WLOG assume that $$$a$$$ is sorted (sort it if it isn't). Assume for some prefix $$$a_1,a_2,\cdots,a_k$$$ the sum is $$$S_k$$$, and it generates all integers in $$$[0,S_k]$$$. Then $$$a_{k+1}$$$ must be equal to or less than $$$S_k+1$$$ for the new prefix $$$a_1,a_2,\cdots,a_{k+1}$$$ to generate all subset sum in $$$[0,S_k+a_{k+1}]$$$. We can prove this by contradiction. If $$$a_{k+1} \ge S_k+2$$$, then we cannot generate the integer $$$S_k+1$$$, and thus $$$a_{k+1}$$$ is at most $$$S_k+1$$$. If $$$a_{k+1} \le S_k+1$$$, then we can trivially generate the integers in the range $$$[a_{k+1},S_k+a_{k+1}]$$$. As the two ranges are continuous, we conclude that we can generate every integer in $$$[0,S_k+a_{k+1}]$$$. Thus, $$$a_{k+1} \le S_k+1$$$ is both necessary and sufficient to generate all subset sums in $$$[0,S_k+a_{k+1}]$$$.