If for any $a$, $g(a, b) \rightarrow 0$ as $b \rightarrow \infty$, then does $\sup_a g(a, b) \rightarrow 0$ when $b \rightarrow \infty$?

Question

Here, $f:A \times \mathbb{R} \rightarrow \mathbb{R}$ is some real-valued function, $A$ is some nonempty set.

motivation: a set of random variables $\{X_\alpha\}$ is uniformaly integrable if $\lim_{K \rightarrow \infty}\sup_{\alpha}E[|X_\alpha| 1_{|X_\alpha| > K}] = 0$, now if I know for any $\alpha$, $\lim_{K \rightarrow \infty}E[|X_\alpha| 1_{|X_\alpha| > K}] = 0$, can I conclude that the set of random variables is uniformly integrable?

Answer 1

Take $g(a,b)=\frac ab$, defined on $\mathbb R_+^2$. Then, for any $a\in\mathbb R_+$, it is easy to see that $$\lim_{b\to\infty}g(a,b)=\lim_{b\to\infty}\frac{a}{b}=a\cdot \lim_{b\to\infty}\frac1b = 0$$

however, for every $b\in\mathbb R_+$, we also have $$\sup_a g(a,b)=\sup_a \frac ab = \frac 1b\cdot \sup_a (a) = \infty$$ which means that the limit

$$\lim_{b\to\infty}\sup_a g(a,b)$$ does not exist.