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.
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!