Based on the question I asked here, if we assume that the independent random variables $U$ and $X$ follow the exponential distribution with parameter $\lambda$ and the general distribution $f_X(x)$, respectively. Therefore, if $U>X>0$, then $V=U-X$ is follows exponential distribution with parameter $\lambda$. Now suppose the random variable $W$ and $W'$ is defined as follows: $$W=U-X+X$$ therefore: $$W'=Q+X$$
where $Q$ is a random variable follows exponential distribution with parameter $\lambda$. Can we say that the probability distribution of $W$ is equal to $W'$ and therefore follows the exponential distribution with parameter $\lambda$?
If not, what is the problem with my argument?
If $P(U>X)=1$ then $\int P(X<u)\lambda e^{-\lambda x} dx=1$. Since $P(X<u) \leq 1$ this implies that $P(X<u)=1$ a.e. (w.r.t Lebesgue measure). But this is implies that $P(X \leq 0)=1$. So you r random varbles do not exist.
Intuitively an inequality like $U>X$ between r.v.'s destroys independence.