Let $(X_i)_{i\geq 1}$ and $(Y_i)_{i\geq 1}$ be sequences of two real-valued random variables, where the support of $X_i$ is bounded. I assume that $X_1, \dots, X_n$ are independent conditionally on $\mathcal{I}_n = \{Y_1, \dots, Y_n\}$. I want to show that $$\displaystyle S_n = \dfrac{1}{n}\sum_{i = 1}^n \left(X_i - \mathbb{E}(X_i|\mathcal{I}_n)\right)$$ converges in probability to $0$.
Is the following proof correct?
We have $\mathbb{E}(S_n) = 0$ and $\mathbb{V}(S_n) = \mathbb{E}_{\mathcal{I}_n} \left(\dfrac{1}{n^2}\sum_{i = 1}^n\sigma^2_{i}\right)$, where $\sigma^2_{i} = \mathbb{V}(X_i|\mathcal{I}_n)$. As $X_i$ have bounded supports, $\sigma^2_{i}$ is bounded and $\mathbb{V}(S_n) \to 0$ as $n$ grows to infinity.
Using Chebyshev's inequality, I come to the result that $$\mathbb{P}(|S_n| > \varepsilon) \leq \dfrac{\mathbb{V}(S_n)}{\varepsilon^2},$$ for any $\varepsilon > 0$.