Variation on the Dominated Convergence Theorem

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,

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

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 $$g$$.
• 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 $x \geq 1$? – Haarlem90 Dec 11 '16 at 17:03
• @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. – Olivier Oloa Dec 11 '16 at 17:06