# 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 $$f_n$$, $$f_n for any $$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?

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

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.