Assume the domain is bounded; we're talking about definite integrals. For convenience, let the domain have a fixed volume of v, as is the case in real or complex vector space.
Again, without uniform convergence the integral of g may not be the limit of the integrals, as shown by fn(x) = 2nx(1-x2)N on the closed interval [0,1]. The nth integral is n/(n+1), with limit 1. Yet the limit function g is 0, with integral 0.
Now assume uniform convergence and watch what happens. To start, let fn and g map into the real numbers. In most applications, each function fn is continuous, hence g is continuous, hence integrable. Choose n so that fn() is within ε/v of g throughout the entire domain. We can do this, thanks to uniform convergence. Now the integral of g-fn is less than ε, hence the integrals of fn approach the integral of g.
If the range is complex, or any other real vector space, uniform convergence propagates down to the component functions. Select n so that all component functions are within ε/v of their limits simultaneously. Now all integrals are within ε of their limits, and the limit of the integrals is again the integral of the limit function.
If f[n] represents a series of functions, rather than a sequence, the integral of the sum is the sum of the integrals, provided the partial sums exhibit uniform convergence. We can often show this using the Weierstrass m test.
Let fn remain nonnegative throughout its domain, and for every x in the domain, fn(x) monotonically increases as it approaches g(x). This is the case when f represents the partial sums of a series of nonnegative functions. For simplicity, let the domain be the unit interval [0,1]. Further assume fn is uniformly convergent on every interval [0,t], for t < 1. Therefore the integral of the limit function g is the limit of the integrals when the domain is restricted to [0,t]. Let's take the limit as t approaches 1.
The integral of g from 0 to t becomes the integral of g on [0,1]. This is a theorem if g is bounded, and it is the definition of "integral" if g is unbounded at 1.
Assume the indefinite integral exists, and is equal to c. The integrals of fn on [0,1] form an increasing sequence that is bounded by c, so let b be the least upper bound. If b = c we are done. If b < c, choose t so that the integral of g from 0 to t exceeds b. Then choose n so that the integral of fn from 0 to t is between b and c. Since fn is never negative, the integral of fn from 0 to 1 is even larger; hence it exceeds b. This is a contradiction, so the limit of the integrals is c.
Conversely, assume the limit of the integrals is b. If the integral of g is unbounded as t approaches 1, choose t so that the integral of g from 0 to t exceeds b. Call it c. Then choose n so that the integral of fn from 0 to t is in between b and c. The integral from 0 to 1 is even larger, which contradicts the fact that all integrals are bounded by b. Therefore the integral of g is well defined. Let this integral be c and invoke the previous paragraph, which asserts b = c. We can exchange limit and integration in this case.