How to apply Chebysev's inequality here? Hi Could anyone just tell me how one has used Chebysev's inequality here1 in theorem 3.5, also they have not defined the notation $\mathbb D$, It would be great if someone can explain to me! 
1Rafał Kapica, Maciej Ślȩczka: Law of large numbers for random iteration; DOI:10.1080/10236198.2017.1332054
 A: The notation $\mathbb D$ is a notation for variance. 
They use the fact that for $n$ large enough, $\left\lvert \mathbb E\left[A_n/n\right]-a(\psi)\right\rvert\lt 1/2$ hence the inclusion 
$$
\left\{\left\lvert \frac{A_n}n-a(\psi)\right\rvert\gt\varepsilon\right\}\subset \left\{\left\lvert \frac{A_n}n-\mathbb E\left[\frac{A_n}n\right]\right\rvert\gt\varepsilon/2\right\}
$$
holds and then they apply Chebyshev's inequality.
A: As Davide has stated, The notation $\mathbb D$ is a notation for the variance. 
Here is my interpretation of what they do:
They first use the triangle inequality: $|\frac{A_n}n - a(\psi)| = |\frac{A_n}n - \mathbb E\left[\frac{A_n}n\right] + \mathbb E\left[\frac{A_n}n\right] - a(\psi)| \leq |\frac{A_n}n - \mathbb E\left[\frac{A_n}n\right]| + |\mathbb E\left[\frac{A_n}{n}\right] - a(\psi)|$, so we have that 
$$
\left\{\left\lvert \frac{A_n}n-a(\psi)\right\rvert\gt\varepsilon\right\}\subset \left\{\left\lvert \frac{A_n}n-\mathbb E\left[\frac{A_n}n\right]\right\rvert\gt\varepsilon/2\right\} \bigcup \left\{\left\lvert \mathbb E\left[\frac{A_n}n\right] -a(\psi)\right\rvert\gt\varepsilon/2\right\}
$$
(If you are summing up two things to be greater than $\epsilon$, one of them has to be greater than $\epsilon/2$)
So, now you can use union bound to bound the probability of the event on the left, and you can use Chebyshev's bound to bound the deviation of $ \frac{A_n}n$ from its expectation.
P.S.: I took the expression and modified it from Davide's answer; it was much cleaner than what I could quickly type. 
