Dual of $L^p$ space avoiding reflexivity and Radon-Nikodym Theorem

Question

Let $(X,\mathscr{F},\mu)$ be a measure space, with $\mu(X)<+\infty$, let $p\in [1,+\infty[$ and $q$ its Hölder-conjugate (that is, $1/p+1/q=1$). If $T\in (L^p(X,d\mu))^*$ is a continuous functional such that for every nonnegative $f\in L^p$ one have $T(f)\geq 0$, we wish to show that there exists a function $g\geq 0$ in $L^q$ such that $$ T(f) = \int_X fg \; d\mu, \quad \forall f\in L^p. $$ The procedure is as follows:

1. Show that $\lambda(A)=T(\chi_A)$ for every $A\in\mathscr{F}$ defines a finite positive measure on $X$.

2. Let $\nu=\mu+\lambda$. Prove that $\mu(A)=0$ iff $\nu(A)=0$ and that $\mu(A)=0\Rightarrow \lambda(A)=0$.

3. Let $S:L^2(X,d\nu)\to\mathbb{R}$ be defined by $$ S(f) = \int_X f \; d\lambda, \quad \forall f\in L^2(X, d\nu). $$ Show that $S$ is a well-defined linear continuouos functional and that $$ |S(f)|\leq \|f\|_{L^2(X,d\nu)}\sqrt{\lambda(X)}. $$

4. Using the Riesz's representation theorem for Hilbert spaces show that there exists a function $h\in L^2(X,d\nu)$ such that $$ \int_X f \; d\lambda = \int_X fh \; d\nu, \quad \forall f\in L^2(X,d\nu). $$

5. By considering $f=\chi_{h<0}$ and $f=\chi_{h\geq 1}$ show that $0\leq h<1$ $\nu$-a.e. and so we can replace $h$ by a function $\nu$-a.e. equal to $h$ taking values in the interval $[0,1[$.

6. Show that for every function $f\geq 0$ in $L^p(X,d\mu)$ $$ T(f) = \int_X fh \; d\mu + \int_X fh \; d\lambda \quad \text{and} \quad T(f(1-h)) = \int_X fh \; d\mu. $$

7. Show that for every positive integer $k$ the function $\min(f/(1-h),k)$ belongs to $L^p(X,d\mu)$ and that $$ T(\min(f/(1-h),k)) = \int_X \min(f/(1-h),k) \; d\mu. $$

8. Take $g=h/(1-h)$ and conclude by making $k\to +\infty$.

I'm stuck in the 7th step of this proof, so if someone can help me with a hint or a proof for this part, I'll be very gratefull. I don't need already known proofs as Brezis (using reflexivity) or Folland (using Radon-Nikodym Theorem) present.

Answer 1

As far as I can see there is a mistake in 7) and that is the only reason you got stuck. Replace $f$ in 6) by $\min (\frac f {1-h},k)$. You will get 7) with an extra factor of $1-h$ on the left. Now let $k \to \infty$.