It's my first time to learn measure theory. it's hard for me. Can you guys give me an idea for this problem? thx.
the problem is :
let $f\in L^1(\mathbb{R}^n)$ and define $f_k(x)= \begin{cases} f(x)&\text{if $|f(x)|\le k$ and $|x|\le k$,}\\ 0 &\text{otherwise}. \end{cases} $ then, how can I prove that : $\lim_{k\to\infty} \int f_k\, d\lambda = \int f\, d\lambda$
Measurable Function problem
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3Hint: try to find a way to use dominated convergence theorem. – 2012-05-06
1 Answers
We can write $f(x)=f(x)[|f(x)|\leq k]\cdot [|x|\leq k]$, where $[P]=1$ if $P$ is realized, $0$ otherwise. We have \begin{align} \int fd\lambda-\int f_kd\lambda&=\int_{\{|x|\geq k\}}(f(x)-f_k(x))d\lambda+ \int_{\{|x|< k\}}(f(x)-f_k(x))d\lambda\\\ &=\int_{\{|x|\geq k\}}(f(x)-f_k(x))d\lambda+\int_{\{|x|< k\}}f(x)(1-[|f(x)|\leq k])d\lambda \end{align} hence \begin{align} \left|\int fd\lambda-\int f_kd\lambda\right|&\leq \int_{|x|>k}|f(x)|d\lambda+ \int_{\{|x|
So this result show that the integral of a integrable function can be approach by the integral of a bounded function, which is non-vanishing at most in a bounded set.
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1@GEdgar I saw also other question of this user without he/she showed an attempt. If I had seen this before, I wouldn't have post this answer (but now if I deleted it we still will be able to see it). – 2012-05-06