Show if $f \in L^1(\Bbb{R})$ then $\int_{\Bbb{R}} \vert f \vert \chi_{\{x \in \Bbb{R}: \vert f(x) \vert > n\}} dm=0$ as $n \rightarrow \infty$ So let $f \in L^1(\Bbb{R})$ and show
$$\lim_{n \rightarrow \infty} \int_{\Bbb{R}} \vert f \vert \chi_{\{x \in \Bbb{R}: \vert f(x) \vert > n\}} dm = 0$$
Is this because $\int_{\Bbb{R}} \vert f \vert dm < \infty$ thus for large $n$, $\chi$ is zero? any $\textbf{Hints}$ greatly appreciated. Not full on solutions. Should I define a sequence of measurable functions
$$f_n(x):=f \cdot\chi(x)$$
which is integrable since $f$ is integrable. Then show the integral of these goes to $0$ as $n$ tends to $\infty$?
 A: The hint is to actually define measurable sets:
$$A_n = \{x \in \mathbb{R}: |f(x)| \geq n \}$$
I'll let you think about why this is a measurable set. How does $A_{n+1}$ relate to $A_n$? Then, define as sequence of non-negative functions:
$$f_n = |f| \chi_{A_n}$$
What is the point-wise limit of this sequence of functions? How does $f_{n+1}$ relate to $f_n$? Use what you determined about the sequence $(A_n)_{n \in \mathbb{N}}$!
Full Solution Ahead, Do Not Read Initially
Define $A_n = \{x \in \mathbb{R}: |f(x)| \geq n \}$. Observe that $A_{n+1} \subseteq A_n$. In particular, $(A_n)_{n \in \mathbb{N}}$ is a decreasing sequence of sets. In particular, if we define a sequence of functions:
$$f_n = |f|\chi_{A_n}$$
then it follows that $f_{n+1} \leq f_n$. Next, let $x \in \mathbb{R}$. Then, there exists an $N \in \mathbb{N}$ such that for all $n \geq N$, $n > |f(x)|$. This implies that for all $n \geq N$, $x \notin A_n$. That is, $f_n(x) = 0$ when $n \geq N$. Hence, $f_n(x) \to 0$. By the Monotone Convergence Theorem:
$$\lim_{n \to \infty} \int_{\mathbb{R}} f_n(x) \ dx = \int_{\mathbb{R}} 0 \ dx = 0$$
as was desired.
