I'm wondering whether the following statement holds:Let f n , f : R → R 0 + be functions with ∫ f n ( x ) d x = ∫ f ( x ) d x = 1 and for every bounded and countinuous function g : R → R the following integral-convergence ∫ g ( x ) ⋅ f n ( x ) d x → n ∫ g ( x ) ⋅ f ( x ) d xholds. Then it follows that f n → f almost everywhere.Intuitively the statement looks false, but I can't find a counterexample. If it doesn't hold: changes the further assumption that the f n , f have to be continuous anything?Kind regards

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I&#039;m wondering whether the following statement holds:Let       f    n    ,  f  :      R    →            R        0    +   be functions with   ∫      f    n    (  x  )  d  x  =  ∫  f  (  x  )  d  x  =  1 and for every bounded and countinuous function   g  :      R    →      R   the following integral-convergence  ∫  g  (  x  )  ⋅      f    n    (  x  )  d  x      →    n    ∫  g  (  x  )  ⋅  f  (  x  )  d  xholds. Then it follows that       f    n    →  f almost everywhere.Intuitively the statement looks false, but I can&#039;t find a counterexample. If it doesn&#039;t hold: changes the further assumption that the       f    n    ,  f have to be continuous anything?Kind regards

Franco Cohen · Accepted Answer

A classical counterexample is      f    n    (  x  )  :=            (        1  +  sin  ⁡  (  2  π  n  x  )            )                  1              {      x      ∈      (      0      ,      1      )      }        ,and  f  (  x  )  :=            1              {      x      ∈      (      0      ,      1      )      }        .We have   ∫      f    n    =  ∫  f  =  1 and, for every bounded continuous   g  :      R    →      R  ,   ∫      f    n    g  →  ∫  f  g as   n  →  ∞ (e.g., by Riemann–Lebesgue lemma). However,   (      f    n        )          n      ≥      1       does not converge at all.

Bruno Pittman · Accepted Answer

Riemann-Lebesgue lemma gives you directly that       ∫    0    1    g  (  x  )  sin  ⁡  (  2  π  n  x  )        d    x  →  0 as   n  →  ∞

I'm wondering whether the following statement holds: Let f n </msub> , f

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