Math Help - Orthonormal sequences in L^2 and almost everywhere convergence

1. Orthonormal sequences in L^2 and almost everywhere convergence

I have an analysis question that's bugging me. In an old qualifying exam, there was the following question:

Let $f_j$ be an orthonormal sequence in $L^{2}([0,1])$. Prove that $S_{n}=\frac{1}{n}\sum_{j=1}^{n} f_{j}$ converges to zero a.e.

Now, I know how to show that $S_{n}$ converges to 0 in $L^{2}$ norm (and hence there's at least a subsequence that converges to zero a.e. and also it converges in measure).

However, I have a vague memory that when I last looked at this problem, there was a counterexample to the question as stated. Can anyone tell me at least whether or not a counterexample exists?

When I tried proving the problem as given, I couldn't think of an approach except for showing that the integral of the limit of the absolute values of the $S_{n}$'s was 0, but that seems like a dead end, since I'm doubtful I could get dominated convergence to work. When I tried making a counterexample, it didn't seem like I had a lot of freedom to do anything interesting. Averages of initial segments of orthonormal sets all seem to look the same.

Help?

2. I realized I was wrong about my memory of a counterexample, and I think I have a solution, but it would be nice if someone could confirm that my reasoning makes sense.
1. Since L2 is complete, it suffices to show the sequence is Cauchy and that the sequence converges to zero in norm.
2. By orthonormality of the f's, the sequence is Cauchy (see below for boring calculations).
3. By similar calculations to those used in step 2, the sequence converges to 0 in norm.

Calculations for 2:
$
\lim_{n,m\to\infty}\left\Vert S_{n}-S_{m}\right\Vert$

$
=\lim_{n,m\to\infty}\left\Vert \frac{1}{n}\sum_{i=1}^{n}f_{i}-\frac{1}{m}\sum_{j=1}^{m}f_{j}\right\Vert$

$
=\lim_{n,m\to\infty}\left\Vert \frac{1}{nm}\left(\sum_{i=1}^{n}mf_{i}-\sum_{j=1}^{m}nf_{j}\right)\right\Vert$

$
=\lim_{n,m\to\infty}\left\Vert \frac{1}{nm}\left(\sum_{i=1}^{m}-\left(n-m\right)f_{i}+\sum_{j=m+1}^{n}mf_{j}\right)\right\ Vert \text{ WLOG assume }n>m$

$
=\lim_{n,m\to\infty}\frac{1}{nm}\sqrt{\int\left(\s um_{i=1}^{m}-\left(n-m\right)f_{i}+\sum_{j=m+1}^{n}mf_{j}\right)^{2}}$

$
=\lim_{n,m\to\infty}\frac{1}{nm}\sqrt{\int\left(\s um_{i=1}^{m}-\left(n-m\right)f_{i}\right)^{2}+\int\left(\sum_{j=m+1}^{n }mf_{j}\right)^{2}+2\int\left(\sum_{j=m+1}^{n}mf_{ j}\right)\left(\sum_{i=1}^{m}-\left(n-m\right)f_{i}\right)}$

$
=\lim_{n,m\to\infty}\frac{1}{nm}\sqrt{\left(n-m\right)^{2}m+m^{2}\left(n-m\right)+2\int\left(\sum_{j=m+1}^{n}mf_{j}\right)\ left(\sum_{i=1}^{m}-\left(n-m\right)f_{i}\right)}\text{ by orthonormality}$

$
=\lim_{n,m\to\infty}\frac{1}{nm}\sqrt{nm\left(n-m\right)+2*0}\text{ by orthogonality}$

$
=\lim_{n,m\to\infty}\sqrt{\frac{n-m}{nm}}$

$
=\lim_{N,M\to0}\sqrt{\frac{1/N-1/M}{\left(1/N\right)\left(1/M\right)}}\text{ if the limit exists}$

$
=\lim_{N,M\to0}\sqrt{M-N}$

$
=0$

3. The above argument is wrong; That gives convergence in L2 again, which I already knew.