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.