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I am trying to check whether the following variation of the Dominated Convergence Theorem is true or false.

A sequence of measurable functions $ {f_n} $ converges pointwise to $f$ in $[0, \infty)$

$ \forall x \geq 1$, $ \forall n \geq 1$: $ \lvert f_n(x) \lvert \leq \frac{1}{x^2}$

Then,

$ \lim_{n \to \infty} \int_{[1, \infty)} f_n = \int_{[1, \infty)} f$

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What you wrote is fine, you have just found a dominating function $g(x):=\dfrac1{x^2}$ such that $$ \int_{[1,\infty)}\lvert f_n \rvert \le \int_{[1,\infty)}g<\infty $$ then the dominated convergence theorem applies here with such a $g$.

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  • $\begingroup$ Thanks, sorry if my question is to basic, but can I still apply the Dominated Convergence Theorem even if $f_n$ converges pointwise to $f$ in $[0, \infty)$ and $ \lvert f_n(x) \lvert \leq \frac{1}{x^2}$ is only true for $x \geq 1$? $\endgroup$ – Haarlem90 Dec 11 '16 at 17:03
  • $\begingroup$ @Haarlem90 Yes, if $f_n \to f$ over $[0,\infty)$ then necessarily $f_n \to f$ over $[1,\infty)$ then you can apply the DCT on $[1,\infty)$ which is licit. $\endgroup$ – Olivier Oloa Dec 11 '16 at 17:06

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