Let f n : R → R be defined by f n ( ω ) = ω ( 1 + 1 n ) and let μ be the Lebesgue measure.I have to show that f n does not converge in measure μ to ω.Here's my attempt: By definition, f n converges in measure to ω if: lim n → ∞ μ ( ω : | f n ( ω ) − ω | ≥ δ ) = 0. Rewriting, it becomes that: lim n → ∞ μ ( ω : | ω | ≥ n ⋅ δ ) = 0.I realized that if δ = 1 for example, then the measure is surely 0 as n → ∞. I need to come up with a value of δ such that the measure is non-zero. I thought about letting δ = 1 n ,, but am unsure if it'll become indeterminate once I set n → ∞ (because I basically get ∞ ⋅ 1 ∞ ) .Does this solution of δ = 1 n truly work as a counterexample? If not, what should I do?

Question

Let       f    n    :      R    →      R   be defined by       f    n    (  ω  )  =  ω  (  1  +      1    n    ) and let   μ be the Lebesgue measure.I have to show that       f    n   does not converge in measure μ to ω.Here&#039;s my attempt: By definition,       f    n   converges in measure to   ω if:            lim              n        →        ∞              μ    (    ω    :          |              f      n        (    ω    )    −    ω          |        ≥    δ    )    =    0.  Rewriting, it becomes that:      lim          n      →      ∞        μ  (  ω  :      |    ω      |    ≥  n  ⋅  δ  )  =  0.I realized that if   δ  =  1 for example, then the measure is surely 0 as   n  →  ∞. I need to come up with a value of   δ such that the measure is non-zero. I thought about letting   δ  =      1    n    ,, but am unsure if it&#039;ll become indeterminate once I set   n  →  ∞ (because I basically get   ∞  ⋅      1    ∞    )  .Does this solution of   δ  =      1    n   truly work as a counterexample? If not, what should I do?

candelo6a · Accepted Answer

It doesn&#039;t matter what   δ is.   μ  (  ω  :      |    ω      |    ≥  n  ⋅  δ  )  =  ∞ for every   n and every   δ.  {  ω  :      |    ω      |    ≥  n  ⋅  δ  }  =  [  n  δ  ,  ∞  )  ∪  (  −  ∞  ,  −  n  δ  ]

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