# Showing that a sequence of functions has no dominant (Dominated convergence theorem)

Define $$u_n: \mathbb{R}^+ \to \mathbb{R}^+$$ as $$u_n(x) = n1_{(0,1/n)}$$. I need to show that this has no dominant. A dominant is defined as an integrable function $$w: X \to \mathbb{R}^+$$ such that $$\forall n, \forall x, |u_n(x)| \leq w(x)$$.

My attempt so far: suppose it has a dominant $$w$$. Then since $$w$$ is integrable we have $$\int|w|d\lambda < \infty$$. Fix $$x$$. Then we have $$|n1_{(0,1/n)}(x)| \leq w(x) \ \ \forall n.$$ This means that $$w(x) \geq n$$ for all $$x \in (0,1/n)$$. Now pick the sequence $$x_n \in (0,\frac{1}{n})$$. Then we have $$w(x_n) \geq n$$ for all $$n$$.

I don't know how to proceed here. Normally I would use continuity and consider $$\lim_{n \to \infty} x_n$$ but we don't have continuity, only integrability of $$w$$.

If $$w(x) \ge u_n(x)$$ for all $$n$$ and $$x$$, then in particular, $$w(x) \ge n$$ on the interval $$[\frac{1}{n+1},\frac{1}{n})$$. So $$\int_{\mathbb{R}^+} w\ d\lambda \ge \sum_{n=1}^\infty \int_{[\tfrac{1}{n+1},\tfrac{1}{n})} n\ d\lambda = \sum_{n=1}^\infty n\left( \frac{1}{n} - \frac{1}{n+1} \right) = \sum_{n=1}^\infty \frac{1}{n+1} = \infty.$$