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Let $F$ and $F_n$ be distribution functions with $\lim_n F_n(x)=F(x)$ for all continuity points $x$ of $F$. In a proof there is the following part:

Block quote [...] choose the finite points $a=x_0<x_1<\cdots<x_r=b$ as continuity points of $F(x)$ and such that $F(a)\leq \varepsilon, 1-F(b)\leq\varepsilon$, and $\lvert x_{j+1}-x_{j}\rvert\leq\varepsilon$ and, for $n\geq N$, $\lvert F(x_j)-F_n(x_j)\rvert\leq\varepsilon$ for all $j$.

My question: Why can we simply choose such finite many continuity points?? For me this falls from sky without any reason.

H. Hawks
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2 Answers2

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The set of continuity points of $F$ ($C_F$) is dense in $\mathbb{R}$ (as the complement of at most countable set of jumps of $F$). Fix $\varepsilon>0$ and consider points $a,b\in C_F$ s.t. $F(a)\le \varepsilon$ and $1-F(b)\le \varepsilon$. Then split $[a,b]$ into $r_{\varepsilon}=2(a-b)/\varepsilon$ intervals of length $\varepsilon/2$. We get $r_{\varepsilon}+1$ equally spaced points $\{x_i\}$. If $x_i\in \mathbb{R}\setminus C_F$ for some $0\le i\le r_{\varepsilon}$, then the denseness of $C_F$ ensures that we can peak a point $x_i'\in B(x_i,\varepsilon/4)\cap C_F$. After handling all such collisions, the resulting set of points satisfies the required condition.

Finally, points $a$ and $b$ exist by the properties of the CDF and the denseness of $C_F$ (similarly).

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    Thanks! By the way: Can we say something about how many continuity points exist? We know that at most countably many discontinuity points exist but is there a smilar result for continuity points? – H. Hawks May 07 '16 at 09:17
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    @H.Hawks $C_F=\mathbb{R}\setminus J_F$, where $J_F$ (a set of discontinuity points) is at most countable –  May 07 '16 at 10:03
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    Hence we have uncountably many continuity points, don't we? – H. Hawks May 07 '16 at 11:05
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I assume by "distribution function" you mean a cumulative distribution function (CDF), and not a generalized function.

In that case, we note that any CDF is nondecreasing so that it admits only simple discontinuities. Can you figure out the rest? Actually, reading your question more carefully, the result does not seem obvious to me. But this is probably the right path to go down.

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parsiad
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  • At most countably many discontinuity points means that we have uncountably many continuity poimts, so in particular we can choose finite many? - If yes: Why with these desired properties? – H. Hawks May 06 '16 at 22:51