For a set of functions $f_n\in F$ my definition of uniform integrability over the range $[-1,1]$ is
$$\forall \epsilon>0, \exists M_{\epsilon}>0 : \sup_{f_n\in F}\int_{\{|f_n|\geq M_{\epsilon}\}}|f_n|\,d\mu<\epsilon$$ where $n=1,2,3...$. My function of interest is confusing though,
$$f_n(x)=\frac{1}{x^2+1/n}$$
because it is integrable over $x\in [-1,1]$ for any finite $n$, but its limit in $n\rightarrow\infty$ is not integrable.
Using the definition of uniform integrability, for any $\epsilon$ and any arbitrarily large $n$, I can find an $M_{\epsilon}$ that makes the statement true. But on the contrary, for any $\epsilon$ and any $M_{\epsilon}$, I can find an $n$ that makes it false. This seems like a problem of which infinity grows faster. How can I resolve whether my test function is uniformly integrable or not?
It is not uniformly integrable.
$\int_{f_n(x)\geq M}f_n(x)dx \geq \int_0^{a_n} \frac 1 {x^{2}+1/n}dx$ if $a_n^{2}+1/n =\frac 1 M$. Hence, $\int_{f_n(x)\geq M}f_n(x)dx \geq \sqrt n \arctan (\sqrt n a_n)\to \infty$ as $n \to \infty$