So I was reading my textbook about Dominated Convergence Theorem:

I have $(X,\mathscr{F},\mu)$ as a measure space

I have $f,f_n,: X\to [-\infty, \infty], g:X\to [0,\infty]$ integrable and it is the part with the condition $|f_n|\leq g$ for all $n$ that makes it dominated.

Is it still possible to have an example where sequence of the functions $f_n$ tend to, say $0$, and $\int f_n\to 0$ without the dominated condition, in other words, without $g$?


If you mean the sequence is not dominated by any integrable function, then yes.

Take, for example, $[0,1]$ with Lebesgue measure and $f_n(x)={1\over x}\cdot\chi_{[{1\over n+1},{1\over n}]}$.

  • $\begingroup$ Can you explain to me why $\int f_n$ is $0$? Is it because of $1/x$? $\endgroup$ – Akaichan Oct 8 '14 at 20:50
  • 1
    $\begingroup$ @Akaichan The limit is $0$. $\int_0^1 f_n(x)\,dx=\int_{1/(n+1)}^{1/n} 1/x\,dx=\ln{n+1\over n}\rightarrow0$. $\endgroup$ – David Mitra Oct 8 '14 at 20:52

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