How prove this integral limit $=f(\frac{1}{2})$ 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. 
 A: $\newcommand{\+}{^{\dagger}}
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$\ds{\bbox[5px,#ffd]{\lim_{n \to \infty}
\int_{0}^{1}\!\!\!\!\cdots\!\!\int_{0}^{1}
\fermi\pars{x_{1} + \cdots + x_{n} \over n}
\,\dd x_{1}\ldots\dd x_{n} = \fermi\pars{\half}}:\ {\Large ?}}$

\begin{align}
&\bbox[#ffd,5px]{%
\lim_{n \to \infty}\int_{0}^{1}\cdots\int_{0}^{1}
\fermi\pars{x_{1} + \cdots + x_{n} \over n}
\,\dd x_{1}\ldots\dd x_{n}}
\\[5mm] = &\
\lim_{n \to \infty}\
\int_{0}^{1}\cdots\int_{0}^{1}
\int_{-\infty}^{\infty}
\\[2mm] &\ \phantom{\lim_{n \to \infty}\,\,\,}
\tilde{\fermi}\pars{k}
\exp\pars{\ic k\,{x_{1} + \cdots + x_{n} \over
n}}
\,{\dd k \over 2\pi}\dd x_{1}\ldots\dd x_{n}
\end{align}
where $\ds{%
\tilde{\fermi}\pars{k} \equiv \int_{-\infty}^{\infty}\fermi\pars{x}
\expo{-\ic k x}\,\dd x}$ is the $\ds{\fermi\pars{x}}$ Fourier Transform. 
\begin{align}
&\bbox[#ffd,5px]{%
\lim_{n \to \infty}
\int_{0}^{1}\cdots\int_{0}^{1}
\fermi\pars{x_{1} + \cdots + x_{n} \over n}
\,\dd x_{1}\ldots\dd x_{n}}
\\[5mm] = &
\lim_{n \to \infty}\int_{-\infty}^{\infty}\tilde{\fermi}\pars{k}
\pars{\int_{0}^{1}\expo{\ic kx/n}\,\dd x}^{n}\,{\dd k \over 2\pi}
\\[5mm] = &\
\lim_{n \to \infty}\int_{-\infty}^{\infty}\tilde{\fermi}\pars{k}
\pars{\expo{\ic k/n} - 1 \over \ic k/n}^{n}\,{\dd k \over 2\pi}
\\[5mm] = &\
\lim_{n \to \infty}\int_{-\infty}^{\infty}
\tilde{\fermi}\pars{k}
\exp\pars{\ic k \over 2}
\braces{\sin\pars{k/\bracks{2n}} \over k/\bracks{2n}}^{n}\,{\dd k \over 2\pi}
\end{align}

\begin{align}
&\bbox[#ffd,5px]{\lim_{n \to \infty}
\int_{0}^{1}\cdots\int_{0}^{1}
\fermi\pars{x_{1} + \cdots + x_{n} \over n}
\,\dd x_{1}\ldots\dd x_{n}}
\\[2mm] = &\
\int_{-\infty}^{\infty}
\fermi\pars{x}\lim_{n \to \infty}
\operatorname{K}_{n}\pars{x - \half}\,\dd x
\end{align}
where
$$
\operatorname{K}_{n}\pars{x} \equiv
\int_{-\infty}^{\infty}
\exp\pars{-\ic k x}
\braces{\sin\pars{k/\bracks{2n}} \over k/\bracks{2n}}^{n}
\,{\dd k \over 2\pi}
$$
Since
$\ds{\lim_{n \to \infty}{\rm K}_{n}\pars{x} = \delta\pars{x}}$, we'll have:
\begin{align}
&\bbox[#ffd,5px]{\lim_{n \to \infty}
\int_{0}^{1}\cdots\int_{0}^{1}
\fermi\pars{x_{1} + \cdots + x_{n} \over n}
\,\dd x_{1}\ldots\dd x_{n}}
\\[5mm] = &\
\int_{-\infty}^{\infty}
\fermi\pars{x}\delta\pars{x - \half}\,\dd x
\end{align}
$\ds{\delta}$ is the
Dirac Delta Function.

