每日一题002 | 证明：分布函数序列点态收敛到一个连续的分布函数，则可以推出一致收敛。（英文）

Question

\(\{F_n(x),n\geq1\}\) is a sequence of c.d.f.'s and \(F_n(x)\rightarrow F(x)\) for each \(x\in(-\infty,\infty)\), where \(F(x)\) is a continuous c.d.f.. Prove that \(\sup\limits_{x\in\{-\infty,\infty\}}|F_n(x)-F(x)|\rightarrow0\), as \(n\rightarrow\infty\), i.e. \(\{F_n(x)\}\) converge to \(F(x)\) uniformly for all \(x\in(-\infty,\infty)\).

翻译：分布函数序列的点态收敛到一个连续的分布函数，则可以推出一致收敛。

Proof

For any \(\epsilon>0\), we first can always find a large enough positive constant \(M(\epsilon)>0\), such that \(1-F(x)<\epsilon/2,\forall x>M\) and \(F(x)<\epsilon/2,\forall x<-M\)，since \(F(x):(-\infty,\infty)\rightarrow[0,1]\) is a continuous c.d.f..

We know \(F(x)\) is uniformly continuous in the close interval \([-M,M]\), so we can find \(k(\epsilon)\in \mathbb{N}^+\) large enough (then \(M/(k-1)\) is small enough for \(\epsilon\)), and let \(x_i=-M+\frac{2M}{k-1}(i-1),i=1,\ldots,k\) be points of division of \([-M,M]\), such that \(F(x_{i+1})-F(x_i)<\epsilon/2,1\leq i<k\). In addition, we denote \(x_0:=-\infty\) and \(x_{k+1}:=\infty\), then

\[F(x_{i+1})-F(x_i)<\epsilon/2,\quad 0\leq i< k+1. \]

For each \(x_i\), since \(F_n(x_i){\rightarrow}F(x_i),n\rightarrow\infty\), we have \(\exist N(\epsilon)>0,\forall n>N\),

\[|F_n(x_i)-F(x_i)|<\epsilon/2,\quad 0\leq i\leq k+1. \]

Now \(\forall x\in(-\infty,\infty)\), \(\exist i\in\{0,\ldots,k\}\), s.t. \(x\in(x_i,x_{i+1}]\), we have

\[F(x_i)-\epsilon/2<F_n(x_i)\leq F_n(x)\leq F(x_{i+1})<F(x_{i+1})+\epsilon/2, \]

where the middle two inequalities are from the non-decreasing property of c.d.f. \(F_n(x)\). And then, continuously by this property, we have

\[F_n(x)-F(x)<F(x_{i+1})+\epsilon/2-F(x)\leq F(x_{i+1})-F(x_i)+\epsilon/2<\epsilon;\\ F_n(x)-F(x)>F(x_i)-\epsilon/2-F(x)\geq F(x_i)-F(x_{i+1})-\epsilon/2>-\epsilon. \]

In conclusion, \(\forall\epsilon>0,\exist N(\epsilon)>0\), s.t. \(\forall n>N,\forall x\in(-\infty,\infty),|F_n(x)-F(x)|<\epsilon\), which completes the proof.