Given Ω = ( 0 , 1 ), A is the Borel Sigma-Algebra restricted to...

glitinosim3

glitinosim3

Answered

2022-07-04

Given Ω = ( 0 , 1 ), A is the Borel Sigma-Algebra restricted to Ω, and μ is the Lebesgue measure, let f ( ω ) = 1 ω and
0 < ω < 1 n
1 n ω < 1
I have to prove that f n converges μ-almost uniformly to f.
From what I learned in class, I have to show:
ϵ > 0 , E ϵ A , μ ( E ϵ c ) < ϵ f n  converges uniformly to  f  on  E ϵ .
My idea was to let E ϵ = [ ϵ 0.000000001 , 1 ). Then E ϵ c = ( 0 , ϵ 0.000000001 ) μ ( E ϵ c ) < ϵ .
To prove uniform convergence, I need:
δ > 0 , N δ n > N δ , ω E ϵ , | f n ( ω ) f ( ω ) | < δ .
Since E ϵ = [ ϵ 0.000000001 , 1 ) , ω ϵ 0.000000001, so if ϵ 0.000000001 1 n , we get | f n ( ω ) f ( ω ) | = 0 < δ .
This holds for all n 1 ϵ 0.000000001 , so we let N δ = 1 ϵ 0.000000001 , proving the almost uniform convergence.
Is this argument correct?

Answer & Explanation

Hayley Mccarthy

Hayley Mccarthy

Expert

2022-07-05Added 19 answers

You can use the configuration of uniform convergence with the sup distance. Consider the partition E m c = ( 0 , 1 / m ) , E m = [ 1 / m , 1 ) for m N . We have that on E m c the convergence is not uniform, but on E m it is as we have sup ω E m | f n ( ω ) f ( ω ) | = ( m 1 ) 1 { n : n < m } ( n ) n 0. Notice this is valid for all m N . Since μ ( ( 0 , 1 / m ) ) = 1 / m, for arbitrary ε > 0 we can choose m > 1 / ε and the condition is satisfied.

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