Subsequence Properties and Lp Spaces Exercise:

Let $E$ be a measurable set in $\mathbb{R}$ under Lebesgue measure, and let $1<p<\infty$.  Suppose $\{f_n\}$ is a bounded sequence in $L^p(E)$ and $f \text{ belongs to } L^p(E) $.  Consider the following 4 properties:
(i) $\{f_n\} \rightarrow f$ pointwise a.e. on $E$.
(ii) $\{f_n\} \rightharpoonup f$ in $L^p(E)$.
(iii) $\{\|f_n\|_p\}$ converges to $\{\|f\|_p\}$.
(iv) $\{f_n\} \rightarrow f$ in $L^p(E)$.
If $\{f_n\}$ possesses two of these properties, does a subsequence posses all four properties?

I'm not quite sure where to start.  My intuition tells me that that this is false.  But I'm not sure as to which properties to start with.  Any thoughts would be helpful.
 A: (Not a full answer.)
First of all, as Daniel Fischer notes in the comments,


*

*$L^p$ convergence implies convergence in measure.  Convergence in measure implies there is some subsequence converging pointwise.

*Strong ($L^p$) convergence implies weak convergence.

*$\newcommand{\norm}[1]{\left\| #1 \right\|_p}
0 \le \left| \norm{f_n} - \norm{f} \right| \le \norm{f_n - f}$ by Triangle inequality.
Therefore, if you want to find a counterexample (which you do), you should steer clear of $L^p$ convergence.
But there is more:


*

*Convergence a.e. and of norms implies strong convergence. (Uses $p < \infty$) (Weaker form: Pointwise convergence and bounded norms imply weak convergence)


Therefore there are two cases left: Convergence of norms + weak convergence,
and Pointwise a.e. convergence + weak convergence.
A: if (iv) is one of the two assumptions, then all four are true for some subsequence.
To check the other three case,
(i)+(ii), for $1<p<\infty$, $f_n$ bounded in $L^p$, pointwise a.e convergence implies weak convergence, thus the only asumption we have is basically $f_n \rightarrow f$ pointwise a.e. This would make the statement false, take $f_n(x) = \chi_{[n,n+1]}$ as a counter example.
Both (i)+(iii) and (ii) + (iii) implies strong convergence; the statement is true. 
It would be a good exercise to check for the case when $p=1$.
