If fn≥0 a.e. and fn→f in measure then f≥0 a.e.

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Shannon Hodgkinson · Accepted Answer

{x:f(x)&amp;lt;0}⊆∪k{x:fnk(x)&amp;lt;0}∪A where A is the set of points where (fnk(x)) does not converge to f(x)&amp;gt;

scomparve5j · Accepted Answer

fn≥0 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 x∈Nc.  Now |∪nNn∪N|=0 and for all x∈(∪nNn∪N)c  we have fnk(x)≥0 and fnk(x)→f(x), this implies f(x)≥0 and hence f≥0 a.e. The later almost everywhere is due to ∪nNn∪N.

If fn≥0 a.e. and fn→f in measure then f≥0 a.e.

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