Given two decreasing measurable positive functions $f,g$ on $(0,\infty)$ which are equimeasurable. Is $f=g$ a.e.? If not, what about the case when $f$ is right-continuous?
My approach is a little bit akward and required the right-continuity.
Take $t\in (0,\infty)$. Assume that $g(t)< f(t)$. Then, by the right-continuity of $f$, we can find an interval $[t,t+\delta]$ such that $f(s)>g(s)$ on $[t,t+\delta]$, which contradicts with the equimeasurability. So, $g(t)\ge f(t)$.
If there are only countably many $t$ such that $g(t)\ne f(t)$, then we are done. If not, then there are uncountably many $t$ such that $g(t)\ne f(t)$.
So, all these $g(t)$ have different values. However, we have uncountably many $t$. Clearly, $g$ can not have uncoutably many discountinuous point $t$. So, some of them are continuous point, which is a contradiction.
