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
2
$\begingroup$
real-analysis
measure-theory
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3Hint: try to find a way to use dominated convergence theorem. – 2012-05-06
1 Answers
1
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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3Why not wait for the OP to come back, read the hint, and make an attempt, before you give a full answer? – 2012-05-06
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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