Let $f_n\to f$ in $L^1((0,+\infty))$ with respect to Lebesgue measure. I am asked to show that $$\lim_{n\to\infty}\int_{n}^{2n}f_n(x)dx=0$$
I think I could use dominated convergence for the sequence $(\chi_{(n,2n)}f_n)_n$ somehow, but I don't really see how to show the hypotesis or where would that lead. Hints?
Let $\varepsilon > 0$. Since $f_n \to f$ in $L^1((0,+\infty))$, there exists a positive integer $N$ such that if $n \ge N$, then $$\int_0^\infty |f_n(x) - f(x)|\, dx < \varepsilon.$$ Thus, if $n\ge N$, $$\left\lvert \int_n^{2n} (f_n(x) - f(x))\, dx\right\rvert \le \int_n^{2n} |f_n(x) - f(x)|\, dx \le \int_0^\infty |f_n(x) - f(x)|\, dx < \varepsilon.$$ Therefore $$\lim_{n\to \infty} \int_n^{2n} (f_n(x) - f(x))\, dx = 0.$$ Since $f\chi_{[n,2n]} \to 0$, $|f(x)\chi_{[n,2n]}(x)| \le |f(x)|$ for all $x$, and $f\in L^1((0,+\infty))$, it follows from the dominated convergence theorem that $$\lim_{n\to \infty} \int_n^{2n} f(x)\, dx = \lim_{n\to \infty} \int_0^\infty f(x)\chi_{[n,2n]}(x)\, dx = \int_0^\infty \lim_{n\to \infty} f(x)\chi_{[n,2n]}(x)\, dx = 0.$$ Hence, $$\lim_{n\to \infty} \int_n^{2n} f_n(x)\, dx = \lim_{n\to \infty} \int_n^{2n} (f_n(x) - f(x))\, dx + \lim_{n\to \infty} \int_n^{2n} f_n(x)\, dx = 0 + 0 = 0.$$