注意到: 如果${\displaystyle f(y_1,\cdots ,y_k)}$在${\displaystyle (c_1,\cdots ,c_k)}$点连续, ${\displaystyle Y_{ni}(1\le i\le k,n\in \mathbb{N})}$是随机变量, 满足${\displaystyle P(\underset{n\to \mathrm{\infty }}{lim}{Y}_{ni}=c_i)=1}$, 则${\displaystyle P(\underset{n\to \mathrm{\infty }}{lim}f(Y_{n1},\cdots ,Y_{nk})=f(c_1,\cdots ,c_k))=1}$.
根据 Kolmogorov强大数定律 , ${\displaystyle P(\underset{n\to \mathrm{\infty }}{lim}{a}_{ni}={\alpha }_i)=1}$. 而${\displaystyle {\mu }_k=\sum _{r=0}^k(-1{)}^{k-r}{\alpha }_r{\alpha }_1^{k-r}}$, 则由 这里的(1.2) 和注意到的事实, ${\displaystyle P(\underset{n\to \mathrm{\infty }}{lim}{m}_{nk}={\mu }_k)=1}$, 故结合${\displaystyle G}$的连续性得证.

设总体分布为$$P_{\theta }(X=1)=1-P_{\theta }(X=0)=\{\begin{array}{rl}& p,0\le p\le 1,p\in \mathbb{Q},\\ & 1-p,0\le p\le 1,p\notin \mathbb{Q},\end{array}$$ ${\displaystyle X_1\cdots ,X_n}$是简单随机样本, 似然函数为$$f(p|T)=\{\begin{array}{rl}& p^T(1-p{)}^{n-T},p\in \mathbb{Q},\\ & p^{n-T}(1-p{)}^T,p\notin \mathbb{Q}.\end{array}(T=\sum _{i=1}^n{X}_i)$$
注意到${\displaystyle \frac{T}{n}}$是有理数; 当${\displaystyle p={\hat{p}}_n=\frac{T}{n}}$时, 似然函数取到最大值 (令${\displaystyle \frac{\partial f}{\partial p}=0}$), 也即${\displaystyle {\hat{p}}_n=\frac{T}{n}}$是${\displaystyle p}$的极大似然估计; 但由总体分布可知${\displaystyle {\hat{p}}_n}$以概率 1 收敛于${\displaystyle p}$或${\displaystyle 1-p}$, 因此当${\displaystyle p\notin \mathbb{Q}}$时${\displaystyle {\hat{p}}_n}$无法以概率 1 收敛于${\displaystyle p}$.