A sequence with no almost everywhere converging subsequence Let $I=[0,1]$ with $\mu$ the Lebesgue measure restricted to the class of Borel subsets of $I$. Suppose we have a sequence $\boldsymbol\varphi=\langle\kern.7mm\varphi_i:i\in\mathbb N_0\kern.4mm\rangle$ of functions $\varphi_i:I\to[-1,1]$ such that $\big\langle\kern.2mm\int_A\varphi_i{\kern.6mm\rm d\kern.6mm}\mu:i\in\mathbb N_0\kern.4mm\rangle\to 0$ for all $A\in{\rm dom\kern.8mm}\mu$. Someone might conjecture that $\boldsymbol\varphi$ then has a subsequence converging to zero a.e. on $I$. I have convinced myself that taking the Rademachers $\varphi_i=\langle\kern.7mm\operatorname{sgn\kern.6mm}(\sin(2^{i+1}\,\pi\,t)):t\in I\kern.7mm\rangle$ gives a counterexample. Here $\operatorname{sgn\kern.3mm}t=t^{-1}|t|$ for $t\not=0$ and $\operatorname{sgn\kern.6mm}0=0$. I have "thought through" the details of the proof that every subsequence of $\boldsymbol\varphi$ converges to an element of $\{\,-1,0,1\,\}$ only on a set of measure zero. A detailed presentation of the proof would be quite tedious, at least to me. Now I ask
 Does anyone have another counterexample to the above conjecture for which the associated (reasonably detailed) proof would be relatively short, or a short proof to my counterexample above. 
Added explanation. After reading the answers, I can point out my "error". Straight from the beginning, when I started considering this matter, I realized that the set of $t$ with $\varphi_i(t)=0$ for some $i$ is countable (and hence has measure zero) and that only at points of this set can any subsequence of $\boldsymbol\varphi$ converge to zero. However, somehow I forgot that I just needed this, and I began to prove that also the sets where a subsequence may converge $1$ or to $-1$ are of measure zero. This is the unnecessary hard work in this context. These sets generally are uncountable but still have measure zero.
 A: If there were a subsequence converging almost everywhere, by dominated convergence it would also converge in $L^1$. But $\int |\varphi_i| = 1$ for all $i$ so no subsequence can converge to 0 in $L^1$.
A: I believe you can find an example where none of the functions are even equal to $0$, which would make the proof trivial that there cannot be a.e. convergence to zero for any subsequence of the functions. Namely, define $\phi_n$ to be $1$ on $[i,i/n)$ when $i$ is even, and $-1$ when $i$ is odd. (And the last interval $[(n-1)/n,1]$ includes both the endpoints).  Then on every interval $A \subset [0,1]$, you get $\lim_{n \to \infty} \int_A \phi_n d \mu= 0$, which should be all you need to extend to all $A \in {\hbox{dom }} \mu$. Unless I don't understand Borel well enough.  I like the other answer better though, because it shows that in fact a subsequene of $\phi_n$ cannot converge almost everywhere at all, regardless of what the specific limit values are for different $x \in [0,1]$. 
A: Following on the answer of user2566092, there is an even easier way to see that your counterexample works. Each $\varphi_i$ is 0 at only finitely many points. So $A := \bigcup_i \{x : \varphi_i(x)=0\}$ is countable and hence has measure 0. For $x$ not in $A$, $\varphi_i(x)$ is 1 or -1 for every $i$, so no subsequence can converge to 0.
