Sequence of Functions Satisfying Conditions Can a sequence $(f_n):[0,1]\rightarrow [0,\infty)$ of functions be constructed so that $f_n(x)\rightarrow 0$ everywhere, $\int_0^1f_n(x)dx\rightarrow 0$, but $\sup_nf_n(x)\notin L^1$?
 A: Here's a reasonably general construction. Let $I_n$ be any pairwise disjoint sequence of nontrivial intervals (or more generally, measurable sets with positive measure) contained in $[0,1]$. Let $a_n$ be any sequence of nonnegative real numbers such that $a_n \to 0$ but $\sum a_n = \infty$. Define
$$f_n(x) = \begin{cases}
\displaystyle\frac{a_n}{|I_n|} & \text{ if }x \in I_n \\
0 & \text{ otherwise}
\end{cases}$$
Then $f_n(x) \to 0$ pointwise (indeed for a given $x$, $f_n(x)$ is nonzero for at most one $n$). Moreover,
$$\int_0^1 f_n(x)\ dx = \int_{I_n}\frac{a_n}{|I_n|}\ dx = a_n \to 0$$
and, because $f_n \geq 0$ and the supports of $f_n$ are pairwise disjoint,
$$\int_0^1 \sup_n f_n(x)\ dx =  \int_0^1 \sum_n f_n(x)\ dx = \sum_n \int_0^1 f_n(x)\ dx = \sum_n a_n = \infty$$
For a concrete example, you can take $I_n = [1/(n+1), 1/n)$ and $a_n = 1/n$, which means that $|I_n| = 1/(n(n+1))$ and
$$f_n(x) = \begin{cases}
n+1 & \text{ if }x \in [1/(n+1), 1/n) \\
0 & \text{ otherwise } \end{cases}$$
