With reference to exercise 3 on page 34 of this: http://courses.mai.liu.se/GU/TATM85/Solomon%20-%20Measure%20Theory%20and%20Lebesgue%20Integration.pdf
The exercise asks to show that if $f_n$ is a series of integrable functions on $E$, then$\int_E |f-f_n| \to 0 \iff \int_E|f_n| \to \int_E|f|$.
Pleas note I am not asking for a solution to this exercise, just an example related to it.
I was wondering if anyone could provide an example of why we need the absolute values on the left, to help me understand it more intuitively. Could someone please provide an example of a sequence of $f_n$ that are measurable on $E$, and some function $f$ where $$\lim_{n\to \infty}\int_Ef_n = \int_Ef$$ but $$\lim_{n\to \infty}\int_E|f_n-f| \neq 0$$ Thank you!
Let $E = \Bbb{R}$, $f_n(x) = \sin(nx) 1_{[-\pi,\pi]}(x)$. (i.e. The sine curve on $[-\pi,\pi]$). By symmetry, $\int_E f_n = 0$ for all $n \in \Bbb{N}$, so take $f \equiv 0$, so that $\lim\limits_{n\to \infty}\int_Ef_n = \int_Ef$. Since $$\int_E |f_n - f| = 2n \int_0^{\pi/n} \sin(nx) dx = 2[-\cos(nx)]_0^{\pi/n} = 4,$$ $\lim\limits_{n\to \infty}\int_E|f_n-f| = 4 \neq 0$.
The above calculations also answers the question in the comment: $$\lim\limits_{n \to \infty} \int_E |f_n| = \lim\limits_{n \to \infty} \int_E |f_n - f| = 4 \ne 0 = \int_E f.$$
Graph on Desmos