I have the following task from my textbook:Let { f n } n ∈ N be a sequence of measurable function on M with f n → f a.s., where f is also a measurable function. Here, a.s. means almost surely (= almost everywhere).Show: if there exists a nonnegative measurable function g satisfying the following conditions: | f n | ≤ g a.s. for all n ∈ N and ∫ M g p d μ &lt; ∞ for one p &gt; 0 ,then ∫ M | f n − f | p d μ → 0 for n → ∞ .These notations remind me a little bit of the Hölder inequality.What I have got:I thought that we have | f n | p ≤ g p a . s .Thus ∫ M | f n | p d μ ≤ ∫ M g p d μ ≤ i n f t yAnd as f n → fthen the rest follows?

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I have the following task from my textbook:Let             {              f                  n                    }              n      ∈              N             be a sequence of measurable function on M with      f          n        →  f   a.s.,  where f is also a measurable function. Here, a.s. means almost surely (= almost everywhere).Show: if there exists a nonnegative measurable function g satisfying the following conditions:      |          f              n              |    ≤  g   a.s. for all   n  ∈      N  and      ∫          M            g          p        d  μ  &amp;lt;  ∞   for one   p  &amp;gt;  0  ,then      ∫          M                  |              f                  n                    −      f      |              p        d  μ  →  0   for   n  →  ∞  .These notations remind me a little bit of the Hölder inequality.What I have got:I thought that we have      |        f    n              |        p    ≤      g    p    a  .  s  .Thus      ∫    M        |        f    n              |        p    d  μ  ≤      ∫    M        g    p    d  μ  ≤  i  n  f  t  yAnd as      f    n    →  fthen the rest follows?

lisicw2 · Accepted Answer

For   p  &amp;gt;  1  ,, the function   x  ↦      x    p   is convex so      |        f    n    −  f            |        p    =      2    p              |                        f          n                2            +                        (          −          f          )                2            |        p    ≤      2          p      −      1        (      |        f    n              |        p    +      |    f            |        p    )so      2          p      −      1        (      |        f    n              |        p    +      |    f            |        p    )  −      |        f    n    −  f            |        p    ≥  0.In the standard way that dominated convergence theorem is proven, just apply Fatou&#039;s lemma.

I have the following task from my textbook: Let { f <

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