Counter-Example (or Proof) to $\int_{0}^{1}f_{n}\;dx\to0$ Implies $f_{n}\to0$ a.e. $x$ Whenever $f_{n}\geq0$. I am dealing with a problem at the moment where the hypothesis can be restated as $\int_{0}^{1}f_{n}\;dx\to0$ and $f_{n}\geq0$.  Under these conditions, I want to conclude that $\lim f_{n}$ exists and is $0$ for almost every $x$ in $[0,1]$.  Without boundedness or monotonicity, the usual convergence theorems are not immedaitely applicable.  Also, if the limit of integrals converged to anything other than $0$, or if the $f_{n}$ were allowed to be signed, or if the set of integration had infinite measure, well known counter-example(s) would be available.  It seems that the finite measure of $[0,1]$, the nonnegativity of $f_{n}$, and the assumption $\int_{0}^{1}f_{n}\;dx\to0$  should force $f_{n}\to0$ by appealing (in some manner) the well known fact that for $g\geq0$ measurable, $\int_{0}^{1}g\;dx=0$ if and only if $g=0$ a.e. $x\in[0,1]$.  The only way out of this is if the $f_{n}$ spiked on sets of small measure; but as $n\to\infty$, these sets where the $f_{n}$ spike must become correspondingly smaller (in measure) since we have $\int_{0}^{1}f_{n}\;dx<\epsilon$ for $n$ large.  In the limit, these "spike sets" should yield to a null set, thus proving the claim.
The question I am referring to is here Limit of Integral of Difference Quotients of Measurable/Bounded $f$ Being $0$ Implies $f$ is Constant

The canonical counterexample is to take the indicator functions of $[0,1]$; $[0,1/2]$,$[1/2,1]$, $[0,1/4]$, $[1/4,1/2]$, $[1/2,3/4]$, $[3/4,1]$ &c.
(If the pattern is not evident: break $[0,1]$ into $2^k$ intervals let $f_n$ be the sequence of indicator functions of each $2^k$ intervals obtained at each step from left to right, let $k=0,1,2,\ldots$) 
In this case $\liminf f_n=0$ while $\limsup f_n=1$ so the sequence of functions converges nowhere, yet $$\int_0^1 f_n\to 0$$ 
