$f(x)$在$[0,1]$连续证明：$$\lim_{n \to \infty}\int_{[0,1]^n}f(\frac{x_1+x_2+\cdots+x_n}{n})\mathop{}\mathrm{d}x_1\mathop{}\mathrm{d}x_2\cdots\mathop{}\mathrm{d}x_n=f(\frac{1}{2})$$
$$I_n=[0,1]^n$$
1.$$\int_{I_n}(x_1+x_2+\cdots+x_n)\mathop{}\mathrm{d}x_1\mathop{}\mathrm{d}x_2\cdots\mathop{}\mathrm{d}x_n=\frac{n}{2}$$
2.$$\int_{I_n}(x_1^2+x_2^2+\cdots+x_n^2)\mathop{}\mathrm{d}x_1\mathop{}\mathrm{d}x_2\cdots\mathop{}\mathrm{d}x_n=\frac{n}{3}$$
3.$$\int_{I_n}(x_1+x_2+\cdots+x_n)^2\mathop{}\mathrm{d}x_1\mathop{}\mathrm{d}x_2\cdots\mathop{}\mathrm{d}x_n=\frac{n}{3}+\frac{n(n-1)}{4}$$
4.求$$A=\lim_{n \to \infty}\int_{[0,1]^n}f(\frac{x_1+x_2+\cdots+x_n}{n})\mathop{}\mathrm{d}x_1\mathop{}\mathrm{d}x_2\cdots\mathop{}\mathrm{d}x_n=f(\frac{1}{2})$$
注：$\mu(S_n(a))=\frac{a^n}{n!}\to 0(n\to 0),\mu(B_{2n}(a))=\frac{a^{2n}\pi^n}{n!}\to 0,\mu(I_n)=1$
记$u_n=\frac{x_1+x_2+\cdots+x_n}{n},I_n=(\lbrace |u_n-\frac{1}{2}|\leqslant\delta \bigcap I_n\rbrace)\bigcup(\lbrace |u_n-\frac{1}{2}|>\delta \bigcap I_n\rbrace)$
$$
\begin{aligned}
&\left | A-f(\frac{1}{2})\right | \\
&=\left | \int_{I_n}(f(u_n)-f(\frac{1}{2})\mathop{}\mathrm{d}x_1\mathop{}\mathrm{d}x_2\cdots\mathop{}\mathrm{d}x_n\right | \\
&\leqslant \int_{I_n}\left |(f(u_n)-f(\frac{1}{2})\right |\mathop{}\mathrm{d}x_1\mathop{}\mathrm{d}x_2\cdots\mathop{}\mathrm{d}x_n\\
&=\int_{I_\delta}\left |(f(u_n)-f(\frac{1}{2})\right |\mathop{}\mathrm{d}x_1\mathop{}\mathrm{d}x_2\cdots\mathop{}\mathrm{d}x_n
&+\int_{I_n\backslash I_\delta}\left |(f(u_n)-f(\frac{1}{2})\right |\mathop{}\mathrm{d}x_1\mathop{}\mathrm{d}x_2\cdots\mathop{}\mathrm{d}x_n
\end{aligned}\\
\forall \epsilon >0,\exists \delta >0,when |x-y|<0,|f(x)-f(y)|<\epsilon\\
\therefore above<\epsilon\mu(I_\delta)+2M\int_{I_n\backslash I_\delta}\mathop{}\mathrm{d}x_1\mathop{}\mathrm{d}x_2\cdots\mathop{}\mathrm{d}x_n\\
\text{下证} \lim_{n\to +\infty}
\int_{I_n\backslash I_\delta}1\mathop{}\mathop{}\mathrm{d}x_1\mathop{}\mathrm{d}x_2\cdots\mathop{}\mathrm{d}x_n=0\\
\begin{aligned}
\int_{I_n\backslash I_\delta}1\mathop{}\mathop{}\mathrm{d}x_1\mathop{}\mathrm{d}x_2\cdots\mathop{}\mathrm{d}x_n&\leqslant \int_{I_n
\backslash I_\delta}\frac{(u_n-\frac{1}{2})^2}{\delta^2}\mathop{}\mathop{}\mathrm{d}x_1\mathop{}\mathrm{d}x_2\cdots\mathop{}\mathrm{d}x_n\\
&\leqslant\frac{1}{\delta^2}\int_{I_n}(u_n-\frac{1}{2})^2\mathop{}\mathop{}\mathrm{d}x_1\mathop{}\mathrm{d}x_2\cdots\mathop{}\mathrm{d}x_n\\
&=\frac{1}{\delta^2}\int_{I_n}\left[\frac{(x_1+x_2+\cdots+x_n)^2}{n^2}-\frac{x_1+\cdots+x_n}{n}+\frac{1}{4}\right]\mathop{}\mathop{}\mathrm{d}x_1\mathop{}\mathrm{d}x_2\cdots\mathop{}\mathrm{d}x_n\\
&=\frac{1}{12n\delta^2}\to 0
\end{aligned}
$$
证明$$\iint_{I^2}(xy)^{xy}\mathop{}\mathrm{d}x\mathop{}\mathrm{d}y=\int_{0}^{1}t^t\mathop{}\mathrm{d}t$$
$$
\left\lbrace
\begin{aligned}
&u=xy &(0\leqslant u\leqslant 1)\\
&v=x &(u\leqslant x\leqslant 1)
\end{aligned}
\right.\\
left=-\iint u^u\frac{1}{v}\mathop{}\mathrm{d}u\mathop{}\mathrm{d}v=-\int_{0}^{1}\mathop{}\mathrm{d}u\int_{u}^{1}u^u\frac{1}{v}=-\int_{0}^{1}u^u\ln u\mathop{}\mathrm{d}u\\
\frac{\mathop{}\mathrm{d}{y^y}}{\mathop{}\mathrm{d}{y}}=y^y(\ln y+1)\\
0=\int_{0}^{1}\mathop{}\mathrm{d}(y^y)=\int_{0}^{1}y^y\mathop{}\mathrm{d}y+\int_{0}^{1}y^y\ln y\mathop{}\mathrm{d}y\\
\therefore -\int_{0}^{1}u^u\ln u\mathop{}\mathrm{d}u=\int_{0}^{1}u^u\mathop{}\mathrm{d}u\\
\therefore \iint_{I^2}(xy)^{xy}\mathop{}\mathrm{d}x\mathop{}\mathrm{d}y=\int_{0}^{1}t^t\mathop{}\mathrm{d}t
$$