Show that $∫_E f=\lim ∫_E f_k $For any measurable set E .

Question

assume $f_1,f_2,...$ are nonnegative functions in $L^1$, that $\lim f_k(x)=f(x)$ exists a.e., that $f \in L^1$, and that $\lim ∫f_K=∫f$ show that $\lim ∫|f_k-f|=0$.

show that this implies that for every measurable set $E$ $$ ∫_E f= \lim∫_E f_k .$$

I showed the first part with LDCT, but I don't know why that implies the other. Help!

why $|∫_E f_k - ∫_E f| \leq ∫ |f_k-f| \rightarrow 0$ implies the second part? in link @ε-δ or Kavi Rama Murthy — Voldemort, May 21 '20 at 23:24

score 1 · Answer 1 · answered May 21 '20 at 18:13

$f_n+ f-|f-f_n|\geq0$ and by Fatou's lemma \begin{aligned} \int 2f\,d\mu&\leq \liminf_n \int (f_n+f -|f-f_n|)\,d\mu\\ &\leq 2\int f\,d\mu +\liminf_n\Big(-\int|f-f_n|\,d\mu\Big)\\ &\leq 2\int f\,d\mu -\limsup_n\int|f-f_n|\,d\mu \end{aligned} This shows that $\limsup_n\int|f-f_n|\,d\mu=0$

Show that $∫_E f=\lim ∫_E f_k $For any measurable set E .

1 Answers1