Lebesgue's Dominated Convergence Theorem questions

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)?
• – derivative Nov 21 '13 at 20:33

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 $\int |f_n|\mathrm d\lambda=2$.
• 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 $\lim_n \int |f_n - f| = \lim_n\int |f_n| = 0$. – Ben Blum-Smith Mar 7 '14 at 19:46