Limit of integration over a cube in $\mathbb{R}^{\infty}$ I want to compute the integral
$$
I := \lim_{n\to\infty}\int_{[0,1]^n}n\left(f(g(x)) - f\left(\frac{1}{2}\right)\right)dx,
$$
where $f:[0,1]\to\mathbb{R}$ is bounded, $f\in C^3([0,1])$, and $g(x_1,x_2,...,x_n)=\frac{1}{n}\sum_{1}^n x_i$ Below is my attempt:

Let $X_1,...,X_n$ be random variables with $X_i\sim \text{Unif}[0,1]$ for all $i$. Then the integral $$ \lim_{n\to\infty}\int_{[0,1]^n} nf(g(x))dx$$ may be written as$$n\cdot\mathbb{E}\left[ f\left(\frac{1}{n}\sum_1^n X_i  \right) \right].$$
By SLLN we have $(1/n)\sum_1^n X_i\to 1/2$ a.s. By continuity of $f$ we also have that $$f\left(\frac{1}{n}\sum_1^n X_i  \right)\to f(1/2).$$

But I am not sure how to proceed. How am I supposed to handle the $n$ outside the integrand? Any of your help will be greatly appreciated! :)
 A: I'll write $g_n := \frac 1n \sum_1^n X_i$.
By Taylor's theorem, $f(g_n)-f\left(\frac 12 \right) = f'\left(\frac 12 \right)\left(g_n - \frac 12\right) + \frac 12 f''\left(\frac 12 \right)\left(g_n - \frac 12\right)^2 + \frac 16 f'''\left(K(x) \right)\left(g_n - \frac 12\right)^3$ for some $K(x)$.  Hence
\begin{align*}
n\mathbb{E}\left[f(g_n)-f\left(\frac 12 \right)\right] &=nf'\left(\frac 12 \right)\mathbb{E}\left[g_n - \frac 12\right]\\
& \qquad + \frac n2 f''\left(\frac 12 \right)\mathbb{E}\left[\left(g_n - \frac 12\right)^2\right] \\
& \qquad + \frac 16 \mathbb{E}\left[f'''\left(K(x) \right)\left(g_n - \frac 12\right)^3\right] \\
&=\frac n2 f''\left(\frac 12 \right)\mathbb{E}\left[\left(g_n - \frac 12\right)^2\right] \\
& \qquad + \frac n6 \mathbb{E}\left[f'''\left(K(x) \right)\left(g_n - \frac 12\right)^3\right]
\end{align*}
because $\mathbb{E}[g_n] = \frac 12$.  For the first term,
\begin{align*}
\mathbb{E}\left[\left(\frac 1n \sum_1^nX_i - \frac 12\right)^2\right] &= \frac 1{n^2} \mathbb{E}\left[ \sum_1^n\left(X_i - \frac 12\right)^2\right] \\
&= \frac{1}{n^2} \sum_1^n \mathbb{E}\left[\left(X_i - \frac 12\right)^2\right] \\
&= \frac{1}{12 n}.
\end{align*}
For the second term, using the boundedness of $f'''$ we have
\begin{align*}
\mathbb{E}\left[\left|f'''\left(K(x) \right)\left(g_n - \frac 12\right)^3\right|\right] &\le C \mathbb{E}\left[\left|\frac 1n \sum_1^n X_i - \frac 12\right|^3\right] \\
&\le \frac{C}{n^3}\sum_1^n \mathbb{E}\left[\left|X_i-\frac 12\right|^3\right] \\
&= \frac{C}{32 n^2},
\end{align*}
so $$\lim_{n \rightarrow \infty}\frac n6 \mathbb{E}\left[f'''\left(K(x) \right)\left(g_n - \frac 12\right)^3\right] = 0.$$
Therefore
$$\lim_{n \rightarrow \infty} n\mathbb{E}\left[f(g_n)-f\left(\frac 12 \right)\right] = \lim_{n \rightarrow \infty} \frac n2 f''\left(\frac 12 \right)\mathbb{E}\left[\left(g_n - \frac 12\right)^2\right] = \frac{1}{24} f''\left(\frac 12 \right). $$
