No problem giving a complete answer. Let's see
As already stated, the sample mean is a function of many random variables, and so the symbol $E$ refers to the expected value with respect to their joint distribution. Denoting $\mathbf X$ the multivariate vector of the $n$ r.v.'s, their joint density can be written as $f_{\mathbf X}(\mathbf x)= f_{X_1,...,X_n}(x_1,...,x_n)$ and their joint support
$D = S_{X_1} \times ...\times S_{X_n}$
The sample mean is a function of this multivariate vector, $\bar X = \frac 1n \sum_{i=1}^{n}X_i = g(\mathbf X)$. Using the Law of Unconcscious Statistician We have
$$E[\frac 1n \sum_{i=1}^{n}X_i] = \int_D g(\mathbf x)f_{\mathbf X}(\mathbf x)d\mathbf x$$.
Under convergence regularity conditions we can decompose the multidimensional integral into an n-iterative integral:
$$E[\frac 1n \sum_{i=1}^{n}X_i] = \int_{S_{X_n}}...\int_{S_{X_1}}\left[\frac 1n \sum_{i=1}^{n}x_i\right]f_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_n $$
and using the linearity of integrals we can decompose into
$$ = \frac 1n\int_{S_{X_n}}...\int_{S_{X_1}}x_1f_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_n \; + ...\\ ...+\frac 1n\int_{S_{X_n}}...\int_{S_{X_1}}x_nf_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_n $$
For each n-iterative integral we can re-arrange the order of integration so that, in each, the outer integration is with respect to the variable that is outside the joint density. Namely,
$$\frac 1n\int_{S_{X_n}}...\int_{S_{X_1}}x_1f_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_n = \\\frac 1n\int_{S_{X_1}}x_1\int_{S_{X_n}}...\int_{S_{X_2}}f_{X_1,...,X_n}(x_1,...,x_n)dx_2...dx_ndx_1$$
and in general
$$\frac 1n\int_{S_{X_n}}...\int_{S_{X_j}}...\int_{S_{X_1}}x_jf_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_j...dx_n =$$
$$=\frac 1n\int_{S_{X_j}}x_j\int_{S_{X_n}}...\int_{S_{X_{j-1}}}\int_{S_{X_{j+1}}}...\int_{S_{X_1}}f_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_{j-1}dx_{j+1}......dx_ndx_j$$
As we calculate one-by-one the integral in each n-iterative integral (starting from the inside), we "integrate out" a variable and we obtain in each step the "joint-marginal" distribution of the other variables. Each n-iterative integral therefore will end up as $\frac 1n\int_{S_{X_j}}x_jf_{X_j}(x_j)dx_j$.
Bringing it all together we arrive at
$$E[\frac 1n\sum_{i=1}^{n} X_i ] = \frac 1n\int_{S_{X_1}}x_1f_{X_1}(x_1)dx_1 +...+\frac 1n\int_{S_{X_n}}x_nf_{X_n}(x_n)dx_n $$
But now each simple integral is the expected value of each random variable separately, so
$$= E[\frac 1n\sum_{i=1}^{n} X_i ] = \frac 1nE(X_1) + ...+\frac 1nE(X_n) = \frac 1nE(X) + ...+\frac 1nE(X)$$
$$= \frac 1n nE(X) = E(X)$$