What is the dual space if the measure is not $\sigma-$finite? Let $(X,M,\mu)$ measure space where $\mu$ is not $\sigma-$finite. What is the dual of $L^1(\mu)$ in this case? For example:

Let $X = \{a,b\}$ and define $\mu(a) = 1$, $\mu(b) = \mu(X) = \infty$, and $\mu(\emptyset) = 0$.

In this case, is $L^{\infty}(\mu)$ the dual of $L^1(\mu)$?
 A: 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 $L^\infty$.
-#-#-#-#-#-#-#-#-#-#-#-#-
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 $a$.

A: Question 1: How did OP create $\mu$?
Answer: By writing it down.
.
Problem 2: Prove $\mu$ is not $\sigma$-finite.
Solution: If $\mu$ were $\sigma$-finite, then $X = \cup_{n=1}^\infty A_n$ with each $A_n$ measurable and $\mu(A_n)<\infty$. Some $A_n$ would have to contain $b$, but it can't since $\mu(A_n) < \infty$. So $\mu$ is not $\sigma$-finite.
.
Claim 3: $\mathbb{C}^2$ and $\mathbb{C}$ are not isomorphic as vector spaces.
Proof: They have different dimensions.
.
Proposition 4: $L^\infty(X) \cong \mathbb{C}^2$ as vector spaces.
Proof: Define $\Phi: L^\infty(X) \to \mathbb{C}^2$ by $\Phi(f) = (f(a),f(b))$. Note $\Phi$ indeed maps into $\mathbb{C}^2$, since $f \in L^\infty$ implies $|f(a)|,|f(b)| < \infty$ since each $a,b$ get positive measure. Clearly $\Phi$ is linear and injective. And it is clearly surjective. So it is an isomorphism.
.
Lemma 5: $L^1(X) \cong \mathbb{C}$ as vector spaces.
Proof: Define $\Phi: L^1(X) \to \mathbb{C}$ by $\Phi(f) = f(a)$. Note $\Phi$ indeed maps into $\mathbb{C}$, since $|f(a)| < \infty$ since $a$ has positive measure. Clearly $\Phi$ is linear and surjective. It is injective since any $f \in L^1(X)$ must have $f(b) = 0$, since $b$ has infinite measure.
.
Fact 6: Let $V$ be a finite dimensional vector space. Then $V^* \cong V$ as vector spaces.
Proof: Google.
.
Corollary 7: $(L^1(X))^* \cong \mathbb{C}$ as vector spaces.
Proof: Use Lemma 5 and Fact 6.
.
Theorem 8: $L^\infty(X)$ is not the dual of $L^1(X)$.
Proof: What this means is that $L^\infty(X)$ is not isomorphic to $(L^1(X))^*$ as vector spaces. If they were isomorphic, then by Proposition 4 and Corollary 7, we'd have $\mathbb{C}^2 \cong \mathbb{C}$ as vector spaces, contradicting Claim 3. $\square$
