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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 : $\lim_{k\to\infty} \int f_k\, d\lambda = \int f\, d\lambda$

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Hint: try to find a way to use dominated convergence theorem. –  Thomas E. May 6 '12 at 11:31

1 Answer 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|<k,|f(x)|\geq k\}}|f(x)|d\lambda\\\ &\leq \int_{|x|>k}|f(x)|d\lambda+ \int_{\{|f(x)|\geq k\}}|f(x)|d\lambda\\\ &=\sum_{j=k+1}^{+\infty}\int_{\{j\leq |x|<j+1\}}|f(x)|d\lambda+\sum_{j=k+1}^{+\infty}\int_{\{j\leq |f(x)|<j+1\}}|f(x)|d\lambda, \end{align} and these two terms converge to $0$ when $k\to \infty$, because it's the remainder of a converging series ($f$ is integrable).

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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Why not wait for the OP to come back, read the hint, and make an attempt, before you give a full answer? –  GEdgar May 6 '12 at 12:34
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@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). –  Davide Giraudo May 6 '12 at 12:56

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