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

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

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.

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

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.

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

Gregory Jones

Answered question

2022-01-12

f_{n} \geq 0

a.e. and

f_{n} \to f

in measure then

f \geq 0

a.e.

Answer & Explanation

Shannon Hodgkinson

Beginner2022-01-13Added 34 answers

{x : f (x) < 0} \subseteq \cup_{k} {x : f_{n_{k}} (x) < 0} \cup A

where A is the set of points where

(f_{n_{k}} (x))

does not converge to

f (x) >

scomparve5j

Beginner2022-01-14Added 38 answers

f_{n} \geq 0

a.e., in other words, for each

n = 1, 2, \dots,

there is a set

N_{n} with | N_{n} | = 0

and that

f_{n} (x) \geq 0 for all x \in {(N_{n})}^{c}

.
Now the subsequence

(f_{n_{k}})

is such that

f_{n_{k}} (x) \to f (x)

a.e., so a set N is such that

f_{n_{k}} (x) \to f (x)

for all

x \in N^{c}

.
Now

| \cup_{n} N_{n} \cup N | = 0

and for all

x \in {(\cup_{n} N_{n} \cup N)}^{c}

we have

f_{n_{k}} (x) \geq 0 and f_{n_{k}} (x) \to f (x)

, this implies

f (x) \geq 0

and hence

f \geq 0

a.e. The later almost everywhere is due to

\cup_{n} N_{n} \cup N

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