In the proof of Theorem 6.16 in Rudin's Real and Complex Analysis, I found a part that is a bit ambiguous for me. It says:
By Linearity it follows that
\begin{equation*}
\Phi(f) = \int_Xfgd\mu
\end{equation*}
holds for every simple measurable $f$, and so also for every $f\in L^\infty(\mu)$, since every $f\in L^\infty(\mu)$ is a uniform limit of simple functions $f_i$. Note that the uniform convergence of $f_i$ to $f$ implies $\|f_i – f\|_p\to0$, hence $\Phi(f_i)\to\Phi(f)$, as $i\to\infty$.
To prove that there exist simple measurable functions $f_i$ satisfying $\|f_i-f\|_p\to0$, I firstly let $f_0\in L^\infty(\mu)$ be a bounded function satisfying $\|f-f_0\|_\infty=0$ and then constructed simple measurable functions $f_i$ that uniformly converge to $f_0$ on $X$, thereby proving $\|f-f_i\|_p \to 0$ as $i\to\infty$. However, it only shows that $f_i$ uniformly converge to $f$ on a set $X – N$ with $\mu(N) = 0$ because $f$ is essentially bounded. So, my question is: Is there another way to construct simple measurable functions $f_i$ that uniformly converge to $f$ on $X$ or is it legitimate to say "uniformly converge to $f$" in this case?
Best Answer
Construction is needed.
Since $f=f^{+}-f^{-}$, we can assume that $f\geq 0$.
For $0\leq f(x)\leq\|f\|_{L^{\infty}}:=M$ a.e. and hence the construction $f_{n}$ that \begin{align*} f_{n}(x)=\sum_{i=1}^{2^{n}}\dfrac{i-1}{2^{n}}\cdot M\chi_{f^{-1}\left(\dfrac{i-1}{2^{n}}\cdot M,\dfrac{i}{2^{n}}\cdot M\right]} \end{align*} is such that for $x\in f^{-1}\left(\dfrac{i-1}{2^{n}}\cdot M,\dfrac{i}{2^{n}}\cdot M\right]$ that \begin{align*} f_{n}(x)=\dfrac{i-1}{2^{n}}\cdot M<f(x)\leq\dfrac{i}{2^{n}}\cdot M, \end{align*} so \begin{align*} 0<(f-f_{n})(x)\leq\dfrac{M}{2^{n}}, \end{align*} but the points do not lie inside $f^{-1}(0,M]$ are of measure zero, and hence \begin{align*} \|f-f_{n}\|_{L^{\infty}}\leq\dfrac{M}{2^{n}}\rightarrow 0 \end{align*} as $n\rightarrow\infty$.