Assume that $f_n \to f$ almost everywhere, with $f_n$ integrable for all $n$ and $g$ is an integrable function such that $\lvert f_n \rvert \le g$.

(A) Then $f$ is integrable and -

$$\int f\,\mathrm d\mu = \lim_n \int f_n\,\mathrm d\mu.$$

And in fact a stronger condition (B) is true -

$$\int \left\lvert f_n - f \right\rvert\,\mathrm d\mu \to 0$$ as $n \to \infty$.

I have two questions on this.

  1. How does (B) imply (A)?
  2. How is (B) stronger than (A)?
  1. (B) implies (A) because for each $n$, $\left|\int f_n\mathrm d\mu-\int f\mathrm d\mu\right|\leqslant\int|f_n-f|\mathrm d\mu$ and the RHS goes to $0$.

  2. Consider $[-1,1]$ with Lebesgue measure, and $f_n:=n\chi_{(0,1/n)}-n\chi_{(-1/n,0)}$: then $\int f_n\mathrm d\lambda=0$, the pointwise limit is $0$, but $\int |f_n|\mathrm d\lambda=2$.

  • 1
    $\begingroup$ I see how 2. works but I don't see how that implies (B) is stronger than (A). I would have thought that to show something is stronger would mean showing that because (B) is stronger we can say some fact about some functions in the case of (B) that we can't say about (A)? $\endgroup$ – sonicboom Nov 22 '13 at 12:51
  • $\begingroup$ Yes, the fact is (B) itself. (A) doesn't imply (B). Davide Giraudo's example 2 illustrates this with $f=0$. Here the functions are going to 0 pointwise, and their integrals are all zero, thus $\lim_n \int f_n = \int f = 0$, but it is not the case that $\lim_n \int |f_n - f| = \lim_n\int |f_n| = 0$. $\endgroup$ – Ben Blum-Smith Mar 7 '14 at 19:46

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