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If $f_n=1_{(n,n+1)}(x)$, where $1_A(x)$ is the indicator function. Why is $f_n \rightarrow0$? Same is true for $f_n=1_{(n,\infty)}$.

i just dont get it.

i thought $f_n$ was always 0 for all n so i think $f_n\rightarrow1$ but its not the case. i try to reason it by the integral which is 1 for all n but then i dont go anyway.

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Take for example $n=4.5$. Then the sequence $(f_n(x))$ goes 0, 0, 0, 1, 0, 0, and it stays at 0 from then on, so it converges to 0. Exactly the same thing will happen for any $x$. –  Robin Chapman May 15 '10 at 18:38
    
Robin Chapman i dont understand you here could you please elaborate. I you saying n=4.5 or n=4 then 5? How do I get 0, 0, 0, 1, 0, 0? What are these numbers? –  Walter May 15 '10 at 19:50
    
That's a misprint: I meant $x=4.5$. So $f_1(4.5)=0$, $f_2(4.5)=0$, $f_3(4.5)=0$, $f_4(4.5)=1$, $f_5(4.5)=0$, etc. –  Robin Chapman May 15 '10 at 20:03

1 Answer 1

up vote 4 down vote accepted

This is just the definition of convergence for sequences of functions : $\forall x \in \mathbb R, \lim_{n \to+\infty} f_n(x) =0$, which is of course the case here, all sequences $(f_n(x))$ being stationary.

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show me how to apply this definition, is it like $|f_n(x)|<\epsilon$, but then $f_n(x)$ can be 1 if $x \in $, so $\epsilon$ cannot be made small as we want. –  Walter May 15 '10 at 19:58
    
Be careful with the quantifiers! The definition is : $\forall x \in \mathbb R, \forall \epsilon >0, \exists N_0=N_0(x,\epsilon)$ such that $\forall n \geq N_0, |f_n(x)|< \epsilon$. And here, you may take $N_0(x,\epsilon) = [x]+1$. –  Henri May 15 '10 at 20:32

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