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 &lt; ω &lt; 1 n 1 n ≤ ω &lt; 1I have to prove that f n converges μ-almost uniformly to f.From what I learned in class, I have to show: ∀ ϵ &gt; 0 , ∃ E ϵ ∈ A , μ ( E ϵ c ) &lt; 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 ) &lt; ϵ .To prove uniform convergence, I need: ∀ δ &gt; 0 , ∃ N δ ⟹ ∀ n &gt; N δ , ∀ ω ∈ E ϵ , | f n ( ω ) − f ( ω ) | &lt; 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 &lt; 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?

Question

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  &amp;lt;  ω  &amp;lt;      1    n        1    n    ≤  ω  &amp;lt;  1I have to prove that       f    n   converges   μ-almost uniformly to   f.From what I learned in class, I have to show:  ∀  ϵ  &amp;gt;  0  ,  ∃      E          ϵ        ∈      A    ,  μ  (      E          ϵ        c    )  &amp;lt;  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    )  &amp;lt;  ϵ  .To prove uniform convergence, I need:  ∀  δ  &amp;gt;  0  ,  ∃      N          δ          ⟹    ∀  n  &amp;gt;      N          δ        ,  ∀  ω  ∈      E          ϵ        ,      |        f    n    (  ω  )  −  f  (  ω  )      |    &amp;lt;  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  &amp;lt;  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?

Hayley Mccarthy · Accepted Answer

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      &amp;lt;      m      }        (  n  )                    →                    n        →        ∞              0. Notice this is valid for all   m  ∈      N  . Since   μ  (  (  0  ,  1      /    m  )  )  =  1      /    m, for arbitrary   ε  &amp;gt;  0 we can choose   m  &amp;gt;  1      /    ε and the condition is satisfied.

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