \[condition 2\] are satisfied with $c_1>(1-c_2)/[c_2(2-c_2)]$ and $\lambda_n\ge c_3 n^{-(1/2-\beta)}$ for some constant 