$$\underset{n\to \infty }{\mathop{\lim }}\,\int_{0}^{1}{\int_{0}^{1}{\cdots \int_{0}^{1}{{{\left( \frac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n} \right)}^{2}}d{{x}_{1}}d{{x}_{2}}\cdots d{{x}_{n}}}}}$$
I start thinking first in the Lebesgue monotone convergent theorem but this leads to closed road is there any shortcut to solve this problem ??
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Suppose $X_1,X_2,\ldots$ are independent random variables having the uniform distribution on $[0,1]$. Then the common expectation of these variables exist and equals $\mu=1/2$.
Define $$\overline X_n=\frac{1}{n}\sum_{k=1}^n X_k$$
By Khintchine's weak law of large numbers, $$\overline X_n\stackrel{P}{\longrightarrow}\mu\quad\text{ as }\quad n\to\infty$$
And by the continuous mapping theorem, $$\overline X_n^2\stackrel{P}{\longrightarrow}\mu^2\quad\text{ as }\quad n\to\infty\tag{1}$$
Moreover, $$0\le X_1,\ldots,X_n\le 1\implies 0\le \overline X_n\le 1\implies 0\le \overline X_n^2\le 1\tag{2}$$
$(1)$ and $(2)$ together imply $$\int_{[0,1]^n}\left(\frac{x_1+\cdots+x_n}{n}\right)^2\mathrm{d}x_1\ldots\mathrm{d}x_n = E\left(\overline X_n^2\right)\stackrel{n\to\infty}{\longrightarrow}\frac{1}{4}$$
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We have $$(x_1+x_2+\cdots +x_n)^2=\sum_i x_i^2+\sum_{i,j\ne i} x_ix_j$$first of all note that $$\int_{0}^{1}{\int_{0}^{1}{\cdots \int_{0}^{1}{{{x_i}^{2}}d{{x}_{1}}d{{x}_{2}}\cdots d{{x}_{n}}}}}=\int_0^1 x_i^2dx_i={1\over 3}$$and $$\int_{0}^{1}{\int_{0}^{1}{\cdots \int_{0}^{1}{{{x_ix_j}{}}d{{x}_{1}}d{{x}_{2}}\cdots d{{x}_{n}}}}}={1\over 4}$$therefore$$\int_{0}^{1}{\int_{0}^{1}{\cdots \int_{0}^{1}{{{\left( {{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{} \right)}^{2}}d{{x}_{1}}d{{x}_{2}}\cdots d{{x}_{n}}}}}=n\cdot {1\over 3}+(n^2-n){1\over 4}$$and by substitution we obtain $$\int_{0}^{1}{\int_{0}^{1}{\cdots \int_{0}^{1}{{{\left( \frac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n} \right)}^{2}}d{{x}_{1}}d{{x}_{2}}\cdots d{{x}_{n}}}}}={1\over 3n}+{n-1\over 4n}$$which obviously shows that the limit is $1\over 4$.
The integral has an equivalent expression $$ E\left[\left(\frac{1}{n}\sum_{i=1}^n U_i\right)^2\right] $$ where $U_i$, $i\le n$ are independently and uniformly distributed random variables on $[0,1]$. This gives $$\begin{eqnarray} E\left[\left(\frac{1}{n}\sum_{i=1}^n U_i\right)^2\right]&=&\frac{1}{n^2}\sum_{i,j=1}^nE\left[U_iU_j\right]\\ &=&\frac{1}{n^2}\sum_{i=1}^nE\left[U_i^2\right] +\frac{1}{n^2}\sum_{i\ne j}E\left[U_iU_j\right]\\ &=&\frac{n}{n^2}\frac{1}{3}+\frac{n(n-1)}{n^2}\frac{1}{4}\to \frac{1}{4} \end{eqnarray}$$ since $E[U_i]=\int_0^1 xdx=\frac{1}{2}$, $E[U_i^2]=\int_0^1 x^2dx =\frac{1}{3}$ and $E[U_iU_j]=E[U_i]E[U_j]=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$ for all $i\ne j$.