Duality between $L^p$ spaces and weak or weak* convergence From Wikipedia about $L^p$ spaces:

If the measure $μ$ on $S$ is sigma-finite, then the dual of $L^1(μ)$ is isometrically isomorphic to $L^∞(μ)$ (more precisely, the map $κ_1$ corresponding to $p = 1$ is an isometry from $L^∞(μ)$ onto $L^1(μ)^∗$).
The dual of $L^∞$ is subtler. Elements of $(L^∞(μ))^∗$ can be identified with bounded signed finitely additive measures on $S$ that are absolutely continuous with respect to $μ$. See ba space for more details. If we assume the axiom of choice, this space is much bigger than $L^1(μ)$ except in some trivial cases. However, there are relatively consistent extensions of Zermelo-Fraenkel set theory in which the dual of $ℓ^∞$ is $ℓ^1$.

Note that $V^*$ above means the continuous dual of $V$, if I am correct.
From Wikipedia about continuous dual space:

there is always a naturally defined continuous linear operator $Ψ : V → V′′$ from a normed space $V$ into its continuous double dual $V′′$, defined by
  $$
    \Psi(x)(\varphi) = \varphi(x), \quad x \in V, \ \varphi \in V'. \, 
$$
  As a consequence of the Hahn–Banach theorem, this map is in fact an isometry, meaning $||Ψ(x)|| = ||x||$ for all $x \in V$.

Note that $V'$ above means the continuous dual of $V$.
My questions are as following. The first quote says when the measure $\mu$ is sigma finite, the continuou dual of $L^1(\mu)$ is $L^\infty(\mu)$. Both $L^1(\mu)$ and $L^\infty(\mu)$ are normed spaces, according to the second quote, the double continuous dual of $L^1(\mu)$ should be $L^1(\mu)$ itself, but according to the first quote, the continuous dual of $L^\infty(\mu)$ may not be $L^1(\mu)$ and may be much bigger than $L^1(\mu)$. So I was wondering what I have misunderstood? 
Thanks and regards!
 A: If the map $\Psi: V \rightarrow V^{\ast\ast}$ is surjective then $V$ is called reflexive. Notice that a non-reflexive vector space can still be isomorphic to its dual, just not via the canonical mapping, there's some discussion and a reference of this here. In general the map $\Psi$ is an linear embedding of $V$ into $V^{\ast\ast}$. 
Here is a somewhat illustrative example. I will work with the sequence spaces $\newcommand{\N}{\mathbb N} \ell^1(\N)$, $\ell^\infty(\N)$ and $\ell^\infty(\N)^\ast$, where $\mathbb N$ is endowed with the counting measure. Consider the subspace $c(\mathbb N)\subset \ell^\infty(\N)$ of convergent sequences. Define a functional $f: c(\mathbb N) \rightarrow \mathbb C$ by 
$$f(x_n)=\lim_{n\rightarrow \infty} x_n.$$
Notice that no element of $\ell^1(\N)$ induces this functional on $c(\mathbb N)$. To see this pick a sequence $y \in \ell^1(\N)$ then 
$$\hat{y}(x)=\sum_{n=1}^\infty y_nx_n,$$
in particular if $y\neq 0$ we may find some $y_n\neq 0$ and $\hat{y}(e_n) \neq 0$ but $f(e_n)=0$. Now we can extend $f$ to a functional $\tilde{f}$ on $\ell^\infty(\N)$ by Hahn Banach. We see that $\tilde{f} \notin \Psi(\ell^1(\N))$ so $\ell^1(\N)$ is not reflexive. 
