# Application of the Dominated Convergence Theorem (probabilistic version).

I am currently working on the following problem and I think I've got the solution more or less, but there is a minor question about the usage of the Dominated Convergence Theorem.

Let $f: [0,1] \to \mathbb R$ be continous. Show that \begin{align*} \lim_{n \to \infty} \int_0^1 \int_0^1 \cdots \int_0^1 f\left(\sum_{i = 1}^n x_i/n\right) \, dx_1 \, dx_2 \cdots dx_n = f(1/2) \end{align*}

Let $X_1, X_2, \ldots$ be independent, uniform$(0,1)$ random variables. Then we have that \begin{align*} \lim_{n \to \infty} \int_0^1 \int_0^1 \cdots \int_0^1 f\left(\sum_{i = 1}^n x_i/n\right) \, dx_1 \, dx_2 \cdots dx_n &= \lim_{n \to \infty} \mathbb E\left[f\left(\sum_{i = 1}^n X_i/n\right)\right]. \end{align*}

Now I want to use the Dominated Convergence Theorem:

Let $(\Omega, \mathcal A, \mathbb P)$ be a probability space, $p \ge 1$ a real number and $(X_n)$ a sequence of real-valued random variables, such that $(X_n)$ converges in probability to a random variable $X$. Let $Y \in L^p$ such that $\forall n \in \mathbb N$ holds $|X_n| \le Y$ $\mathbb P$-almost everywhere. Then $X_n, X \in L^p$, $\forall n \in \mathbb N$, $(X_n)$ converges in $L^p$ to $X$ and $\mathbb E[X_n] \xrightarrow{n \to \infty} \mathbb E[X]$.

By the Dominated Convergence Theorem we get \begin{align*} \lim_{n \to \infty} \int_0^1 \int_0^1 \cdots \int_0^1 f\left(\sum_{i = 1}^n x_i/n\right) \, dx_1 \, dx_2 \cdots dx_n &= \mathbb E\left[\lim_{n \to \infty} f\left(\sum_{i = 1}^n X_i/n\right)\right] \\ &= \mathbb E\left[f\left(\lim_{n \to \infty} \sum_{i = 1}^n X_i/n\right)\right]. \end{align*} By the Strong Law of Large Numbers follows \begin{align*} \lim_{n \to \infty} \int_0^1 \int_0^1 \cdots \int_0^1 f\left(\sum_{i = 1}^n x_i/n\right) \, dx_1 \, dx_2 \cdots dx_n = \mathbb E\left[f(\mathbb E[X_i])\right] = f(\mathbb E[X_i]) = f(1/2). \end{align*}

My question is: Why exactly am I allowed to apply the Dominated Convergence Theorem here? Thanks in advance.

Define $Y_n:=f\left(\frac 1n\sum_{j=1}^nX_i\right)$: since $X_i\in [0,1]$, $\frac 1n\sum_{j=1}^nX_j\in [0,1]$. The function $f$ is continuous on a compact set hence bounded, so $|Y_n|$ is bounded by a constant independent on $n$ (and $\omega$).