Law of large number integral asymptotic results Let $(X_n)$ be a i.i.d. random variables, following a uniform distribution. Using the weak law of large numbers,
$$\lim_{n +\infty} \int_0^1 ... \int_0^1 \frac{\phi(x_1) + ... + \phi(x_n)}{\psi(x_1) + ... + \psi(x_n)} dx_1 ... dx_n = \frac{\int_0^1 \phi(x) dx}{\int_0^1 \psi(x) dx}$$
where $\phi$ and $\psi$ are continuous and strictly positive functions.
My thoughts so far:

*

*I looked at Fubini's theorem, which has the same functional form as what I'm trying to prove. However I am not supposed to know any of measure theory, so it is not the right way to approach the problem.

*The 0-1 bounds makes me think that the uniform distribution has parameter 0 and 1.

*The use of the law of large numbers can be intuited since it is a convergence theorem and we want to prove an asymptotic result.

But I don't know where to start.
 A: The l.h.s. is
$$
\lim_{n +\infty} \int_0^1 ... \int_0^1 \frac{\phi(x_1) + ... + \phi(x_n)}{\psi(x_1) + ... + \psi(x_n)} dx_1 ... dx_n = \lim_{n +\infty} \mathbb E\left[ \frac{\phi(X_1) + ... + \phi(X_n)}{\psi(X_1) + ... + \psi(X_n)} \right] 
$$
$$
=\lim_{n +\infty} \mathbb E\left[ \frac{\phi(X_1) + ... + \phi(X_n)}{n}\cdot\frac{n}{\psi(X_1) + ... + \psi(X_n)} \right] 
$$
Random variables $\phi(X_1),\phi(X_2),\ldots$ are independent and identically distributed with finite expectation since continuous function $\phi$ on $[0, 1]$ is bounded. And the same holds for  r.v.'s $\psi(X_1),\psi(X_2),\ldots$.
By weak LLN,
$$
 \frac{\phi(X_1) + ... + \phi(X_n)}{n}\xrightarrow{p} \mathbb E[\phi(X_1)], 
$$
$$
\frac{n}{\psi(X_1) + ... + \psi(X_n)} \xrightarrow{p} \frac{1}{\mathbb E[\psi(X_1)]},
$$
so
$$\frac{\phi(X_1) + ... + \phi(X_n)}{n}\cdot\frac{n}{\psi(X_1) + ... + \psi(X_n)} \xrightarrow{p} \frac{\mathbb E[\phi(X_1)]}{\mathbb E[\psi(X_1)]}=\frac{\int_0^1 \phi(x)\,dx}{\int_0^1 \psi(x)\,dx}
$$
The only question is why the expectation converges to the same limit. And the ratio
$$
 \frac{\phi(X_1) + ... + \phi(X_n)}{n}\cdot\frac{n}{\psi(X_1) + ... + \psi(X_n)} 
$$
is strictly positive and bounded above. So, by dominance convergence theorem, expectations converge to the ratio of integrals at the r.h.s. If you knows some theorems in probability about conditions of convergence of expectations, you can invoke uniform integrability which follows from boundedness.