Finally,
\begin{align}
&\bbox[#ffd,5px]{\lim_{n \to \infty}
\int_{0}^{1}\cdots\int_{0}^{1}
\fermi\pars{x_{1} + \cdots + x_{n} \over n}
\,\dd x_{1}\ldots\dd x_{n}}
\\[5mm] = &\
\bbox[5px,border:1px groove navy]{\fermi\pars{\half}} \\ &
\end{align}
A: 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,
$$
\lim_{n\to\infty}\int_\mathbb{R}f(x)\psi_n(x)\,\mathrm{d}x=f(1/2)
$$
A: We can take advantage of a probabilistic interpretation. Let $X_1,\ldots,X_n$ be uniformly distributed random variables on $[0,1]$ and $\bar X=\frac{1}{n}(X_1+\cdots+X_n)$. Then
$$\lim\limits_{n\to \infty}\int_0^1\cdots\int_0^1f\left(\frac{x_1+\cdots+x_n}{n}\right)dx_1\cdots dx_n =\lim\limits_{n\to\infty}E[f(\bar X)]=E\left[f\left(\frac12\right)\right]=f\left(\frac12\right)$$
since $\bar X$ converges in distribution to $\frac12$ as $n\to \infty$. 
A: Starting with OP's hint,
$\displaystyle \lim\limits_{n\to\infty}\int_{0}^{1}\cdots\int_{0}^{1} \left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{2}$, 
Now, we try to show, $\displaystyle \lim\limits_{n\to\infty}\int_{0}^{1}\cdots\int_{0}^{1} \left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)^k dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{2^k}$
If we count the number of terms in the multinomial expansion of $\bigg(\sum\limits_{i=1}^n x_i\bigg)^k$, that contains all variables with power not exceeding $1$, is $n(n-1)(n-2)\cdots(n-k+1) = n^k + O(n^{k-1})$, 
and the number of terms that has atleast one $x_i$ term with powers exceeding $1$ is not more than $n.n^{k-2} = n^{k-1}$.
So, combining all the terms we get,
 $\displaystyle \int_{0}^{1}\cdots\int_{0}^{1} \left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)^k dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{2^k} + O\bigg(\frac{1}{n}\bigg)$
Thus the above limit is true for polynomials.
Using Weierstrass Approximation theorem, we can choose a polynomial $P$, such that $|f(x) - P(x)| < \epsilon/3$ for all $x \in [0,1]$.
Thus, we can find an $N \in \mathbb{N}$, such that $\forall n >N$, 
$\displaystyle \left| \int_{0}^{1}\cdots\int_{0}^{1} P\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) dx_{1}dx_{2}\cdots dx_{n} - P\left(\frac12\right)\right| < \epsilon/3$
Therefore, for $n > N$,
$\displaystyle \left| \int_{0}^{1}\cdots\int_{0}^{1} f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) dx_{1}dx_{2}\cdots dx_{n} - f\left(\frac12\right)\right| < \left| \int_{0}^{1}\cdots\int_{0}^{1} f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) - P\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) dx_{1}dx_{2}\cdots dx_{n}\right| + \left| \int_{0}^{1}\cdots\int_{0}^{1} P\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) dx_{1}dx_{2}\cdots dx_{n} - P\left(\frac12\right)\right| + \left|f\left(\frac12\right) - P\left(\frac12\right)\right| $
$\le \displaystyle \int_{0}^{1}\cdots\int_{0}^{1} \left| f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) - P\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)\right| dx_{1}dx_{2}\cdots dx_{n} + 2\epsilon/3 < \epsilon$
Thus, $\displaystyle \lim\limits_{n\to\infty} \int_{0}^{1}\cdots\cdots\int_{0}^{1}f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)dx_{1}dx_{2}\cdots dx_{n}=f\left(\dfrac{1}{2}\right)$.
