Can $L^{2}$ be represented as a space of functions (not equivalence classes)? Let $X$ be the vector space of all Lebesgue-measurable functions $f:\left[a,b\right]\rightarrowℝ$ such that $\int^{b}_{a}\left|f\left(x\right)\right|^{2}dx<\infty$ (Lebesgue integral). Then we can define an equivalence relation on $X$ as follows: $f \cong g$ if $f(x)=g(x)$ almost everywhere on $\left[a,b\right]$.  Then we construct equivalence classes $\tilde{f}=\{g\in X:f\cong g\}$, and the vector space of these equivalence classes is $L^{2}[a,b]$, on which we define the norm $||\tilde{f}||_{1}=\sqrt{\int^{b}_{a}\left|f\left(x\right)\right|^{2}dx}$ (Lebesgue integral).  Now some of these equivalences classes are rather special: they contain a continuous function in them, so this is the natural choice for a representative of the equivalence class.  Let 
$D\subseteq L^{2}[a,b]$ be the subspace containing these special equivalence classes.  My basic question is, if we assign the equivalence classes in $D$ their continuous representatives, what are the natural representatives of the other equivalence classes?  
We can make this more precise.  Let $C[a,b]$ be the vector space of continuous functions $f:\left[a,b\right]\rightarrowℝ$, endowed with a norm $||f||_{2}=\sqrt{\int^{b}_{a}\left|f\left(x\right)\right|^{2}dx}$ (Riemann or Lebesgue integral).  Then the norm-completion of this space is in fact $L^{2}[a,b]$.  The upshot of all this is that $D$ is dense in $L^{2}[a,b]$, and we have a norm-respecting isomorphism $T:(D,||\cdot||_{1})\rightarrow(C[a,b] , ||\cdot||_{2})$ defined by $T(\tilde{f})\in \tilde{f}$ (assigning each element of $D$ its continuous representative).  So now the question becomes, does there exist a continuous linear extension $S$ of $T$ defined on all of $L^{2}[a,b]$ such that $S|_{D}=T$ and $S(\tilde{f})\in \tilde{f}$ ?  Well, $T$ is a bounded linear transformation (with operator norm 1) defined on a dense subspace, so it meets all the conditions of the BLT theorem other than the fact that its codomain is not a Banach space.  Thus we have to expand $C[a,b]$ to a larger subspace of $X$, so that the codomain of $T$ becomes complete.  
There are two potential ways to do this, depending on whether we define the norm $||\cdot||_{2}$ in terms of Riemann or Lebesgue integrals.  If we use Riemann integrals, we would need a subspace of $X$ consisting of Riemann-integrable functions, so we would have to answer the following in order to establish completeness: if $f_{n}\rightarrow f$ with respect to the the $||\cdot||_{2}$ (where $f$ need not be continuous), is $f$ necessarily Riemann integrable?  (My first instinct is no, because Riemann-integrability requires boundedness, and you can have a sequence of continuous functions with ever-increasing bounds, so that the limit is unbounded).  If we use Lebesgue integrals, we would need to ensure that two distinct elements of the subspace cannot have zero distance, so we would have to answer the following: if $f_{n}\rightarrow f$ and $g_{n}\rightarrow g$ with respect to the $||\cdot||_{2}$ norm (where $f$ and $g$ need not be continuous) and $f(x)=g(x)$ almost everywhere on $[a,b]$, then are $f$ and $g$ necessarily the same function?  (Again I fear the answer is no, because perhaps you can have a sequence of continuous functions that converges to a function with a removable discontinuity).
I know I've included a lot of convoluted detail, but my fundamental question is relatively simple: can we replace the equivalence classes in $L^{2}[a,b]$ with natural representative functions, using continuous representatives where possible?  Or to put it another way: does there exist a subspace $Y$ of $X$ containing $C[a,b]$, on which we can define a norm which will make it isomorphic to $L^{2}[a,b]$?
EDIT: As Gerald has pointed out, a simpler way to phrase my question is that I want a lifting of $L^{2}[a,b]$ or more generally $L^{2}(ℝ^{3})$.
Any help would be greatly appreciated.
Thank You in Advance.
 A: A "lifting" is exactly a choice of one element of each equivalence class.  When done on $L^\infty$, you want not only linear combinations of representatives to be representatives, but also products.  There is a literature on this question.  For example:  
Topics in the Theory of Lifting (Ergebnisse der Mathematik und ihrer Grenzgebiete. 2. Folge) by Alexandra Ionescu Tulcea and C. Ionescu Tulcea  
Also found in the book: in a certain precise sense (which I don't remember) lifting is impossible for $L^p$ with $p<\infty$.  
A: One way to partially answer your last question might be the following. To each $f\in L^2(a,b)$, first associate its Lebesgue primitive $F(x)=\int_a ^x f(t)dt$, then define $Tf$ as one of the four Dini derivatives of $F$, e.g.
$$ Tf(x)=\limsup _{h\to 0^+}h^{-1}(F(x+h)-F(x)).$$
Then $Tf=Tg$ everywhere if $f=g$ almost everywhere, $Tf=f$ almost everywhere, and $Tf$ is continuous if $f$ is equivalent to a continuous function. Thus the map $T$ associates to all members of a class of equivalence in $L^2$ the same function, which is the continuous representative of the class when it exists. An additional advantage is that the method is 'constructive'.
A: Well, there is a general sense in which your question can be answered in the affirmative. X = L^2 is a Banach space, and every Banach space X can be represented linearly and isometrically as a subspace of the continuous functions on a compact Hausdorff space K. The points of K are the continuous linear functionals on X. You deal with point functopns, not equivalence classes, but you have greatly extended the space of points.
A: Yes and no. 
The yes part is the Zorn lemma: consider the set of all subspaces $L\supset C[a,b]$ in the vector space of measurable square integrable functions such that no two functions in $L$ are equivalent partially ordered by inclusion. Since the union of any linearly ordered chain of such subspaces is such subspace again, we have a maximal such subspace $L$. It is easy to check that each square integrable function $f$ is equivalent to some function in $L$ (otherwise $\text{span\,}(L,f)$ is a bigger subspace).
The no part has been spelled out by Simon: no such subspace is any more reasonable or easier to put one's hands on than the Hamel basis of $\mathbb R$.
