We have $f: \left[ 0,1 \right] \rightarrow \mathbb{R}$, a measurable function, and $U_n, n \in \mathbb{N}$ independent random variables and $U_n \sim Uniform([0,1])$. $$I_n = \frac{1}{n} \sum_{l=1}^{n}f(U_l)~~ ~ \mathrm{and} ~~~I = \int_0^1 f(x)~\mathrm{d}x$$ We also know that $\int_0^1 |f(x)| ~ \mathrm{d}x < \infty ~\mathrm{and} ~\int_0^1 |f(x)|^2 ~\mathrm{d}x < \infty$. In previous exercises I proved that $I_n \rightarrow I$ in probability. Now I have to choose $\varepsilon > 0 $ and bound $$\mathbb{P} (|I - I_n| \geq \frac{\varepsilon}{\sqrt{n}}) $$ using Chebyshev's inequality.
Can I simply do this?: $$\mathbb{P}(|I - I_n| \geq \frac{\varepsilon}{\sqrt{n}}) \leq \frac{Var(I_n)}{\frac{\varepsilon^2}{n}}$$, because $\mathbb{E}(I_n) = I$. $$\frac{Var(I_n)}{\frac{\varepsilon^2}{n}} = \frac{n \cdot \frac{1}{n^2}\sum_{l=1}^n Var(f(U_l))}{\varepsilon^2} = \frac{\frac{1}{n}\sum_{l=1}^n Var(f(U_l))}{\varepsilon^2}$$ Now let $\varepsilon^2 = \sum_{l=1}^n Var(f(U_l))$. This is positive and finite, because $$Var(f(U_l)) = \mathbb{E}(f(U_l)^2)-\mathbb{E}(f(U_l))^2 \leq \mathbb{E}(f(U_l)^2) = \int_0^1 f(x)^2 ~\mathrm{d}x < \infty$$ Because then we have: $$\mathbb{P}(|I - I_n| \geq \frac{\varepsilon}{\sqrt{n}}) \leq \frac{1}{n}$$ where $\frac{1}{n}$ is our upper bound.