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).
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1$\begingroup$ 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 ${\Pr}_{F_2}(F_1 \ne F_2) \gt 0$? $\endgroup$– whuber ♦Commented Jan 22, 2021 at 15:50
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1$\begingroup$ Thank you, reformulation of statement helps. $\endgroup$– NikolayCommented Jan 22, 2021 at 16:08
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1 Answer
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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 $y<x$ such that $F_2(y) < F_2(x)$ and $F_2(y) > F_1(x)$. Then $F_2(x) \neq F_1(x)$ on $[y,x]$ (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.
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$\begingroup$ +1: the final remark is insightful. $\endgroup$– whuber ♦Commented Jan 22, 2021 at 16:15