I'm trying to develop some intuition for the concepts of tightness and uniform integrability, in a probabilistic context.
(i) Does uniform integrability imply tightness?
(ii) If not, is $X_n(x)=I_{[n, n+1]}$ uniformly integrable?
If we take the measure space composed of Borel $\sigma$-algebra on $\mathbb{R}$ with the Lebesgue measure, $(\mathbb{R}, \mathcal{B}, \lambda)$, then is $f_n(x)=I_{[n, n+1]}$ uniformly integrable? It is certainly not tight, but $\mathbb{E}[|f_n|I_{[|f_n| \geq K]}] \leq \varepsilon$ seems to satisfy the uniform integrability condition.
Best Answer
(i) In fact, even a sequence $(X_n)_{n\geqslant 1} $ which is bounded in $\mathbb L^1$ (that is, $\sup_n\mathbb E|X_n| <\infty$) is tight. This can be seen from the inequality $$\mathbb P\{|X_n|\gt R \}\leqslant \frac 1R\mathbb E[|X_n|]\leqslant \frac 1R\sup_j\mathbb E[|X_j|].$$
(ii) When the underlying measure space is not finite, we rather use the following definition of uniform integrability: the family $\left(f_i\right)_{i\in I}$ is uniformly integrable if for each positive $\varepsilon$, there exists an integrable function $g$ such that $\sup_{i\in I}\int_{\{|f_i|\geqslant g }|f_i|<\varepsilon$. With the classical definition in the context of probability space, we would find that the sequence $ (\mathbf 1_{(0,n)})_{n\geqslant 1}$ is uniformly integrable, which is not acceptable because it is not even bounded in $\mathbb L^1$.