$\lim_n \int_{\frac{1}{n}\leq |f|\leq n} |f|\,d\mu=\int |f| \,d\mu$. Let $(X,\tau,\mu)$ a measure space and $f\in L^1$.
a) Show that $\lim_n \int_{\frac{1}{n}\leq |f|\leq n} |f|\,d\mu=\int |f|\, d\mu$.
b) Deduce for all $\epsilon>0$ $\exists A\in \tau$ such that $\mu(A)<\infty$,
$\sup_{x\in A }|f(x)|<\infty$  and $\int_{A^c} |f|\,d\mu<\epsilon$
For a). Should the monotone convergence theorem be applicable?
 A: $a$ is not quite correct : either on the left hand side you have $f$ instead of $|f|$ (in the integrand), or on the right hand side you have $|f|$ instead of $f$. I will make the former correction : the latter follows by replacing $f$ by $|f|$ in the former.
When you see a difference of integration of domain, just add in characteristic functions so that you can apply theorems of your choice.
E.g. $\displaystyle\int_{\frac 1n \leq |f| \leq n} f d \mu = \int f \chi_{\{\frac 1n \leq |f| \leq n\}} \, d \mu$.
Now, $\chi_{\frac 1n \leq |f| \leq n} \to \chi_{\{f \neq 0\}}$ pointwise, and the functions $f \chi_{\{\frac 1n \leq |f| \leq n\}}$ are not necessarily increasing, since $f$ can now take negative values. So MCT does not apply.  It will apply, though, if $f$ is non-negative, since in that case these functions are actually increasing.
However, dominated convergence theorem applies, since these functions are bounded by the integrable function $|f|$. In that case, we get $\displaystyle \lim_n \int_{\frac 1n \leq |f| \leq n} f \,d \mu \to \int f \chi_{f \neq 0} \,d \mu = \int f \,d \mu$.

For part $b$, use the definition of limit : for any $\epsilon > 0$, there exists $N$ such that $\int_{\frac 1n \leq f \leq N} |f| d \mu \geq \int |f| d \mu - \epsilon$ (and this is true for all larger $n  > N$ , but we will not need that).
Consequently, let $A = \{\frac 1N \leq |f| \leq N\}$. Then $f$ is bounded on $A$ by definition, and $\mu(A) < \infty$, otherwise $\int_A |f|\,d \mu \geq \int_A \frac 1N d \mu$, and this won't be finite if $A$ has infinite measure, and we cannot have finite greater than positive infinite. The complement condition is satisfied by choice of $N$.
Hence, we are done.
A: Let $A_n = \{1/n \leqslant |f| \leqslant n\}$ then by (a) there exists $n$ such that
$$0 \leqslant \int_{A_n^c} |f| \, d\mu = \int |f| \, d\mu - \int_{A_n} |f| \, d\mu < \epsilon$$
We have $\sup_{x \in A_n} |f| \leqslant  n < \infty$ and $\infty > \int |f| \, d\mu > \int_{A_n} |f| \, d\mu \geqslant \frac{1}{n} \mu(A_n) \implies \mu(A_n) < \infty.$
