Strong consistency of estimator 2*mean on uniform distribution > $(X_1, ..., X_n)$ are uniformly distributed iid on $R[0, \theta]$. Find if the estimator $Y = 2\overline{X}$ is consistent. I suppose I have to use Chebyshev's inequality: $P(|Y-\theta|^2 \geq \epsilon^2) \leq \frac{E(Y - \theta)^2}{\epsilon^2}\\\$. $E(Y - \theta)^2 = EY^2 - 2\theta EY + E \theta^2 = 4E\overline{X}^2 - 4\theta E \overline{X} + \theta^2 = \theta^2 - 2\theta^2 + \theta^2 = 0$. So the estimator is consistent. But how to show the strong consistence of this estimator (or its absense...)?

$$E(Y - \theta)^2 = E(Y-EY)^2=Var(Y) = Var \left(2\overline X\right) = 4 Var \left(\dfrac{X_1+\ldots+X_n}{n}\right) = \dfrac{4}{n^2}\left(Var (X_1)+\ldots+ Var (X_n)\right) = \dfrac{4}{n^2} \cdot n\cdot Var (X_1) = \dfrac{4}{n}\cdot \dfrac{\theta^2}{12} = \dfrac{\theta^2}{3n} $$ In your equations $4E\overline{X}^2 \
eq \theta^2$ and $E(Y - \theta)^2 \
eq 0$. Moreover, if $$P(|Y-\theta|\geq \varepsilon^2)\leq 0 \text{ for all } \varepsilon>0,$$ then $Y=\theta$ a.s. But $Y=2\overline X$ is not a constant: it's distribution is absolutely continuous with some p.d.f. greater than zero in $(0,2\theta)$.

It is a possible way to prove consistency by definition using Chebyshev inequality. And you can use LLN unstead: $$ 2Y=2 \dfrac{X_1+\ldots+X_n}{n} \xrightarrow{p} 2EX_1=2\frac{\theta}{2} = \theta. $$

For strong consistency, use Strong Law of Large Numbers SLLN: for i.i.d. $X_1,\ldots,X_n$ with finite expectation holds: $$\overline X\xrightarrow{a.s.}EX_1\text{ as } n\to\infty.$$