Let's consider functions defined on the interval $[1, \infty)$. Let $f_n(x) = 1/n^2$ for $1 \le x \le n^2 $, and $f_n(x) = 0$ for $x > n^2$. Clearly $f_n$ converges to the constant function $0$ pointwise (actually uniformly). Moreover, for each $n$, $f_n$ is dominated by $g(x) = 1/x^2$ which is integrable. So, by the Lebesgue dominated convergence theorem, the integral sign and the limit on $n$ should be interchangable. But $$\lim_{n \to \infty}\int_1^\infty f_n = \lim_{n \to \infty}\frac{n^2 - 1}{n^2} = 1 \ne 0 = \int_1^\infty \lim_{n \to \infty} f_n. $$ I think I must have made some silly mistakes or have some misconceptions. What is wrong?

  • 1
    $\begingroup$ You have $f_n (x) \leq 1/x $, not $f_n (x) \leq 1/x^2$. $\endgroup$
    – PhoemueX
    Jan 24, 2016 at 12:56
  • $\begingroup$ Oops! Really silly mistake. PhoemueX, please make your comment as an answer if you want. I'll accept it. $\endgroup$
    – leeto
    Jan 24, 2016 at 13:39

1 Answer 1


Because of $f_n (x)=1/n^2$ for $1\leq x \leq n^2$, we only have $f_n (x) \leq 1/x$ (with equality at $x=n^2$) and not $f_n (x)\leq 1/x^2$.


You must log in to answer this question.

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