I'm revising laws of large numbers, and I came across such proposition:
Let $ f: [0, 1] \to \mathbb{R} $ be continuous function. Then $$ \lim_{n \to \infty} \int^1_0 \cdots \int^1_0 f\left(\frac{x_1 + \cdots + x_n}{n}\right) dx_1 \dots dx_n = f(\mathbb{E}X_1) = f\left(\frac{1}{2}\right)$$
Why does it hold? Can I generalize it to integrals with different limits (and potentially different distribution), like
$$ \lim_{n \to \infty} \int^\infty_0 \cdots \int^\infty_0 f\left(\frac{x_1 + \cdots + x_n}{n}\right)e^{-(x_1 + \cdots + x_n)} dx_1 \dots dx_n,\quad f(x) = \frac{1}{1+x}?$$
The second integral looks a lot like exponential distribution with $ \lambda = 1 $.
Answer for the first part: let $\{X_i\}$ be i.i.d random variables with uniform distribution in $(0,1)$. Then, by Strong Law of Large Numbers $\frac {X_1+X_2+\cdots+X_n} n \to EX_1=\frac 1 2$ almost surely. In particular the convergence holds in distribution and this is enough to say that $Ef(\frac {X_1+X_2+\cdots+X_n} n) \to f(\frac 1 2)$ for every continuous function $f$. Also, $Ef(\frac {X_1+X_2+\cdots+X_n} n) =\int_0^{1}\int_0^{1}\cdots \int_0^{1} f(\frac {x_1+x_2+\cdots+x_n} n)dx_1dx_1\cdots dx_n$. This proves the first assertion.
Answer for the second part. An identical argument using exponential distribution shows that the limit in this case $f(1)$ for any bounded continuous function $f$ on $(0,\infty)$.