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How to evaluate the following limit- $$\lim_{n \to \infty} \int_0^1 \int_0^1\cdots\int_0^1 f \bigg(\frac{x_1 + x_2 + \cdots + x_n}{n} \bigg) dx_1 dx_2....dx_n$$.

Here $f()$ is a continuous function $f:[0,1] \to \mathbb{R}$. Is there any bounds for this integration or this will strictly evaluate to some value?

whuber
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edison
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    One method: use Chebyshev's Inequality to show that when $n$ is large enough, almost all the time the argument of $f$ is close to $1/2.$ It's unclear what you might mean by "any bounds," because you have explicitly given $[0,1]^n$ as the limits of the integral and obviously all these iterated integrals lie between the extremes of $f$ (which are finite because $f$ is continuous and $[0,1]$ is compact). – whuber Mar 22 '21 at 19:16
  • @whuber can you please explain how to apply Chebyshev's Inequality to prove the argument is close to 1/2. – edison Mar 22 '21 at 19:29
  • Because the edits by @Zhanxiong made the formula more readable, I have rolled the question back to that version. – whuber Mar 22 '21 at 20:09
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    @Zhanxiong I am new to stack exchange. I mistakenly clicked rollback. – edison Mar 22 '21 at 20:10
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    Given any $\epsilon\gt 0,$ Chebychev's Inequality assures us that for sufficiently large $n,$ the volume of the region where $|(x_1+x_2+\cdots+x_n)/n-1/2|\le \epsilon$ is very close to $1.$ Assuming $|f|$ is bounded on $[0,1],$ this implies the integral is an average of the values of $f$ within this neighborhood of $1/2$ plus a vanishingly small term. Thus, provided $f$ has a well-defined limiting average value at $1/2$ (which its continuity there assures), the integrals must be approaching that average. – whuber Mar 22 '21 at 20:44

1 Answers1

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Clearly, the integral can be rewritten as $E[f(Y_n)]$, where $Y_n = \frac{1}{n}(X_1 + \cdots + X_n)$, and $X_1, \ldots, X_n \text{ i.i.d.} \sim U(0, 1)$. By (weak) law of large numbers, we have $Y_n \to_d \frac{1}{2}$. This implies, by portmanteau lemma for any bounded and continuous function $h$, we have $E[h(Y_n)] \to E[h(1/2)] = h(1/2)$ as $n \to \infty$.

The $f$ in your problem satisfies the boundedness and continuous condition (extend its domain to $\mathbb{R}$ if you want more rigor), so the answer is $f(1/2)$.

$\newcommand{\eps}{\varepsilon}$

As @whuber suggested, there is a non-probabilistic argument, which goes as follows.

For arbitrary given $\eps > 0$, since $f$ is continuous at $1/2$, there exists $\delta > 0$ such that $|f(x) - f(1/2)| < \eps$ whenever $|x - 1/2| < \delta$. Also, since $f$ is continuous on $[0, 1]$, there exists $M > 0$ such that $|f| \leq M$ for all $x \in [0, 1]$. (Indeed, from the proof below we can see that the original continuity condition may be weakened to $f$ is continuous at $1/2$ and it is sequentially integrable).

For notational conciseness, for $0 \leq x_1, \ldots, x_n \leq 1$, denote $x_1 + \cdots + x_n$ by $s_n$, the region $[0, 1] \times \cdots \times [0, 1]$ by $V_n$, and the region $\{(x_1, \ldots, x_n): |n^{-1}s_n - 1/2| \geq \delta\}$ by $V_{n,\delta}$. Also denote $dx_1\cdots dx_n$ by $dx$, then \begin{align*} &\left|\int_{V_n} f(n^{-1}s_n) dx - f(1/2)\right| \leq \int_{V_n} |f(n^{-1}s_n) dx - f(1/2)| dx \\ =& \int_{V_{n, \delta}} |f(n^{-1}s_n) - f(1/2)| dx + \int_{V_{n, \delta}^c} |f(n^{-1}s_n) - f(1/2)| dx \\ <& \int_{V_{n, \delta}} |f(n^{-1}s_n) - f(1/2)| dx + \eps \leq 2M\int_{V_{n, \delta}} dx + \eps. \end{align*} by the setting up. So it remains to show the volume of $V_{n, \delta}$ can be made arbitrarily small when $n$ is sufficiently large.

To this end, Chebyshev's inequality (or repeating its proof essence) implies that \begin{align*} & \int_{V_{n, \delta}} dx \leq \delta^{-2}\int_{V_n}(n^{-1}s_n - 1/2)^2 dx \\ =& \delta^{-2}\int_{V_n}\left(n^{-2}s_n^2 - n^{-1}s_n + \frac{1}{4}\right) dx \\ =& \delta^{-2}\left(n^{-2}\sum_{i = 1}^n \int_{V_n}x_i^2 dx + 2n^{-2}\sum_{1 \leq i < j \leq n}\int_{V_n}x_ix_j dx - n^{-1}\sum_{i = 1}^n\int_{V_n}x_i dx + \frac{1}{4}\right) \\ =& \delta^{-2}\left(\frac{1}{3n} + 2n^{-2} \times \frac{n(n - 1)}{2} \times \frac{1}{4} - n^{-1}\times \frac{n}{2} + \frac{1}{4}\right) \\ =& \frac{1}{12n\delta^2} \to 0 \end{align*} as $n \to \infty$, and this is what we want to show.

Zhanxiong
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