Let f n and f functions of L 1 R n such that lim n → ∞ f n = f almost always in R n and ∫ R n | f n | d x → ∫ R n | f | d x. Show that for every set A ⊂ R n lebesgue measurable ∫ A f n d x → ∫ A f d x .How can I solve this problem?

Question

Let       f    n   and   f functions of       L    1              R        n   such that       lim          n      →      ∞            f    n    =  f almost always in             R        n   and       ∫                            R                n                  |        f    n        |       d  x  →      ∫                            R                n                  |    f      |       d  x. Show that for every set   A  ⊂            R        n   lebesgue measurable      ∫          A            f    n       d  x  →      ∫          A        f     d  x  .How can I solve this problem?

engaliar0l · Accepted Answer

Note that                                          |                          f                      n                                    |                                                  =                              |                          (                      f                          n                                −          f          )                +        f                  |                                                  ≤                              |                      f                          n                                −          f          |                +                  |          f          |                .            Therefore,       |        f          n            |    −      |    f      |    −      |        f          n        −  f      |    ≤  0  ≤  2      |    f      |  . On the other hand,                                          |                      f                          n                                −          f          |                                                  ≤                              |                          f                      n                                    |                +                  |                f                  |                ,            re-arranging terms yields       |        f          n            |    −      |    f      |    −      |        f          n        −  f      |    ≥  −  2      |    f      |  . It follows that      |          |              f              n                    |        −          |        f          |        −          |              f              n              −    f          |        |    ≤  2      |    f      |    .Observe that       |        f          n            |    −      |    f      |    −      |        f          n        −  f      |    →  0 pointwisely a.e.. By Lebesgue Dominated Convergence Theorem, we have that  ∫      (          |              f              n                    |        −          |        f          |        −          |              f              n              −    f          |        )    d  m  →  0.It is given that   ∫      (          |              f              n                    |        −          |        f          |        )    d  m  →  0, so   ∫      |        f          n        −  f      |    d  m  →  0. In particular, for any measurable set   A,       |          ∫              A                    f              n              d    m    −          ∫              A              f    d    m    |    ≤      ∫          A            |        f          n        −  f      |    d  m  ≤  ∫      |        f          n        −  f      |    d  m  →  0

EnvivyEvoxys6 · Accepted Answer

Here is a hint to get you started: Apply Fatou&#039;s lemma to      |        f    n        |    +      |    f      |    −      |        f    n    −  f      |    ≥  0to find       f    n    →  f in       L    1  .

Let f n </msub> and f functions of L 1 </msub>

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