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I have a challenge in doing the following proof: if $\{f_n\}$ is a sequence defined by

$f_n(x)=\begin{cases}1, & \text{if}\;n\leq x\leq n+1,\\0 & \text{otherwise},\end{cases}$ then I want to prove that

  1. $f_n\to f\equiv 0 \;\;\text{as}\;n\to \infty.$
  2. $\liminf\limits_{n\to \infty}\int f_n(x)dx\neq \int f(x)dx.$

The truth is, I don't see the solution coming soon but I believe there are people here who can help me do justice to it. Any help would be highly regarded.

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  • $\begingroup$ Do You have problems with 1 or 2 or both? $\endgroup$ – Logic_Problem_42 May 8 '18 at 5:06
  • $\begingroup$ @ Logic_Problem_42:Thanks for being prompt at these times! Actually, I have a problem with both. I'm new into analysis. $\endgroup$ – Omojola Micheal May 8 '18 at 5:08
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  1. Let $x$ be arbitrary but fixed. By definition, we have for all $n>x$: $f_n(x)=0$. This implies $f_n(x)\to 0$. Since $x$ was arbitrary, we have $\lim_n f(x)=0$ for all $x$.

  2. $\int_{\infty}^{\infty} f_n(x)dx=\int_n^{n+1}1dx=1$- this holds for all $n$. Therefore the liminf of $\int_{\infty}^{\infty} f_n(x)dx$ is $1$.

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For each $x$ eventually (i.e if $n > x$) $f_{n}(x) = 0$. So $f(x) = \lim_{ n\rightarrow \infty} f_{n}(x)= 0$ for all $x$.

$\int f_{n}(x)\,dx = \int_{n}^{n+1}1\,dx = 1$. So $\liminf_{n \rightarrow \infty} \int f_{n}(x)\,dx = \lim_{n \rightarrow \infty} \int f_{n}(x)\,dx = 1$.

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For any $x$, there exists $N>x$ so that $f_n(x) = 0$ for all $n>N$. So $f_n\to f \equiv 0$.

Note that $\displaystyle{\int f_n dx = 1}$ for all $n$. So $\displaystyle{\liminf_{n\to\infty} \int f_n dx = 1}$. However, $\displaystyle{\int f dx = 0}$.

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