# Cumulative distribution function equals almost surely

Let $$F_1, F_2$$ - two continuous CDF.

if $$F_1 = F_2\quad F_2$$ almost surely (i.e. probability of $$x$$ where $$F_1(x)\neq F_2(x)$$ is zero with respect to probability with CDF $$F_2$$).

Then $$F_1 = F_2$$ (everywhere).

• Hint: consider the contrapositive. If the $F_i$ are continuous and there exists $x$ for which $F_1(x)\ne F_2(x),$ can you show ${\Pr}_{F_2}(F_1 \ne F_2) \gt 0$?
– whuber
Jan 22 at 15:50
• Thank you, reformulation of statement helps. Jan 22 at 16:08

By contrapositive, if exists $$x$$ such that $$F_1(x) \neq F_2(x)$$
if $$F_2(x) < F_1(x)$$, then choose $$y>x$$ such that $$F_2(y) > F_2(x)$$ and $$F_2(y) < F_1(x)$$. Then $$F_2(x) \neq F_1(x)$$ on $$[x,y]$$ (by the monotony of F_1) and $${\Pr}_{F_2}([x,y]) \ge F_2(y) - F_2(x) > 0$$.
if $$F_2(x) > F_1(x)$$, then choose $$y such that $$F_2(y) < F_2(x)$$ and $$F_2(y) > F_1(x)$$. Then $$F_2(x) \neq F_1(x)$$ on $$[y,x]$$ (by the monotony of F_1) and $${\Pr}_{F_2}([y,x]) \ge F_2(x) - F_2(y)$$ > 0.
Note I use only continuity of $$F_2$$ and statement true if $$F_1$$ is arbitrary.