0
$\begingroup$

Question

$(X, m, \mu)$ be the measure space. Let $f: X \to [0, \infty]$ be a measurable function and assume that $$ \int_X f\ d\mu < \infty. $$ For any $n \in \mathbb{N}$, let $A$ and $f_n$ be as follows. $$ A(n) = \{x \in X \mid f(x) \leq n \}, f_n = f \chi_{A(n)} $$ Show that the following holds. $$ \lim_{n\to\infty} \int_X |f-f_n|d\mu = 0 $$

What I know

From the definition of $f_n$, $f_n<f_{n+1}$ for any $n\in \mathbb{N}$.

Using the monotonic convergence theorem$(*)$, we have \begin{align} \lim_{n\to\infty} \int_X |f-f_n|d\mu &= \lim_{n\to\infty} \int_X |f|(1-\chi_{A(n)})d\mu\\ &=\lim_{n\to\infty} \left(\int_X |f|d\mu - \int_X |f\chi_{A(n)}| d\mu \right) \\ &=\int_X |f|d\mu - \lim_{n\to\infty} \int_X |f_n|d\mu \\ &=\int_X |f|d\mu - \int_X \left( \lim_{n\to\infty} |f_n| \right) d\mu\ (\because *)\\ &=\int_X |f|d\mu - \int_X |f|d\mu = 0. \end{align}

Is it correct?

$\endgroup$
2
  • $\begingroup$ You can justify the interchange of limit and inetgral using either the Monotone Convegence Theorem or DCT. $\endgroup$ Commented Dec 20, 2022 at 12:31
  • $\begingroup$ @geetha290krm I knew it. I would appreciate it if you could check the formula deformation as it is described. $\endgroup$
    – ytnb
    Commented Dec 20, 2022 at 12:33

1 Answer 1

0
$\begingroup$

Monotone convergence is not needed here. Since $f$ is nonnegative and by assumption has finite integral, for any $\varepsilon>0$ we may choose a positive integer $N$ such that $\mu(X\setminus A(N))<\frac\varepsilon N$. Then for $n\geqslant N$ we have $$ \int_X |f-f_n|\ \mathsf d\mu = \int_{A(N)}|f-f_n|\ \mathsf d\mu + \int_{X\setminus A(N)}|f-f_n|\ \mathsf d\mu< 0 + N\cdot\frac\varepsilon N = \varepsilon, $$ so that the limit is zero.

$\endgroup$
1
  • $\begingroup$ I understand what you are writing, but on the other hand, is the solution I wrote incorrect? $\endgroup$
    – ytnb
    Commented Dec 21, 2022 at 2:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .