# Prove that $\lim_{n\to \infty} \int_A f_n d\mu = \int_A fd\mu$

Let $(X, \mathcal{F}, \mu)$ be a measure space, $A\in \mathcal{F}$ and $f: A\to \overline{R}$ is a non-negative measurable function. For each $n\ge 1$, we define $f_n: A\to \overline{R}$ as $$f_n(x) = \begin{cases} f(x) &\text{if } f(x)\le n \\ n^2 &\text{if } f(x) > n \end{cases}.$$ Show that $$\lim_{n\to \infty} \int_A f_n d\mu = \int_A fd\mu.$$

I cannot apply either Monotone Convergence Theorem or Dominated Convergence Theorem here, so I wonder if there exists any counter-example for this problem.

Thank you very much.

• why can't you use monotone convergence? Jun 14, 2017 at 17:32
• Why do you write "show that" if you think it might be false?
– zhw.
Jun 14, 2017 at 17:36
• @pwerth Since $(f_n)$ is not a monotone sequence. Jun 14, 2017 at 17:36
• @zhw. It's a problem in my exam. Jun 14, 2017 at 17:37
• A better way to write it would be "here's a problem from my exam, but I am wondering if the result is true." When you write "show that" you can send people off on a wild goose chase.
– zhw.
Jun 14, 2017 at 17:45

Hint: Consider $f(x) = 1/\sqrt x$ on $(0,1).$