Let $X_i$ be i.i.d. nonnegative random variables such that $E[X_1]<\infty$.
Is is true that
$$m_n:=\frac 1n \sum_{i=1}^n X_i1_{\{X_i \geq \sqrt n\}}$$
converges almost surely to $0$ ?
Since $$E\left[X_i1_{\{X_i \geq \sqrt n\}}\right] = E\left[X_11_{\{X_1 \geq \sqrt n\}}\right] \overset{n\to\infty}{\longrightarrow} 0$$ it is immediate that $m_n$ converges in $L^1$ to $0$, hence in probability as well. Since $X_i$ has finite first moment, I know that $P(X_i \geq \sqrt n) = o(1/\sqrt n)$.
Let $R$ be a fixed integer; for each $n\geqslant R$, $$ 0\leqslant m_n\leqslant\frac 1n\sum_{i=1}^nX_i\mathbf{1}_{\left\{X_i\geqslant \sqrt R\right\}} $$ hence by the strong law of large numbers, the following inequality takes place almost surely: $$0\leqslant \limsup_{n\to\infty}m_n\leqslant\mathbb E\left[X_1 \mathbf{1}_{\left\{X_1\geqslant \sqrt R\right\}}\right]. $$ Since $R$ is arbitrary, we can conclude.
Note that it would also work if we replace $m_n$ by $m'_n$ defined by $$ m'_n=\frac 1n\sum_{i=1}^nX_i\mathbf{1}_{\left\{X_i\geqslant R_n\right\}} $$ where $\left(R_n\right)_{n\geqslant 1}$ is a sequence of positive numbers such that $R_n\to\infty$, that is, $m'_n\to 0$ almost surely.