How would you show that if $f : [0,1] \rightarrow \Bbb R$ is continuous, then

$$\lim_{n\rightarrow \infty}\int_0^1\int_0^1 \cdots \int_0^1 f\left( \frac{x_1+x_2+\cdots+x_n}{n} \right)~dx_1~dx_2\cdots dx_n = f\left( \frac{1}{2} \right) $$


$$\lim_{n\rightarrow \infty}\int_0^1\int_0^1 \cdots \int_0^1 f((x_1 x_2 \cdots x_n)^{\frac{1}{n}})~dx_1~dx_2\cdots dx_n = f\left(\frac{1}{e}\right) $$

  • $\begingroup$ What's $x_k$?$\quad$ $\endgroup$ Apr 16, 2016 at 21:42
  • $\begingroup$ Is $f$ is continuous? $\endgroup$
    – zhw.
    Apr 16, 2016 at 23:05
  • 1
    $\begingroup$ @EricTowers: The integration limits directly say that $0\le x_k\le 1$. $\endgroup$ Apr 17, 2016 at 12:46

3 Answers 3


Note that the LHS is in the form of expectation of some function.

Let $(X_n)_n$ be a sequence of i.i.d random variables, which follow uniform distribution on $[0,1]$.

For the first case: by strong law of large numbers, $\displaystyle S_n = \frac{1}{n}\sum_{i=1}^n X_i \overset{a.s.}{\longrightarrow} \mathbb E(X_1) = \frac{1}{2}$.

As $f$ is continuous, $f(S_n) \to f(\frac{1}{2})$. As $S_n \in [0,1]$ compact, $f(S_n)$ is bounded by a constant (thus integrable on $[0,1]$). So by dominated convergence theorem, $\mathbb E [f(S_n)] \overset{a.s.}{\longrightarrow} \mathbb E [f(\frac{1}{2})] = f(\frac{1}{2})$.

For the second case, $\displaystyle (x_1\cdots x_n)^{1/n} = \exp(-\frac{1}{n} \sum_{i=1}^n \ln\frac{1}{x_i})$. Let $Y_i = \ln\frac{1}{X_i}$, then by strong law of large numbers, $\displaystyle V_n = \frac{1}{n}\sum_{i=1}^n Y_i \overset{a.s.}{\longrightarrow} \mathbb E(Y_1) = 1$. Again, as $f(e^{-V_n})$ is continuous and bounded by a certain constant (thus integrable on $[0,1]$), $\mathbb E [f(e^{-V_n})] \overset{a.s.}{\longrightarrow} \mathbb E [f(e^{-1})] = f(\frac{1}{e})$.

BONUS: using the same approach, if moreover, we restrict $f: [0,1] \to [0,1]$, then $$\lim_{n \to \infty} \int_0^1 ... \int_0^1 \frac{f(x_1)+...f(x_n)}{n}dx_1...dx_n = \int_0^1 f(x) dx$$

  • $\begingroup$ The 1st Q is equivalent to showing that for each $d>0$, the $n$-dimensional volume of $S_n(d)$ tends to $0$ as $n\to \infty,$ where $S_n(d)=$ $\{(x_1,...,x_n)\in [0,1]^n :$ $ |1/2-(1/n)\sum_{i=1}^nx_i|>d\}.$ Good answer. $\endgroup$ Apr 17, 2016 at 1:22
  • $\begingroup$ I do not understand your computation in the second case; are you sure you have the correct definition for $Y_i$? $\endgroup$
    – guest
    Apr 17, 2016 at 19:58
  • $\begingroup$ Ah yes, you are correct. It should be $\ln\frac{1}{X_i}$ instead. Corrected! $\endgroup$
    – SiXUlm
    Apr 17, 2016 at 20:08

For the second problem, recall that Weierstrass tells us that polynomials are dense in $C[0,1].$ So it's enough to prove the result for polynomials, and for this it's enough to prove it for each $x^k.$ For $x^k$ the $n$-fold integral works out nicely to be $1/(1+k/n)^n.$ This has limit $1/e^k = (1/e)^k,$ which is the desired answer for this function.

Weiersrtass can also be used for the first problem as well, although it's not quite as simple.


The first result can be shown introducing a Dirac delta (and its integral representation) into the game. We have

$$ \lim_{n\rightarrow \infty}\int_0^1\int_0^1 \cdots \int_0^1 f \left( \frac{x_1+x_2+\cdots +x_n}{n} \right)~dx_1~dx_2\cdots dx_n = $$ $$ \lim_{n\rightarrow \infty}\int_{\mathbb{R}} dt\ f(t)\int_0^1\int_0^1\cdots \int_0^1 \delta\left(t-\frac{1}{n}\sum_{k=1}^n x_k\right)~dx_1~dx_2\cdots dx_n= $$ $$ \lim_{n\rightarrow \infty}\int_{\mathbb{R}}dt\ f(t)\int_{\mathbb{R}}\frac{dz}{2\pi} e^{i tz} \int_0^1\int_0^1 \cdots \int_0^1\ e^{-iz\sum_{k=1}^n x_k/n} ~dx_1~dx_2\cdots dx_n= $$ $$ \lim_{n\rightarrow \infty}\int_{\mathbb{R}}dt\ f(t)\int_{\mathbb{R}}\frac{dz}{2\pi}e^{i tz}\left[\int_0^1 dx\ e^{-izx/n}\right]^n=\lim_{n\rightarrow \infty} \int_{\mathbb{R}}dt\ f(t)\int_{\mathbb{R}}\frac{dz}{2\pi}e^{i tz}\left[\frac{i n \left(-1+e^{-\frac{i z}{n}}\right)}{z}\right]^n $$ and using $$ \lim_{n\to\infty}\left[\frac{i n \left(-1+e^{-\frac{i z}{n}}\right)}{z}\right]^n=e^{-i z/2} $$ the result is $$ \int_{\mathbb{R}}dt f(t)\int_{\mathbb{R}}\frac{dz}{2\pi}e^{iz(t-1/2)} =\int_{\mathbb{R}}dt\ f(t)\delta(t-1/2)=f(1/2)\ . $$


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