Suppose that $X_n \xrightarrow{\mathrm{prob}} 0$, does it hold that $\frac{X_1 + \cdots + X_n}{n} \xrightarrow{\mathrm{prob}} 0$? If not, what are the additional requirements?
P.S. It seems that the technique used in the proof for the non-probabilistic counterpart
If $\lim_{n \to \infty} x_n \to 0$ then $\lim_{n\to \infty} (x_1 + \cdots + x_n)/n \to 0$.
no longer works here.
Surprisingly, this is not true.
Let $\Omega = [0,1]$ with Lebesgue measure, and let $\{X_n\}$ be a modified typewriter sequence: \begin{align*} X_1(\omega) &= 2\times 1_{[0, 1/2]}(\omega) \\ X_2(\omega) &= 2\times 1_{[1/2,1]}(\omega) \\ X_3(\omega) &= 4\times 1_{[0,1/4]}(\omega) \\ X_4(\omega) &= 4\times 1_{[1/4,1/2]}(\omega) \\ &\vdots \end{align*} For any $\varepsilon > 0$, $\mathbb{P}(|X_n| > \varepsilon) \rightarrow 0$, so $X_n \rightarrow 0$ in probability.
Observe that $X_1 + X_2 = 2$ almost surely, $X_3 + X_4 + X_5 + X_6 = 4$ almost surely, and so on. Thus, in particular, for any $k \geq 1$ $$\sum_{i=1}^{2^{k+1} -2} X_i = 2^{k+1} - 2,$$ and in fact, we have the almost sure limit $$\lim_{n\rightarrow\infty} \frac{\sum_{i=1}^n X_i}{n} = 1.$$