I'm trying to understand some implications concerning the Weak Laws of Large Numbers, and I'm stuck in this problem.
If $f:[0,1]\to\mathbb{R}$ is a measurable function with $\int_0^1\mid f(x)\mid dx<\infty$ and $U_1,...,U_n,...$ is a sequence of independent uniformly distributed random variables, consider the following average:
\begin{equation} I_n = \frac{f(U_1) + ... + f(U_n)}{n} \end{equation}
Then, $I_n\to \int_0^1f(x)dx$ in probability.
I was trying to apply some kind of Weak Law of Large Numbers, since we can compute using the uniform distribution that:
\begin{equation} E[f(U_j)] = \int_0^1f(x)dx \end{equation}
And hence:
\begin{equation} E[I_n] = I \end{equation}
However, I was not able to get some bound on the variance $var[f(U_j)]$ in order to apply the Weak Law of Large Numbers.
So instead I also tried to prove $L^1$ convergence, but got stuck as well: assume $I=0$ wlog, and then I got the following chain of inequalities:
\begin{equation} E[\mid I_n \mid] \leq \frac{1}{n}\sum_{j=1}^{n}E[\mid f(U_j)\mid] = \int_0^1 \mid f(x) \mid dx < \infty \end{equation}
Which is far for enough to prove $L^1$ convergence. When trying to estimate the actual probability convergence, I had no luck either, since my only idea was to use Chebyshev's Inequality, which would be useful only if I knew about some $L^p$ convergence.
What would be a better approach here?
Hint: recall that simple functions are dense in $L^1,$ so it is enough to prove the result for the indicator function of an interval $[a, b].$