Prove that $\lim \limits_{n\to \infty}\int_{n}^{\infty} f_{n}(x)dx\neq \int_{n}^{\infty} f(x)dx $

Question

Let $\{f_n\}$ be a sequence defined by

$$f_n(x)=\begin{cases}1, & \text{if}\;x\geq n,\\0 & \text{if}\;x< n,\end{cases}$$ Prove:

1. $\{f_n\}$ is monotone decreasing and nonnegative.
2. $f_n\to f\equiv 0 \;\;\text{as}\;n\to \infty,$ pointwise.
3. $\lim \limits_{n\to \infty}\int_{n}^{\infty} f_{n}(x)dx\neq \int_{n}^{\infty} f(x)dx.$

Here is my trial:

Solution

2. $\exists\,N>x,$ such that $f_n(x)=0,\;\forall \;n\geq N.$ This implies, $f_n\to f\equiv 0, \text{as}\; n\to\infty.$

3. $$\int_{n}^{\infty} f_{n}(x)dx=\int_{n}^{\infty} (1)dx=\infty.$$ So, $$\lim \limits_{n\to \infty}\int_{n}^{\infty} f_{n}(x)dx=\infty.$$ However, $$\int_{n}^{\infty} f(x)dx=0.$$ So, $$\lim \limits_{n\to \infty}\int_{n}^{\infty} f_{n}(x)dx\neq\int_{n}^{\infty} f(x)dx.$$

My question is, how do I show 1.? If I may ask, are my solutions right?

Answer 1

You answer is correct. For 1), just notice that if $n\leq m$ then $f_n(x)=f_m(x)=1$ if $x\geq m$ and $f_n(x)\geq f_m(x)=0$ if $x<m$. Therefore, $f_n\geq f_m$ if $n\leq m$. Draw the graphics to see what is going on.