Comparison of the consequences of uniform convergence between the real and complex variable cases, In the real variable case, I think that uniform convergence preserves continuity and integrability, i.e., for an integral of a sequence of continuous (or integrable) functions, which converge uniformly to some function over a set E, then we know that this function is continuous (or integrable) -- and then the limit of the integrals is equal to the integral of the limit function.  (the stronger version of this integration theorem, the dominated convergence theorem, only requires pointwise convergence of the functions, in order to take the limit inside the integral.)
What else can we get from uniform convergence in the real variable case?  Does it preserve differentiability?  Or, in general it does not?
And, I think in the complex variable setting, uniform convergence preserves all of continuity, integrability, and differentiability.
...anything else to be aware of?
Thanks in advance,

You need stronger conditions than uniform convergence to ensure that 
$$f_n(x) \to f(x) \implies f_n'(x) \to f'(x).$$
Here is a standard theorem found in virtually all real analysis books. 
Suppose $(f_n)$ is a sequence of differentiable functions that converges pointwise at some point in $[a,b]$ and $(f_n')$ converges uniformly on $[a,b]$ to $g.$ Then $(f_n)$ converges uniformly to a differentiable function $f$ and $f_n'(x) \to g(x)$ for all $x \in [a,b].$
The following is an example where uniform convergence of $f_n \to f$ alone does not ensure $f_n' \to f'$.
Consider the sequence of functions $f_n:[a,b] \to \mathbf{R}$ with 
$$f_n(x) = \frac{x}{1+nx^2}.$$
It's easy to show that $f_n$ has a maximum at $x = 1/\sqrt{n}$. Hence, on $[0,1]$,
$$f_n(x) \leqslant f_n\left(\frac1{\sqrt{n}}\right)= \frac1{2\sqrt{n}}.$$
Consequently  $f_n(x) \rightarrow f(x) \equiv 0$ uniformly on $[0,1]$. 
Also, $f_n$ is differentiable with 
$$f'_n(x) = \frac{1-nx^2}{(1+nx^2)^2}.$$
However, $f'_n(0) \rightarrow 1 \neq f'(0).$
