The dual of $L^p(X,\mu)$, where $1Note. The functions are intended to have real values.
Theorem. Let $(X,\mathcal{A},\mu)$ a $\sigma-$ finite measure space and let $p\in [1,\infty)$. For all $F\in \left(L^p(X)\right)^*$ exists a unique function $f\in L^q(X)$, where $q$ is the conjugate exponent of $p$, such that $$F(g)=\int_X gf\;d\mu\quad\forall g\in L^p(X);$$ moreover we have $$\rVert F \lVert_{(L^p(\mu))^*}=\rVert f \lVert_q.$$

If $\mu$ is only a positive measure, then the above theorem also holds for $1<p< \infty$. I want to prove this using the following steps.

Definition A set $E\in\mathcal{A}$ is said $\sigma-$ finite if exists a sequence $\{E_n\}\subseteq \mathcal{A}$ such that $E=\cup E_n$ and $\mu(E_n)<\infty$ for all $n\in\mathbb{N}$.
Let $$\Sigma=\left\{E\in\mathcal{A}\;|\; E\;\text{is}\;\sigma-\text{finite}\right\}$$

Claim $1$. Let $F\in (L^p)^*$, for all $E\in\Sigma$ exists a unique function $f_E\in L^q$ with $f_E|_{X\setminus E}=0$ such that $F(g)=\int_X gf_E\;d\mu$ for all $g\in L^p(X)$.

My proof for claim $1$.
Consider the measure $\mu_E(A):=\mu(A\cap E)$ for $A\in\mathcal{A}$, then $\mu_E$ is $\sigma-$ finite measure on $\mathcal{A}\cap E$ and for the Theorem, exists a unique $f\in L^q(E)$ such that $F(g)=\int_E gf\;d\mu_E$ for all $g\in L^p (E)$. Defining $$f_E(x)=f(x)\chi_E$$ we have that $$F(g)=\int_X gf_E\;d\mu\quad \forall g\in L^p(X)$$

Claim 2. If $E,E'\in\Sigma$ and $E\subseteq E'$, then $f_E=f_E'$ a.e in $E$.

My proof for claim $2$. It doesn't exist because I can't come to a conclusion.
Let $$\lambda (E)=\int_X\lvert f_E \rvert^q\;d\mu\quad\forall E\in\Sigma:$$

Claim 3. $\lambda$ is an increasing function with respect to inclusion.

My proof for claim $3$. For the Claim $2$ we have that $$\lambda(E)=\int_X\lvert f_E\rvert^q\;d\mu=\int_E \lvert f_E\rvert^q\;d\mu=\int_E \lvert f_E'\rvert^q\;d\mu\le \int_{E^{'}} \lvert f_E\rvert^q\;d\mu=\lambda(E')$$

Claim 4. $\lambda$ it is limited from above

My proof for claim $4$. It doesn't exist because I can't come to a conclusion.
We choose a sequence $\{E_n\}\in\Sigma$ such that $\lambda(E_n)\to m:=\sup_\Sigma \lambda(E)$:

Claim 5. $H=\bigcup_{n=1}^\infty E_n\in \Sigma$ and then $\lambda(H)=m$.

My proof for claim $5$. It doesn't exist because I can't come to a conclusion.

Claim 6. Defining $f=f_H$ we have that $f=f_E$ a.e in $E$ for all $E\in\Sigma$.

My proof for claim $6$. It doesn't exist because I can't come to a conclusion.

Claim 7. If $g\in L^q$, defininf $N=\{x\in X\;|\;g(x)\ne 0\}$, result that $N\in\Sigma$ and $F(g)=\int_X gf_{N\cup H}\;d\mu=\int_X gf\;d\mu.$

My proof for claim $7$. It doesn't exist because I can't come to a conclusion.

Question. I am aware of the fact that I have done little, but I would be grateful for any valuable hints you will give me. Thank you.

