# 分配函數與不連續點

Show that a distribution function has at most countably many discontinuities.

### 2 Answers

- Scharze spaceLv 71 decade agoFavorite Answer
Since distribution function is montone decreasing

so it has at most countably many discontiouous points

2008-03-17 00:51:43 補充：

Let f be the distribution function defined on an interval I

Let S={c€I:f is discontinuous at c}

then we have lim(x->c+)f(x)<lim(x->c-)f(x)

so we can choose a rational number c(r) such that

lim(x->c+)f(x)<c(r)<lim(x->c-)f(x)

If c ,d in S so that c<d

then r(d)<lim(x->d-)f(x)<=r[(c+d)]/2<=lim(x->c+)f(x)<r(c)

=>r(c)≠r(d)

so r:S->Q be a one to one mapping

Since Q is countable, so is its subset

=>S is countable

2008-03-17 00:54:01 補充：

應該是r(c)才

2008-03-18 12:48:46 補充：

應該是f((c+d)/2)

Source(s): me - 99Lv 41 decade ago
分配函數是nondecreasing吧？！

怎麼從 "nondecreasing" 推得 "at most countably many discontiouous points" 呢？