TITLE: Integrable functions converge uniformly to integrable function
QUESTION [3 upvotes]: This proof is probably wrong but I can't see why. 
Let X be a finite measure space.  
If integrable Fn converge uniformly to F then F is integrable (lebasgue wise). 
Proof: since the functions are integrable, each is infinite only on a null set, take the union of the null sets (which gives us a null set, since n is countable)  hence G(x) = sup{ |Fn(x)|} is integrable since it only equals infinity on a null set. 
By the domination theorem on G the result follows.

REPLY [3 votes]: Let $\varepsilon>0$ be given and let $n$ be so large that for all $x\in X,$ $|F_{n}(x)-F(x)|<\varepsilon,$ the existence of which is guaranteed by uniform convergence. $F$ is the uniform limit of integrable, and therefore measurable, functions, so it is measurable. Using the triangle inequality, we have: $$\int_{X}|F(x)|\mu(\mathrm{d}x)\leq \int_{X} |F_{n}(x)|+|F_{n}(x)-F(x)|\mu(\mathrm{d}x)< \int_{X} |F_{n}(x)|\mu(\mathrm{d}x)+\varepsilon\mu(X)<+\infty,$$ since $F_{n}$ is integrable and $X$ is a finite measure space.
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