 A: *

*For claim 2, you have $L^p(E) \subset L^p\left(E'\right)$ so for all $g\in L^p(E)$, \begin{align}F(g) &= \int_X g f_E\mathrm d \mu\end{align} and since $g$ is also a member of $L^p\left(E'\right)$, \begin{align}F(g) &= \int_X g f_{E'}\mathrm d \mu = \int_X \left(g\chi_E\right) f_{E'}\mathrm d \mu = \int_X g\left(f_{E'}\chi_E\right) \mathrm d \mu\end{align}
By unicity of $f_E$ you have $f_E = f_{E'}\chi_E$ a.e. The claim 2 follows immediately.

*

*For claim 4, let $g = \left|f_E\right|^{q-1}$, since \begin{align}\int_E \left|g\right|^{p} \mathrm d \mu &= \int_E \left|f_E^{q-1}\right|^{p} \mathrm d \mu\\
&= \int_X \left|f_E\right|^{(q-1)p} \mathrm d \mu\\
&= \int_X \left|f_E\right|^{q} \mathrm d \mu = \lambda(E) < \infty\end{align} so $g\in L^p(E)$ and $\lambda(E) = \displaystyle\int_{X} \left|f_E\right|^q = F(g) \le \left\|F\right\|_{p,*}\left\|g\right\|_p = \left\|F\right\|_{p,*}\lambda(E)^{\frac1p}$ this implies that $\lambda(E)^{\frac1q} \le \left\|F\right\|_{p,*}$ independently of the set $E$ and the claim 4 follows immediately.


*For claim 5, since $E_n \subset H$ for $n\in \mathbb N$ and $\lambda$ is increasing with respect to inclusion then, $\lambda \left(E_n\right) \le \lambda (H)$. We know also that $H\in\Sigma$ so $$m \ge \lambda (H) \ge \lim\limits_{n\to \infty} \lambda\left(E_n\right) = m.$$ The claim 5 follows immediately.


*For claim 6, let $G = E\cup H\in \Sigma$, by claim 2, $f_G = f_E$ a.e. in $E$. Moreover, \begin{align}m \ge \lambda (G) &= \int_G \left|f_G\right|^q\mathrm d \mu\\
&=\int_H \left|f_G\right|^q\mathrm d \mu + \int_{G\backslash H} \left|f_G\right|^q\mathrm d \mu\\
&=\int_H \left|f_H\right|^q\mathrm d \mu + \int_{G\backslash H} \left|f_G - f_H\right|^q\mathrm d \mu\\
&= \lambda(H) + \int_{G}\left|f_G - f_H\right|^q\mathrm d \mu\\
&= m + \int_X \left|f_G - f_H\right|^q\mathrm d \mu
\end{align}
So $\displaystyle \int_X \left|f_G - f_H\right|^q\mathrm d \mu \le 0$. Hence, $f_G = f_H$ a.e. in $X$. The claim 6 follows immediately.

*

*For claim 7, let $g\in L^p(X)$ and $N_n = \left\{x \in X; \left|g(x)\right|^p > \frac1n\right\}$ then $N = \bigcup_{n} N_n$ and \begin{align}\mu\left(N_n\right) &= \int_X \chi_{N_n}\mathrm d \mu < n \int_X \left|g\right|^p\chi_{N_n}\mathrm d \mu \le n \int_X \left|g\right|^p\mathrm d \mu < \infty\end{align} This proves that $N\in\Sigma$. Using the claim 6 $f_{N\cup H} = f$ a.e. in $N\cup H$ and \begin{align} F(g) &= \int_X gf_{N\cup H}\mathrm d\mu = \int_X gf\mathrm d\mu.\end{align}
A: The theorem is theorem 6.15 on page 190 of "Real Analysis: Modern Techniques and Their Applications" by Folland:

If you are allowed to assume the theorem for $\sigma$-finite measures, then all you need is the last paragraph of the proof:

