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Let $(F_n)$ be a sequence of cumulative distribution functions (CDFs*). Let $\mu$ be a finite measure on $(\mathbb{R}, \mathcal{B})$, and assume $\mu$ is equivalent to Lebesgue measure (notice that any CDF lies in $L^2(\mu)$). Assume $\Vert F_n - F\Vert_{L^2(\mu)}\rightarrow 0$, for some $[F]\in L^2(\mu)$. Can I find some $H\in[F]$ such that $H$ is also a CDF? In other words: is the set of CDFs closed in $L^2(\mu)$?

*A function $F:\mathbb{R}\rightarrow\mathbb{R}$ is said to be a cumulative distribution function if:

  1. $F$ is nondecreasing.
  2. $F$ is right-continuous.
  3. $\lim_{x\rightarrow-\infty}F(x) = 0$.
  4. $\lim_{x\rightarrow+\infty}F(x) = 1$.
user127022
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1 Answers1

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No. You still need some generalized notion of tightness to conclude that $F$ is a CDF. For example, take $\mu$ to be lebesgue measure on $[0,1]$, and $F_n$ to be the CDF of $X_n=n$. Then $\|F_n-0\|_{L^2(\mu)}=0$

Alex R.
  • 32,771
  • Remark: the example remains true if we substitute the above $\mu$ by any finite measure since $\Vert F_n - 0\Vert = \mu\left[n,+\infty\right)$ – user127022 Nov 28 '16 at 23:46