Let $F$ and $G$ be two distributions with the same mean. $F$ is said to second order stochastically dominate (SOSD) $G$ if $$\int u(x)\mathrm dF(x)\ge \int u(x)\mathrm dG(x)\tag{1}$$ for all increasing and concave $u(\cdot)$.
This above definition is equivalent to
$$\int_{-\infty}^x F(t)\mathrm dt\le \int_{-\infty}^xG(t)\mathrm dt,\qquad\forall x\in\mathbb R.\tag{2}$$
I was told that the requirement for $F$ and $G$ to have the same mean is not really necessary. Suppose $F$ and $G$ do not have the same mean. Can we then still have the equivalence between $(1)$ and $(2)$?
N.B. I was able to show $(2)\Rightarrow (1)$ without the same mean condition, but not the other way around.