Let $x,y$ be cadlag functions on $[0,\infty)$. Suppose for each $t$, either $x(t)=y(t)$ or $x(t)=y(t-)$, then how does this imply that $x=y$? I am not sure how to show that for the case $x(t)=y(t-)$, we have in fact $x(t)=y(t)$.


1 Answer 1


Càdlàg functions are everywhere right-continuous, and have left limits everywhere.

Let $$ A = \{ t \mid \text{$y$ is continuous at $t$}\} $$ For all $t \in A$ we have $y(t-) = y(t)$, and consequently $x(t) = y(t)$.

In Cardinality of set of discontinuities of cadlag functions it is shown that càdlàg functions have at most countably many discontinuities, so that the complement of $A$ is at most countable.

Therefore, for $t_0 \notin A$ we can find a sequence $(t_n)$ in $A$ with $t_n > t_0$ and $t_n \to t_0$. Then $x(t_n) = y(t_n)$ for all $n$, and the right continuity implies that $x(t_0) = y(t_0)$.


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