Suppose E is a measurable set and m ( E ) &lt; + <mi mathvariant="normal

tripes3h

tripes3h

Answered question

2022-07-12

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Suppose E is a measurable set and m ( E ) < + . { f k } is a sequence of non-negative measurable functions. If
lim k + E f k ( x ) 1 + f k ( x ) d x = 0 ,
prove that f k converge to 0 in measure.

If we can prove f k ( x ) 1 + f k ( x ) 0 a.e. x E, then the problem is easy to solve. Is it correct? If so, how to prove that?
Appreciate any hint or help!

Answer & Explanation

Nathen Austin

Nathen Austin

Beginner2022-07-13Added 14 answers

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document E f k ( x ) 1 + f k ( x ) d x E ( ϵ ) f k ( x ) 1 + f k ( x ) d x
where E ( ϵ ) = { x E : f k ( x ) > ϵ }. Hence E f k ( x ) 1 + f k ( x ) d x ϵ 1 + ϵ m ( E ( ϵ ) ). This shows that m ( E ( ϵ ) ) 0 as k for any ϵ > 0 which means f k 0 in measure on E.
I have used the fact that y 1 + y is an increasing function on ( 0 , ) so y > ϵ implies y 1 + y > ϵ 1 + ϵ .

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