Up till now, I thought saying $u \in L^2([0,1])$ is the same as saying $u \in L^2((0,1))$, because I see people emphasizing "$u$ is Lebesgue integrable over $[0,1)$".
I thought the whole point of the Lebesgue integral is that null sets don't matter, so it doesn't matter whether the endpoint is included or not. What do I miss?
On
It depends on how one defines $L^2(X)$. One way of defining $L^2(X)$ is as the set of all measurable functions $f:X\rightarrow\mathbb{R}$ such that $|f|^2$ is integrable over $X$. It is, however, more standard to view $L^2(X)$ to be the set of equivalence classes of functions whose domain is $X$ and are square-integrable over $X$. Two functions $f$ and $g$ are deemed equivalent if $f$ and $g$ differ only on a set of measure zero. We will take this second definition.
Now, let $Y$ be a measure zero subset of $X$ and consider $L^2(X\setminus Y)$. Even though the equivalence classes of functions in $L^2(X)$ and $L^2(X\setminus)$ are different (they are equivalence classes from different sets of functions).
There is, however, a natural bijection $L^2(X)\simeq L^2(X\setminus Y)$. The map $L^2(X)\rightarrow L^2(X\setminus Y)$ is the image of the restriction of a function on $X$ to $X\setminus Y$ under the equivalence relation. The map in the other direction takes an equivalence class in $L^2(X\setminus Y)$ with representative $f$ to the equivalence class of the function $$ \widetilde{f}(x)=\begin{cases}f(x)&x\in X\setminus Y\\0&x\in Y\end{cases}. $$ One can show that this map is well-defined (does not depend on the choice of $f$) and $\widetilde{f}$ is square-integrable on $X$. These two maps are also inverses of each other.
Note that the choice of $0$ for $x\in Y$ is arbitrary, since $Y$ is a set of measure zero, so the value on $Y$ is irrelevant. Also, some people take this natural bijection as an identification and allow $L^2(X)$ to be equivalence classes of functions which are defined almost everywhere on $X$ and are equivalent if they agree on full-measure subsets.
Expanding on @MichaelBurr's comment: $L^2[0,1]$ does not consist of functions, but rather of equivalent classes of functions. We say that two functions are in the same equivalence class if the set they disagree on is null.
As such, when we say $1/\sqrt[4]{x}$ is a member of $L^2[0,1]$, we are saying it is in an equivalence class that contains a function that is square integrable.
One such function would be $$f(x) = \left\{ \begin{array}{ll} 1/\sqrt[4]{x} & x \neq 0\\ 0 & x=0\end{array}\right.$$
Since according to Lebesgue measure $\{0\}$ and $\{1\}$ are both null sets, it does not make a difference when considering $L^2(0,1)$, $L^2[0,1)$ or $L^2[0,1]$. They are literally different, since the functions are defined on different domains. However, since the difference in their domains is null there is no qualitative difference.
If we were talking about another measure, such as $\delta_0$, we would have to be more careful. $\delta_0(A) = 1$ if $0 \in A$ and $\delta_0(A)=0$ if $0\not\in A$. Now lets consider the measure $\mu = m + \delta_0$ (the sum of Lebesgue measure and $\delta_0$).
Now the spaces $L^2( (0,1], \mu)$ and $L^2( [0,1], \mu)$ are different not just literally, but qualitatively. This is because the set $\{0\}$ is no longer null.