I was wondering what the dual space of $L^2(\mathbb{R},L^1(0,1))$ is? (equipped with Lebesgue measures) Formally, one would suspect that it is just $L^2(\mathbb{R},L^{\infty}(0,1))$. But this may be a bit too quick, as there many references that suggest that this formally replacing each space by its dual is only possible in the reflexive case.
Here, is already a good answer at Math.stackexchange, but it only works for finite measure spaces: click me.
"Finite measure" is no restriction. Either change the measure on $\mathbb R$ to an equivalent finite one, or write $L^2(\mathbb R)$ as the $l^2$-direct sum of spaces $L^2([n,n+1])$.
As in the link you give, since $L^\infty(0,1)$ fails the Radon-Nikidym property, the dual of $L^2(\mathbb R,L^1(0,1))$ is strictly larger than $L^2(\mathbb R,L^\infty(0,1))$.
When you start with a bounded linear functional on $L^2(\mathbb R, L^1(0,1))$, you come up with a vector measure with values in $L^\infty$. If that vector measure has a Radon-Nikodym derivative, it is the member of $L^2(\mathbb R,L^\infty(0,1))$ that you want. But if not, then your functional does not correspond to an element of $L^2(\mathbb R,L^\infty(0,1))$.
$\def\rmd{{\kern.4mm\rm d\kern.4mm}}\def\sp{\kern.4mm}$For $1\le p<+\infty$ and $I=[\sp 0\sp,1\sp]$ and $F=L^p(\mathbb R,L^1(I))\sp$, a simple example of a continuous linear functional on $F$ that is not given by any element in $L^{p^*}(\mathbb R,L^{+\infty}(I))$ is $\vec x\mapsto\int_0^1\int_0^t x(t,s)\rmd s\rmd t$ when $x:\mathbb R\times I\to\mathbb K$ represents the element $\vec x$ of $F\sp$. The proof is left an exercise to the reader.
This is mainly an additional comment to Gerald Edgar's answer.
Your question has been partially answered in that it has been shown that the naive version is false. However, there is a concrete representation of the dual which might be of interest. The result becomes more transparent in a more general setting which might also be of interest to you. Consider the space $L^p(\mu,E)$ where $\mu$ is a finite measure, $1\leq p<\infty$ and $E$ is a Banach space. Then its dual is naturally identifiable with $L^q(\mu,E')$ if and only if $E'$ has RNP ($q$ the conjugate of $p$). This is treated in detail in the beginning of chapter IV of "Vector measures" by Diestel and Uhl. However, there is a representation which works for any Banach space. I have not found this result explicitly in the literature but the basic idea is due to L. Schwartz (Sem. d'anal. fonctionelle (1974-75), Exp. 4---readily available online). I will state the result for the case of separable $E$: The dual is the space $L^q_{w^\ast}(\mu,E')$, where the subscript $w\ast$ refers to the fact that we are using functions which are weak star rather than norm measurable. Thus in your case, the dual is $L^2_{w\ast}(\mu,L^\infty)$. (As pointed out above there are standard methods to extend such results to the case of $\sigma$-finite measures, in particular the real line with Lebesgue measure).
