If f_{n} \geq 0 a.e. and f_{n} \rightarrow f in

Gregory Jones

Gregory Jones

Answered question


If fn0 a.e. and fnf in measure then f0 a.e.

Answer & Explanation

Shannon Hodgkinson

Shannon Hodgkinson

Beginner2022-01-13Added 34 answers

{x:f(x)<0}k{x:fnk(x)<0}A where A is the set of points where (fnk(x)) does not converge to f(x)>


Beginner2022-01-14Added 38 answers

fn0 a.e., in other words, for each n=1,2,, there is a set Nn with |Nn|=0 and that fn(x)0 for all x(Nn)c.
Now the subsequence (fnk) is such that fnk(x)f(x) a.e., so a set N is such that fnk(x)f(x) for all xNc.
Now |nNnN|=0 and for all x(nNnN)c
we have fnk(x)0 and fnk(x)f(x), this implies f(x)0 and hence f0 a.e. The later almost everywhere is due to nNnN.

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