Limit of Integration of continuous function 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?
 A: 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.
