The dual of $\mathcal{L}_{\mathbb{R}}^{1}$ is $\mathcal{L}_{\mathbb{R}}^{\infty}$? Let $(E,\mathcal{A},\mu)$ be a finite measure space. I have a problem understanding the differences between spaces $\mathcal{L}_{\mathbb{R}}^{1}$ (the space of all classes  functions $f\colon E\to\mathbb R$ with $\int_E|f|\,d\mu<\infty$) and ${L}_{\mathbb{R}}^{1}$ (the space of all classes of $\mu$-equivalent functions $f\colon E\to\mathbb R$ with $\int_E|f|\,d\mu<\infty$).
We know that the dual of ${L}_{\mathbb{R}}^{1}$ is ${L}_{\mathbb{R}}^{\infty}$.
Can we say that the dual of $\mathcal{L}_{\mathbb{R}}^{1}$ is $\mathcal{L}_{\mathbb{R}}^{\infty}$?
 A: Loosely the duals are the same (in some appropriate sense). To see this is a matter
of unwinding definitions and noting that being a continuous functional with respect to the seminorm topology on $L^1(\mathbb{R})$ imposes restrictions on
the functional.
Note that $\| \cdot \|_1$ is a seminorm on $L^1(\mathbb{R})$. The topology the seminorm 
induced on this space has a local base $B_1(0,\epsilon) = \{ x | \|x\|_1 < \epsilon \}$.
Write $ x \sim y $ iff $\|x-y\|_1 = 0$. Note that if $x \sim y$ then $x-y \in B(0,\epsilon)$ for all $\epsilon>0$.
The important thing to note is that if $\phi$ is a continuous functional on 
$L^1(\mathbb{R})$ then we have $\phi(x-y) = 0$ or equivalently,
$\phi(x) = \phi(y)$. In particular, any $\phi \in L^1(\mathbb{R})^*$ is
'insensitive' to changes of measure zero.
We can define the map $J: \phi \in L^1(\mathbb{R})^* \to {\cal L}^1 (\mathbb{R})^*$ by
$(J(\phi))(\overline{x}) = \phi(x)$, where $x \in \overline{x}$.
Note that given $\overline{\phi} \in {\cal L}^1 (\mathbb{R})$, the map $\phi \in L^1(\mathbb{R})^*$ defined  by
$\phi(x) = \overline{\phi}(\overline{x})$ satisfies $J \phi = \overline{\phi}$.
(It is straightforward to show that the map $x \mapsto \overline{x}$ is continuous.)
In particular, $J$ is an isomorphism.
We have $J(L^1(\mathbb{R})^*) = J(L^\infty(\mathbb{R})) = {\cal L}^1 (\mathbb{R})^* = {\cal L}^\infty (\mathbb{R})$.
