Let ( X , μ ) be a probability measure space. Suppose f n → f pointwise and f n is dominated by some μ-integrable function. Let ( A n ) n be a sequence of measurable sets converging to some measurable S with μ ( S ) = 1.Question. Is it true that ∫ X f n 1 A n d μ → ∫ X f d μ ?A half-baked ideaDefine g n := f n 1 A n . In case,Conjecture. 1 A n → 1 S pointwise,which I suspect to be true, then | g n | is dominated by a μ-integrable function and so the Dominated Convergence Theorem gaurantees that ∫ X g n d P → ∫ X f 1 S d P = ∫ X f d P since P ( S ) = 1.

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Let   (  X  ,  μ  ) be a probability measure space. Suppose       f    n    →  f pointwise and       f    n   is dominated by some   μ-integrable function. Let   (      A    n        )    n   be a sequence of measurable sets converging to some measurable   S with   μ  (  S  )  =  1.Question. Is it true that       ∫    X        f    n        1                  A        n                d  μ  →      ∫    X    f    d  μ ?A half-baked ideaDefine       g    n    :=      f    n        1                  A        n            . In case,Conjecture.       1                  A        n              →      1    S   pointwise,which I suspect to be true, then       |        g    n        |   is dominated by a   μ-integrable function and so the Dominated Convergence Theorem gaurantees that       ∫    X        g    n      d  P  →      ∫    X    f      1    S      d  P  =      ∫    X    f    d  P since   P  (  S  )  =  1.

radcas87gex5r · Accepted Answer

Another way to show this is a   2  ϵ-argument. First note that for any integrable function   f, we have that      ∫    X    f  d  μ  =      ∫    S    f  d  μ  +      ∫          X      ∖      S        f  d  μ  =      ∫    S    f  d  μ  .As   X  ∖  S has measure zero. Then we have that                              |                          ∫          X                          f          n                          1                                    A              n                                      d        μ        −                  ∫          X                f        d        μ                  |                                    =                  |                          ∫          X                          f          n                          1                                    A              n                                      d        μ        −                  ∫          X                f                  1                      S                          d        μ                  |                                                  =                  |                          ∫          X                          f          n                          1                                    A              n                                      d        μ        −                  ∫          X                f                  1                                    S              n                                      d        μ        +                  ∫          X                          f          n                          1                      S                          d        μ        −                  ∫          X                          f          n                          1                      S                          d        μ                  |                                                  ≤                  |                          ∫          X                          f          n                          1                                    A              n                                      d        μ        −                  ∫          X                          f          n                          1                      S                                    |                +                  |                          ∫          X                f                  1                      S                          d        μ        −                  ∫          X                          f          n                          1                      S                          d        μ                  |                                                  ≤                  |                          ∫          X                          f          n                (                  1                                    A              n                                      −                  1                      S                          )        d        μ                  |                +                  |                          ∫          X                (                  f          n                −        f        )                  1                      S                          d        μ                  |                                                  ≤        μ        (                  A          n                −        S        )        ⋅                  |                          |                          f          n                          |                                      |                                ∞                          +                  |                          |                          1          S                          |                                      |                                ∞                                    |                          |                          f          n                −        f                  |                                      |                                              L              1                                      →        0            As       f    n    →  f pointwise and   f is uniformely bounded, and       A    n    →  S.

Let ( X , μ ) be a probability measure space. Suppose f

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