The meaning of equality in a complete orthogonal basis expansion of $L^2$ and point-wise equality. Given an element $u\in L^2(0,1)$ and a complete orthonormal basis $(e_n)_n$ we have the convergence of the 'Fourier-Bessel' series
$$
u_m= \sum^m_{n=0}\langle u,e_n\rangle e_n \to \sum^\infty_{n=0}\langle u,e_n\rangle e_n
$$
in $L^2(0,1)$. In particular we know that $||u-u_m|| \to 0$ as $m\to\infty$, and what's more is that since the basis $(e_n)_n$ is maximal we can actually prove that $||u-\sum^\infty_{n=0}\langle u,e_n\rangle e_n||=0$.
Therefore, I think that the equality $u=\sum^\infty_{n=0}\langle u,e_n\rangle e_n$ is equivalent to the following:
$$
u(x)=\sum^\infty_{n=0}\langle u,e_n\rangle e_n(x) \quad \text{for almost all } x\in(0,1) \tag1
$$
which would make sense since $L^2(0,1)$ is actually the set of equivalence classes of square integrable functions which differ up to a set of measure zero.
Is this true?
If yes, does this mean that the RHS of (1) converges point-wise a.e. on $(0,1)$?
If yes, is it possible to replace $(e_n)_n$ with square integrable functions $(\phi_n)_n$ such that $[e_n]=[\phi_n]$ (i.e. so they are equal in $L^2(0,1)$) and we have the point-wise equality
$$
u(x)=\sum^\infty_{n=0}\langle u,e_n\rangle \phi_n(x) \quad \text{for all } x\in(0,1)\tag2
$$
and if so is it possible to explicitly construct such a sequence $(\phi_n)_n$?
Help with answering any of the above would be appreciated.
 A: When you write down an infinite series, you always implicitly assume in which sense the series converges. In the case of the series you wrote, it is implicitly assumed that the series converges in $L^2$, not a.e. And in general, $L^2$ convergence does not imply convergence a.e.
Now, this case (Fourier-Bessel series) is not the most general case of $L^2$ convergence, so one could in principle still hope to have convergence a.e. In fact, it may happen that for some complete basis one in fact has convergence a.e. A very famous example is that of Fourier series, see  Carleson’s theorem .
For more general orthonormal basis, I think convergence a.e. is not true in general (I guess… simply because Carleson’s theorem would be trivial at that point in the $L^2$ case). Maybe you could try to find a counterexample yourself.
To answer to the last question, keep in mind that a.e. convergence of a (countable) sequence of $L^2$ functions does not depend on the representatives you may choose in the equivalence classes (simply because a countable union of sets of measure zero still has measure zero). Therefore, even if you modify the basis with functions that agree a.e., you cannot obtain convergence a.e. with the new basis $(\phi_n)_n$ if you don’t have convergence a.e. with the original basis $(e_n)_n$ in the first place.
