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anonymous
 5 years ago
given assumptions:{f_n} n=1 to infinty be a sequence of realvalued functions defined on [a,b] and suppose fn of functions converges uniformly on E to a function f which is defined on [a,b].fn is integrable on [a,b] then f is also integrable and limn>infnty[integral a to b fn(x)dx]=integral a to b(f(x)dx).now i am writing how i solved:since the sequence fn(x) is uniformly convergent to f i.e fnf <epsilon then integral{a to x}(fnf) <integral{a to x}(epsilon ) because of the absolute value <=0 (left side) and <epsilon(xa) <=0 for the rhs since x in [a,b]now taking limit of both sid
anonymous
 5 years ago
given assumptions:{f_n} n=1 to infinty be a sequence of realvalued functions defined on [a,b] and suppose fn of functions converges uniformly on E to a function f which is defined on [a,b].fn is integrable on [a,b] then f is also integrable and limn>infnty[integral a to b fn(x)dx]=integral a to b(f(x)dx).now i am writing how i solved:since the sequence fn(x) is uniformly convergent to f i.e fnf <epsilon then integral{a to x}(fnf) <integral{a to x}(epsilon ) because of the absolute value <=0 (left side) and <epsilon(xa) <=0 for the rhs since x in [a,b]now taking limit of both sid

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