Let $X_1, Y_1, X_2, Y_2, \ldots$ be an i.i.d. sequence whose members have a uniform distribution on $[0,1]$ and let $f:[0,1] \to [0,1]$. Define $Z_i = \mathbb{1}_{\{ f(X_i) > Y_i \}}$.
I want to prove that \begin{align} \mathbb{E}\bigg[\bigg(\frac{1}{n} \sum_{i=1}^n Z_i - \int_0^1 f(x)\ dx\bigg)^2\bigg] \leq \frac{1}{4n}. \end{align}
Therefore I first want to show that \begin{align} \frac{1}{n}\sum_{i=1}^n Z_i \xrightarrow{\text{a.s.}} \int_0^1 f(x)\ dx. \end{align} In other words, we want to show that for all $\epsilon >0$ \begin{align} \mathbb{P}\bigg(\exists N: \forall n \geq N: \bigg|\frac{1}{n} \sum_{i=1}^n Z_i - \int_0^1 f(x)\ dx\bigg| < \epsilon \bigg) = 1. \end{align} However, I can't come up with an expression for $\frac{1}{n} \sum_{i=1}^n Z_i$ using the fact that the $X_i,Y_i$'s are uniformly distributed. Any suggestions?
Note each $Z_i$ is iid $Bernoulli(p)$ where $p=P[f(X_1)>Y_1]=\int_0^1\int_0^{f(x)}1dydx=\int_0^1 f(x)dx$.
Hence $E(Z_i)=\int_0^1 f(x)dx$ for each $i$ so $E\left(\dfrac{1}{n}\sum_{i=1}^n Z_i\right)=\int_0^1f(x)dx$.
Now note that $E\left[\dfrac{1}{n}\sum_{i=1}^nZ_i-\int_0^1f(x)dx\right]^2=Var(\dfrac{1}{n}\sum_{i=1}^n Z_i)=\dfrac{Var(Z_1)}{n}\leq \dfrac{1}{4n}$ as if $A\sim Bernoulli(p)$ then $Var(A)=p(1-p)\leq \dfrac{1}{4}$.