1. ## Integral limit

Hi !

Prove that for any continuous function $\displaystyle f ~:~ [0,1] \to \mathbb{R}$ :
$\displaystyle \lim_{n \to \infty} \int_0^1 \dots \int_0^1 f \left(\frac{x_1+x_2+\dots+x_n}{n} \right) ~dx_1 dx_2 \dots dx_n=f \left(\frac 12\right)$

It may be very easy for several of you

2. Originally Posted by Moo

Hi !
Prove that for any continuous function $\displaystyle f ~:~ [0,1] \to \mathbb{R} : \ \ \lim_{n \to \infty} \int_0^1 \dots \int_0^1 f \left(\frac{x_1+x_2+\dots+x_n}{n} \right) ~dx_1 dx_2 \dots dx_n=f \left(\frac 12\right)$

It may be very easy for several of you
one idea is to first prove the claim for $\displaystyle f(x)=x^k, \ k=0,1,2, \cdots,$ which will obviously prove the problem for any polynomial. then use the fact that in any closed interval the set of polynomials

is dense in the set of continuous functions (Weierstrass theorem) to complete the proof. i haven't done all the details but i think this idea will work.

3. Okay, actually, this is something where my TA said "and with this example, we can see that probability surpasses analysis"

Let's consider a sequence of independent random variables $\displaystyle (X_n)_{n \geq 1}$ uniformly distributed over $\displaystyle [0,1]$

For any continuous function $\displaystyle f ~:~ [0,1]\to \mathbb{R}$, and hence bounded (since continuous over a compact set), we have :

$\displaystyle \mathbb{E}f \left(\frac{X_1+\dots+X_n}{n}\right)=\int_{\mathbb {R}^n} f \left(\frac{x_1+\dots+x_n}{n}\right) \bold{1}_{\{x_1 \in [0,1]\}}\dots \bold{1}_{\{x_n \in [0,1]\}} ~dx_1\dots dx_n$

where 1 is the indicator function.
This formula is what we call "the transfer formula" (it may be derived from the Radon-Nikodym theorem - okay, it's an analysis theorem xD)

This gives :

$\displaystyle \mathbb{E}f \left(\frac{X_1+\dots+X_n}{n}\right)=\int_0^1 \dots \int_0^1 f \left(\frac{x_1+\dots+x_n}{n} \right) ~dx_1\dots dx_n$

And we know from the weak of large numbers (which can be applied here) that :

$\displaystyle \frac{X_1+\dots+X_n}{n} \underset{n \to \infty}{\longrightarrow} \mathbb{E}(X_1)$ in probability, and thus in distribution.

But $\displaystyle \mathbb{E}(X_1)=\int_0^1 x ~dx=\frac 12$

Convergence in distribution implies that for any bounded & continuous function f, we have :
$\displaystyle \mathbb{E}f \left(\frac{X_1+\dots+X_n}{n}\right) \underset{n \to \infty}{\longrightarrow} \mathbb{E}f \left(\frac 12\right)=f \left(\frac 12\right)$

And this completes the proof.

Sidenotes :
- it's long because I tried to explain it as much as possible
- if you don't like probability, you surely wouldn't like this proof
- it is possible to prove, similarly, that :
$\displaystyle \lim_{n \to \infty} \sum_{k=0}^n \frac{n!}{k!(n-k)!} p^k (1-p)^{n-k} f \left(\frac kn\right)=f(p)$
where $\displaystyle p \in [0,1]$ and f is a continuous function $\displaystyle [0,1] \to \mathbb{R}$

$\displaystyle \lim_{n \to \infty} \sum_{k \geq 0} e^{-\lambda n} \frac{(\lambda n)^k}{k!} f\left(\frac kn\right)=f(\lambda)$
where $\displaystyle \lambda \in (0,\infty)$ and f is a real valued continuous and bounded function defined over $\displaystyle \mathbb{R}^+$