Is Cesaro convergence still weaker in measure?

I've encountered a question I couldn't answer, and I would appreciate any help: Is it true that $$f_n \xrightarrow{m}0\,\,\implies \,\,\frac{1}{n} \sum_{k=1}^{n}f_k \xrightarrow{m}0$$ where $(X,{\mathcal{B}},m)$ is a probability space, and $\xrightarrow{m}$ means: $\forall \epsilon>0,\, m\left(|f_n|\geq \epsilon\right) \rightarrow 0.$

This is similar to Cesaro convergence, which is weaker than regular convergence, but it is not the case here. I've tried to prove the proposition, or to give a counter example, but didn't manage to succeed in either. Any insights will be appreciated!

No, $f_n \stackrel m \to 0$ does not imply $\frac 1 n \sum_{k = 1} ^ n f_k \stackrel m \to 0$. In fact, if $P(X_n = n) = \frac 1 n$ and $X_n = 0$ otherwise, with $X_1, X_2, ...$ independent, it is a nice little exercise to show that $\frac{\sum_{k=1}^n X_k}{n}$ converges in distribution to a nondegenerate random variable.

Just to explicitly give a counterexample, consider $P(X_n = x_n) = \frac 1 n$ and $X_n = 0$ otherwise with $x_n$ increasing. Then $X_n \to 0$ in probability. But for even $n$, $$\sum_{k = 1} ^ n X_k \ge x_{n / 2} I(X_k \ne 0 \mbox{ for some $k \ge \frac n 2$})$$ since $x_n$ is increasing. So $$\sum_{k = 1} ^ k X_k \ge x_{n / 2}$$ with at least probability $$P(X_k \ne 0 \mbox{ for some $k \ge \frac n 2$}) = 1 - \prod_{k = n/2} ^ n \frac{k - 1}{k} = \frac 1 2 + \frac 1 n.$$ So, $\frac{\sum_{k = 1} ^ n X_k}{n} \ge \frac {x_{n / 2}} n$ with at least probability $\frac 1 2$. If $\frac{x_{n / 2}} n$ doesn't go to $0$ then this shows $\frac{\sum_{k = 1} ^ n X_k} n$ won't go to $0$ in probability; for example, if $x_n = n$ this shows that $\frac{\sum_{k=1}^n X_k}{n} \ge \frac 1 2$ with probability at least $\frac 1 2$ for even $n$.