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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.

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  • $\begingroup$ why can't you use monotone convergence? $\endgroup$
    – pwerth
    Jun 14, 2017 at 17:32
  • $\begingroup$ Why do you write "show that" if you think it might be false? $\endgroup$
    – zhw.
    Jun 14, 2017 at 17:36
  • $\begingroup$ @pwerth Since $(f_n)$ is not a monotone sequence. $\endgroup$ Jun 14, 2017 at 17:36
  • $\begingroup$ @zhw. It's a problem in my exam. $\endgroup$ Jun 14, 2017 at 17:37
  • $\begingroup$ 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. $\endgroup$
    – zhw.
    Jun 14, 2017 at 17:45

1 Answer 1

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

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