Suppose that X, Y and Z are random variables on the same probability space and that Y and Z are independent and have equivalent distribution. Can it happen that $X+Y$ and $X+Z$ have non-equivalent distributions?
By equivalent measures I understand measures that vanish exactly on the same sets.
Concretely: Let the underlying probability space be $[0,1]$ with the Borel measure. Let $Y$ be the random variable which takes value $0$ on $[0, 1/2]$ and 1 elsewhere; let $Z$ be the random variable which takes value $0$ on $[1/4, 3/4]$ and $1$ elsewhere. Then $Y, Z \sim \mathrm{Bernoulli}(1/2)$ and they are independent. Now let $X$ be the random variable which takes the value 1 on $[0, 1/4]$ and the value 0 elsewhere. Then $X + Y \sim \mathrm{Bernoulli}(3/4)$, while $X + Z$ can take the value 2 with positive probability (and is thus definitely is not a Bernoulli distribution).
Something that also works very often, is to take $X = Y$. Then $X + Y = 2Y$, while $X + Z = Y + Z$. It will almost never be the case that $2Y$ and $Z + Y$, are identically distributed for $Y, Z$ identically distributed.