Passage of the limit under the integral not equivalent with uniform integrability I'm having some trouble with a real analysis study problem. Given a sequence $(f_n)_n$ of integrable functions on E, $\mu(E) < \infty$, that converges pointwise almost everywhere on $E$ to $f = 0$, show that $lim_{n->\infty}\int_Ef_n = 0$ is not equivalent with $(f_n)_n$ being uniformly integrable over $E$.
I know that on a space of finite measure, uniform integrability implies passage of the limit, so $lim_{n->\infty}\int_Ef_n = \int_E lim_{n->\infty} f_n = \int_E 0 = 0$, but I can't imagine a sequence which is convergent to $0$, has the limit of the integral $=0$, but isn't uniformly integrable. There's a theorem in the textbook I'm using that states that for a nonnegative sequence that converges to $0$, passage of the limit is equivalent with uniform integrability, so the function I'm looking for must be negative at some point. I think my trouble with this question is a general result of my lack of comfort with uniform integrability in general, since I can't really imagine a function that's convergent and isn't uniformly integrable. Any help would be superb.
 A: First make sure that you can find a sequence which converges pointwise almost everywhere to 0, but which isn't uniformly integrable. Since you say that you have problems with the concept of uniform integrability, let's come back to this. 
Secondly, you adjust each function to make the value of the integral close to 0. Uniform integrability is a property that says that each function is 'badly behaved' on some (vanishingly) small portion of the domain. So we should be able to adjust the function on some other part of the domain, without breaking the property of uniform integrability.

Notice that you made two mistakes in your original formulation of the question (I've edited one): you confused uniform integrability with uniform convergence, and you referred to 'a function' that's convergent and isn't uniformly integrable', not a series of functions. Were these just slips and do you definitely understand the distinction in both cases? If not, you might want to post further questions to make sure you understand all the terms before you try to work on the question. 

Now, let's try to find a sequence which is convergent almost everywhere to 0 but which isn't uniformly integrable. Uniformly integrable says that for a given $\varepsilon$, there is some $M$, such that 
$$ \int_{|f| > M} |f| < \varepsilon $$
holds for every function $f$.
The opposite of this is that there is some $\varepsilon$, which might as well be $1$, and we can find a sequence of functions $f_n$ and numbers $M_n$ such that 
$$ \int_{|f_n| > M_n} |f_n| $$ is always bigger than $1$, and $M_n \rightarrow \infty$.
We can also probably try $M_n = n$*. So we are looking for a sequence of functions, where
$$ \int_{|f_n| > n} f_n > 1$$
You should be able to check pretty quickly that a function which is equal to $n+1$ on a piece of the domain (interval or other measurable set) of measure at least $\frac{1}{n+1}$ will satisfy this condition.
Notice that we don't have any requirements about what the rest of the function does. All we need to make sure of is that the pointwise limit is 0 a.e. If you take the function to be 0 outside of the small piece where it is $n+1$, it's not hard to arrange those small pieces so that you get the pointwise limit. The first thing you try (do try something explicitly, don't just expect the pieces to come together by themselves) will either work or fail in a way that suggests a solution.
Now go back to my second paragraph and see if you can get the third condition, the limit of the integrals being 0. 

*Disclaimer: Here when we had to take a small number, we took 1, and when we had to take a sequence tending to $\infty$, we took the naturals. These are the most obvious examples and should be the first things to try, but in many harder cases they won't work and we'll have to do more work.
