# An counterexample for the monotone convergence theorem and dominated convergence theorem

Let $\mu$ denote the Lebesgue measure. Let $f_n(x)=\frac{1}{\sqrt{1+nx^2}}$.

I know that the sequence of functions $\{f_n(x)\}_{n\geq1}$ doesn't satisfy the hypothesis of Monotone convergence theorem because it is not an increasing sequence.

As for dominated convergence theorem, how do we prove that there does not exist a function which is integrable, i.e., $g \in \mathcal{L}$, such that $|f_n(x)| \leq g(x)$ $n=1,2,3,...,x\in\mathbb{R}$.

And how to show that $\int_\mathbb{R}f_nd\mu=\infty$?

• For each $x$, $\sup_n|f_n(x)|=1$. So if $g(x)\geqslant \sup_n|f_n(x)|=1$ then $g$ is not integrable. – Math1000 Apr 9 '18 at 1:49
• @Math1000 $\sup_n|f_n(x)|$ is not equal to one. – YHH Apr 9 '18 at 3:05
• @Math1000 You'd really want something more like $\limsup$ instead of $\sup$ though, right? A dominating function doesn't need to actually dominate all the $f_n$ terms -- just all but the first $n$ of them. – Aaron Montgomery Apr 9 '18 at 19:52

## 1 Answer

The crux of the result would seem to be to show your last claim -- namely, that $\int_{\mathbb R} f_n \, \textrm d \mu = \infty$. If you can show that, then clearly there is no dominating function $g$ -- after all, such a function would have $\int_{\mathbb R} g \, \textrm d \mu \geq \int_{\mathbb R} f_n \, \textrm d \mu$.

A hint for how to show that $f_n$ is not integrable on the real line: for $x > 0$, $$\frac{1}{\sqrt{1 + nx^2}} \geq \frac{1}{\sqrt{1 + 2 \sqrt n x + nx^2}} = \frac{1}{\sqrt{(1 + \sqrt n x)^2}} = \frac{1}{1 + \sqrt n x}$$ so from here, you can either just directly find an antiderivative or use some other lower-bound estimates.

• Thanks. Your answer really helps. – YHH Apr 9 '18 at 19:22