Let $f_n:[1,\infty) \rightarrow \mathbb{R}$ be defined by $f_n(x)=\frac{1}{x}\chi_{[n,\infty)}(x)$. Does $\int f_n \to \int f$?

Question

Let $f_n:[1,\infty) \rightarrow \mathbb{R}$ be defined by $f_n(x)=\frac{1}{x}\chi_{[n,\infty)}(x)$. Does $\int f_n \to \int f$?

I know it is an application of DCT but I really don't know to come up with a bound that would satisfy the hypothesis of DCT. Something like $\frac{1}{x^2}$ should work but $\frac{1}{x} > \frac{1}{x^2}$.

Answer 1

as stated in the comments, the first thing to see is that $f_n\to 0$ pointwise everywhere as $n\to\infty$. Also $\int_{[n,\infty)}f_n=+\infty$ no matter what is $n\in[1,\infty)$ so if you suppose that the DCT holds true, then by intechanging limit and integrals you would obtain $\lim\int_{[n,\infty)}f_n=+\infty=\int_{\mathbb{R}}\lim f_n=0$ which is not possibly true.