Measure Theory - Is my proof too complicated?

Question

Let (X, M, m) be a mesure space and let f_n be a sequence of integrable complex valued functions that converges uniformly to a function f. If m(X) < oo, then, f is integrable and lim Int f_n dm = Int f dm. I found a somewhat different proof for this, but some pepole says its too complicated. Do u agree?
Fix some eps >0. Since f_n --> f uniformly, there exists k such that n >= k => |f_n - f_k| < eps => |f_n| < |f_k| + eps, which implies the k tail of f_n is dominated by |f_k| + eps. Since |f_k| is integrabel (because f_k is) and m(X) < oo, then Int (|f_k| + eps) dm = Int |f_k| dm + eps m(X) < oo, so that |f_k| + eps is integrable. Since the k tail of f_n also converges to f, it follws from the Dominated Convergence Theorem that f is integrable and lim(n >= k) f_n dm = Int f dm, which shows the k tail of Inf f_n dm --> Int f dm. Hence, lim Int f_n dm = Int f dm as stated.

There's another proof for this theorem that doesnt rely on the D. C. Theorem.

Answer 1

No, your proof is not too complicated, and is almost exactly the proof that I would use.

Answer 2

I would use your reasoning to prove the case f=0, then the result trivially generalizes to any uniform limit of integrable complex valued functions. This immediately simplifies the proof.