Limits of integrals How would you show that if
$f : [0,1] \rightarrow \Bbb R$ is continuous, then
$$\lim_{n\rightarrow \infty}\int_0^1\int_0^1 \cdots \int_0^1 f\left( \frac{x_1+x_2+\cdots+x_n}{n} \right)~dx_1~dx_2\cdots dx_n = f\left( \frac{1}{2} \right) $$
and
$$\lim_{n\rightarrow \infty}\int_0^1\int_0^1 \cdots \int_0^1 f((x_1 x_2 \cdots x_n)^{\frac{1}{n}})~dx_1~dx_2\cdots dx_n = f\left(\frac{1}{e}\right) $$
 A: The first result can be shown introducing a Dirac delta (and its integral representation) into the game. We have
$$
\lim_{n\rightarrow \infty}\int_0^1\int_0^1 \cdots \int_0^1 f \left( \frac{x_1+x_2+\cdots +x_n}{n} \right)~dx_1~dx_2\cdots dx_n =
$$
$$
\lim_{n\rightarrow \infty}\int_{\mathbb{R}} dt\ f(t)\int_0^1\int_0^1\cdots \int_0^1  \delta\left(t-\frac{1}{n}\sum_{k=1}^n x_k\right)~dx_1~dx_2\cdots dx_n=
$$
$$
\lim_{n\rightarrow \infty}\int_{\mathbb{R}}dt\ f(t)\int_{\mathbb{R}}\frac{dz}{2\pi} e^{i tz} \int_0^1\int_0^1 \cdots \int_0^1\ e^{-iz\sum_{k=1}^n x_k/n} ~dx_1~dx_2\cdots dx_n=
$$
$$
\lim_{n\rightarrow \infty}\int_{\mathbb{R}}dt\ f(t)\int_{\mathbb{R}}\frac{dz}{2\pi}e^{i tz}\left[\int_0^1 dx\ e^{-izx/n}\right]^n=\lim_{n\rightarrow \infty} \int_{\mathbb{R}}dt\ f(t)\int_{\mathbb{R}}\frac{dz}{2\pi}e^{i tz}\left[\frac{i n \left(-1+e^{-\frac{i z}{n}}\right)}{z}\right]^n
$$
and using
$$
\lim_{n\to\infty}\left[\frac{i n \left(-1+e^{-\frac{i z}{n}}\right)}{z}\right]^n=e^{-i z/2}
$$
the result is 
$$
\int_{\mathbb{R}}dt f(t)\int_{\mathbb{R}}\frac{dz}{2\pi}e^{iz(t-1/2)} =\int_{\mathbb{R}}dt\ f(t)\delta(t-1/2)=f(1/2)\ .
$$
A: For the second problem, recall that Weierstrass tells us that polynomials are dense in $C[0,1].$ So it's enough to  prove the result for polynomials, and for this it's enough to prove it for each $x^k.$ For $x^k$ the $n$-fold integral works out nicely to be $1/(1+k/n)^n.$ This has limit $1/e^k = (1/e)^k,$ which is the desired answer for this function.
Weiersrtass can also be used for the first problem as well, although it's not quite as simple.
A: Note that the LHS is in the form of expectation of some function.
Let $(X_n)_n$ be a sequence of i.i.d random variables, which follow uniform distribution on $[0,1]$. 
For the first case: by strong law of large numbers, $\displaystyle S_n = \frac{1}{n}\sum_{i=1}^n X_i \overset{a.s.}{\longrightarrow} \mathbb E(X_1) = \frac{1}{2}$.
As $f$ is continuous, $f(S_n) \to f(\frac{1}{2})$. As $S_n \in [0,1]$ compact, $f(S_n)$ is bounded by a constant (thus integrable on $[0,1]$). So by dominated convergence theorem, $\mathbb E [f(S_n)] \overset{a.s.}{\longrightarrow} \mathbb E [f(\frac{1}{2})] = f(\frac{1}{2})$.
For the second case, $\displaystyle (x_1\cdots x_n)^{1/n} = \exp(-\frac{1}{n} \sum_{i=1}^n \ln\frac{1}{x_i})$. Let $Y_i = \ln\frac{1}{X_i}$, then by strong law of large numbers, $\displaystyle V_n = \frac{1}{n}\sum_{i=1}^n Y_i \overset{a.s.}{\longrightarrow} \mathbb E(Y_1) = 1$. Again, as $f(e^{-V_n})$ is continuous and bounded by a certain constant (thus integrable on $[0,1]$), $\mathbb E [f(e^{-V_n})] \overset{a.s.}{\longrightarrow} \mathbb E [f(e^{-1})] = f(\frac{1}{e})$.
BONUS: using the same approach, if moreover, we restrict $f: [0,1] \to [0,1]$, then
$$\lim_{n \to \infty} \int_0^1 ... \int_0^1 \frac{f(x_1)+...f(x_n)}{n}dx_1...dx_n = \int_0^1 f(x) dx$$
