I am self-learning the Analysis II book from Tao [1], and stopped with the following problem (8.2.4): For each $n=1,2,3...$ let $ f_n : \mathbb{R} \rightarrow \mathbb{R}$ be the function $ f_n(x) = \chi(x)_{[n,n+1)} - \chi(x)_{[n+1, n+2)}$, where $\chi()$ denotes the indicator function, i.e., $f_n(x) = 1, \forall x \in [n,n+1)$, $f_n(x) = -1, \forall x \in [n+1,n+2)$, and zero otherwise. Show that $$ \int_{\mathbb{R}}\sum_{n=1}^{\infty}f_n(x) \neq \sum_{n=1}^{\infty}\int_{\mathbb{R}}f_n(x)$$
My attempts:
(1) since $f_n(x)$ is not non-negative, then the integrals and sums cannot be exchanged. Period (?)
(2) Defining $f(x) = \sum_{n=1}^{\infty}f_n(x)$, then $\int_{\mathbb{R}}f(x) = sup \{\int_{\mathbb{R}}s\}$ for $s$ being a simple function that minorizes $f(x)$. Since $f(x) \in \{-1,0,1\} \forall x$, and $\int_{\mathbb{R}}s = \sum_{n=1}^{\infty}a_nm([n,n+1)-b_nm([n+1,n+2))$, where $m()$ is the Lebesgue measure. Sice $\lim_{n \rightarrow \infty}f_n(x) = 0$, at the limit point the measure is zero. This results in $\int_{\mathbb{R}}s = \sum_{n=1}^{\infty}(-1)^n$, and so, the integral is not convergent. On the other side, $\int_{\mathbb{R}}f_n(x) = 0$, for all $n=1,2..$, therefore $\sum_{n=1}^{\infty}\int_{\mathbb{R}}f_n(x)=0$.
Do these attempts make any sense? Did I make some mistakes here? Thanks for all the possible hints/explanations.
