What is the dual space if the measure is not $\sigma-$finite?

Nope, it isn't.

You can see this by showing that, in this setting, $L^\infty \cong \mathbb{C}^2$ and $[L^1]^* \cong \mathbb{C}$ (as vectorial spaces). Since $\mathbb{C}^2$ and $\mathbb{C}$ have different dimensions, they cannot be isomorphic, so the dual of $L^1$ is not $L^\infty$.

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To see the isomorphisms, try to look at

$$\begin{array}{rrcl} \phi: & L^\infty &\longrightarrow & \mathbb{C}^2 \\ & f &\longmapsto& (f(a),f(b)) \end{array}$$ and $$\begin{array}{rrcl} \psi: & L^1&\longrightarrow & \mathbb{C} \\ & f &\longmapsto& f(a). \end{array}$$ Two more hints are:

• Any finite dimensional space is isomorphic to its dual;
• In your setting, a function $f\in L^1$ is completely determined by its value in $a$.