If $S_n = \sum_{j = 1}^n X_j$ is $\mathcal{F}_n$ measureable is $S_n^2 - \mathbb{V}[S_n]$ also $\mathcal{F}_n$ measureable?

Question

I want to argue that if $S_n$ is $\mathcal{F}_n$ measurable then $S_n^2 - Var[S_n]$ is also, however, I am not sure what the rights arguments would be. I know that $X_n \in \mathcal{L}^2(P)$ and that they are independent. Then as $x \mapsto x^2 - k$ where $k$ is some constant is a continuous, thus a measurable transformation, then $S_n^2 - Var[S_n]$ is also $\mathcal{F}_n$ measureable? Is this sufficient?