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

Question

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.

Answer 1

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