Show that $\lim_{n \to \infty} (\int_{0}^{\infty}f_n(x)dx) \not= \int_0^\infty f(x)dx$

Question

There is a sequence of functions, $f_{1},f_{2},f_{3},\ldots$, whose domain is $[0, \infty)$, such that $$f_{n}(x)=\begin{cases}
\frac{1}{n} & \text{if } x \in [0, n) \\
\frac{-1}{n}x+1+\frac{1}{n} & \text{if } x \in [n, n+1) \\
0 & \text{if } x \in [n+1, \infty) \\
\end{cases}$$

Show that $\lim_{n \to \infty} (\int_{0}^{\infty}f_n(x)dx) \not= \int_0^\infty f(x)dx$. So they should not be the same as the integral of the limit. As to how I can solve it, I tried but is not successful. Actually, I try to convert this is improper integral, to normal integral. I am stuck though.

Answer 1

Notice that $$\lim_{n\to \infty}f_n(x)=f(x)\equiv 0 $$ so $$\int_0^\infty f(x)dx=0 $$ on the other hand, $$I_n=\int_0^\infty f_n(x)dx =\int_0^n f_ndx+\int_n^{n+1}f_ndx+\int_{n+1}^\infty f_ndx$$ so $$I_n= \int_0^n \frac{1}{n} dx-\int_n^{n+1} \frac{1}{n}x-1-\frac{1}{n} dx+\int_{n+1}^\infty 0dx$$ then $$ I_n = 1+\frac{1}{2n}$$ so we see taking $\lim_{n\to\infty} I_n=1\neq 0$. And the result follows.