I have a little problem with this proof (I'm using Royden), can you help me?

Let $F$ be a bounded linear functional on $L^p$, $1 \leqslant p \leqslant \infty$. Then there is a function $ge \in L^q$ such that $$F(f) = \int fg $$

Proof. Let $I_ s$ be the characteristic function of the interval $[0,s]$. Define a function $\phi$ on $[0,1]$ by $\phi (s) = F(I_s )$ $\phi$ is absolutely continuous so there is an integrable function $g$ on $[0,1]$ such that $$\phi (s) = \int _0^s g $$ thus $$F(I_s) = \int _0^1 g I _s$$ If $\psi$ is a step function we know that $\psi = \sum c_i I _{s_i}$ (except in a finite number of points) so $F(\psi)= \int g \psi$. Ok we got the result for step function.

Let $f$ a mensurable function on $[0,1]$. There is a sequence of step function $\{\psi n_ \}$ such that $\psi _n$ converge for $f$. Now I need use the Lebesgue convergence theorem, for this need show that $g \psi _n$ is less that $|g|$ Royden make two observations, maybe can help (but I don't see how): $|f-\psi _n| -> 0$ and $||f-\psi _n||_p ->0$

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    $\begingroup$ Royden starts with a bounded measurable $f$. In that case you can find $\psi_{n}$ that is uniformly bounded and converges pointwise a.e. to $f$. Then the Lebesgue bounded convergence theorem can be applied to show that $\lim_{n}\|f-\psi_{n}\|_{p}=0$. Then you use the boundedness of $F$ to conclude that $|F(f)-F(\psi_{n})| \le \|F\|\|f-\psi_{n}\|\rightarrow 0$. Then you use the newly-found representation $F(\psi_{n})=\int_{0}^{1}g\psi_{n}\,dt$ and the dominated convergence theorem to conclude that $F(f)=\int_{0}^{1}gf\,dt$ for bounded measurable $f$ (this works because $|g\psi_{n}|\le M|g|$.) $\endgroup$ – DisintegratingByParts Nov 14 '14 at 12:02

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