Dominated convergence theorem for expanding sequence of set

Question

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

Answer 1

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 $\{u_n\}\to u$.

Answer 2

$HINT$

Take $u_n(x)=u(x)1_{A_n}(x)$

Then $|u_n(x)| \leq |u(x)| 1_{X}=|u(x)| \in L^1$

Answer 3

You cannot make the dominating fucntion dependent on $n$. $|u|I_{A_n} \leq |u|$ and $|u|$ is your dominating function.