# showing that a sequence is uniformly integrable

I am currently reading the new edition of Royden and I've gotten to a part where the book made some comments without justification and I'm trying to verify these facts on my own. I want your help in doing that.

Let $A = \{\emptyset, E, X \setminus E, X\}$ be a $\sigma$-algebra with $\mu(X) =1, \mu(E) = \mu(X\setminus E) = 0.5, \mu(\emptyset) = 0$. Let $f_n = n\,1_E - n\, 1_{X\setminus E}$ for any natural $n$. Then claim is that $f_n$ is uniformly integrable and tight and converges pointwise on $X$ to $f$ that takes the value $+\infty$ on $E$ and $-\infty$ on $X\setminus E$.

I know what I have to show for uniform integrability and tightness.

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Each $f_n$ is integrable hence uniformly integrable. The sequence $(f_n)_n$ is NOT uniformly integrable since, for every $a$, $\mathrm E(|f_n|;|f_n|\geqslant a)\to+\infty$ when $n\to\infty$. – Did Mar 8 '12 at 16:37
And tight means? – julien Jan 28 at 13:07