Breaking up subgradients into pieces

Question

I was giving this post some thought and could not help but wonder the following. Suppose that $A_1,\dots,A_k$ are pairwise disjoint open sets and $g_1,\dots,g_k$ are smooth functions, then $1_{A_j}g_j$ (where $1_{A_j}$ is the indicator function on $A_j$) is sub-differentiable but is is true that any subgradient $G$ of $\sum_{j=1}^k I_{A_j} g_j$ can be written as $$ G = \sum_{j=1}^n 1_{A_j} \nabla g_j? $$ (since $g_j$ is continuously differentiable)