TITLE: How prove this integral limit $=f(\frac{1}{2})$
QUESTION [18 upvotes]: Let $f$ be a continuous function on the unit interval $[0,1]$. Show that
$$\lim_{n\to\infty}\int_{0}^{1}\cdots\int_0^1\int_{0}^{1}f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)dx_{1}dx_{2}\cdots dx_{n}=f\left(\dfrac{1}{2}\right)$$
This problem is from Selected Problems in Real Analysis. But the author doesn't include a solution. Maybe there is a method for this sort of problem?
Maybe we can use this:
$$\int_0^1\cdots\int_{0}^{1}\int_{0}^{1}\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{2}?$$
I also found a similar problem in this question.

REPLY [4 votes]: Define
$$
\psi_n(x)=\overbrace{\left(n\chi_{[0,1/n]}\right)\ast\cdots\ast\left(n\chi_{[0,1/n]}\right)}^{\text{convolution of $n$ copies}}
$$
Then
$$
\begin{align}
&\int_0^1\cdots\int_0^1f\left(\frac1n\sum_{k=1}^nx_k\right)\,\mathrm{d}x_1\dots\,\mathrm{d}x_n\\
&=n^n\int_0^{1/n}\cdots\int_0^{1/n}f\left(\sum_{k=1}^nx_k\right)\,\mathrm{d}x_1\dots\,\mathrm{d}x_n\\
&=\int_\mathbb{R}f(x)\psi_n(x)\,\mathrm{d}x
\end{align}
$$
The Central Limit Theorem says that since $n\chi_{[0,1/n]}$ has mean $\frac1{2n}$ and variance $\frac1{12n^2}$, the convolution of $n$ copies tends to a normal distribution with mean $\frac12$ and variance $\frac1{12n}$
$$
\psi_n\sim\sqrt{\frac{6n}\pi}\ e^{-6n(x-1/2)^2}
$$
which is an approximation of the Dirac delta function at $x=1/2$. That is,
$$
\lim_{n\to\infty}\int_\mathbb{R}f(x)\psi_n(x)\,\mathrm{d}x=f(1/2)
$$