In my homework I am trying to show that for a measurespace $(X,\mathcal{A},\mu)$ and $u \in \mathcal{L}^1$ then:
$\lim_{n\rightarrow \infty} \int_{A_n} u\,d\mu=\int_X u\, d\mu$ where $A_n=\{|u|\leq n\}$ for all $n\geq 1$
My idea was to use the dominated convergence theorem:
So that $|u_n(x)| \leq w(x)$ where $w(x)=n \cdot {1}_{A_n}$ such that $w \in \mathcal{L}^1$
We also have $\lim_{n \rightarrow \infty}u_n = u$ for all x and so we can set:
$\lim_{n \rightarrow \infty} \int_{A_n} u d\mu=\lim_{n \rightarrow \infty} \int_X u_n d\mu=\int_X \lim_{n \rightarrow \infty} u_n d\mu=\int_X u d\mu$
My doubt is that if it possible to set $w(x)=n \cdot {1}_{A_n}$ because n varies and therefore I cannot use DCT for this?
Any hint/input would be appreciated
Best Answer
If you set $$ u_n=u\cdot 1_{A_n} $$ then $$ \int_{A_n}u\,d\mu=\int_{X}u_n\,d\mu $$ and $u_n\to u$ pointwise and $|u_n|\le |u|\in L^1$, and hence Lebesgue Dominated Convergence Theorem applies for $\{u_n\}\to u$.