Miscellany

If f_n --> f a.e. and f_n --> g a.e. then f = g almost everywhere.   Proof:  f_n --> f for all x except in a set of measure zero K1 and f_n --> g for all x except in a set of measure zero K2.  So for x not in (K1 Union K2) (a set of measure zero), f = g.  So f=g a.e.  Would this be a valid proof or am I oversimplifying the problem?