For context, it is assumed that $\mu$ is a measure on a locally compact Hausdorff space $X$ which has all properties that follow from Riesz Representation Theorem (regularity, etc.) in the book (if I'm not wrong, this means $\mu$ is a Radon measure).
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The author begins by assuming $0 \le f < 1$ and concludes the first part of the proof by noting $0\le f \le 1$? Though both seem to work, for the sake of consistency one of the two must be a typographical error – and I believe it's the first one. Could you confirm?
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Why is $2^{-n}h_n(x) = t_n$ except in $V_n – K_n$?
I know that $h_n(x) = 1$ for all $x\in K_n$ and $h_n(x) = 0$ for all $x\notin V_n$. Also, $0\le h(x) \le 1$ for $x\in V_n-K_n$. -
How is the assumption of compactness dropped? If $\mu(A) < \infty$ then for every $\epsilon_0 > 0$ there is compact $K\subset A$ such that $\mu(A-K) < \epsilon_0$. I'm thinking about $\tilde f = \chi_K f$, and since $K$ is compact, it is clear that we can find a continuous real-valued function $g$ such that $\mu(\{x: g(x) \ne \tilde f(x)\} < \epsilon_0$, for every $\epsilon_0 > 0$. How do I use this to approximate $f$ by $g$ in the sense of Lusin's theorem?
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Next they consider $f$ to be complex measurable, $B_n = \{x: |f(x)| > n\}$ and claim $\bigcap_{n=1}^\infty B_n = \varnothing$. To see this, assume $x\in \bigcap_{n=1}^\infty B_n$, thus $x \in B_n$ for all $n$, i.e. $|f(x)| > n$ for all $n$. Taking limits as $n\to\infty$, $|f(x)| \ge \infty$, that is $|f(x)| = \infty$. This is not possible, since $\infty$ is not in $\mathbb R$. Seems fine? Thus, $\mu(B_n)\to 0$ as $n\to\infty$. $f$ coincides with $\tilde f_n = (1 – \chi_{B_n}) f$, which is bounded. So we can find a continuous $g$ such that $\mu(\{x: g(x) \ne \tilde f_n(x)\}) <\epsilon/2$. Since $\tilde f_n$ and $f$ agree except on a set whose measure approaches zero as $n\to\infty$, it is clear that we can find $n$ large enough so that $\mu(\{x: \tilde f_n(x) \ne f(x)\}) <\epsilon/2$. What can I say about $\mu(\{x: f(x) \ne g(x)\})$?
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I can see that $g_1 = \varphi\circ g$ satisfies $(2)$, but how do I show that it satisfies $(1)$, given that $g$ satisfies $(1)$?
Proof attached for reference:
Apologies in advance for posting pictures of the proof (it is huge to type, and the pictures are only for reference). I have written all else in MathJax. Thank you!
Best Answer
That's most likely a typo. Anyway if $0\leq f\leq 1$, then $0\leq f/2<1$, so we get the same proof. What we really need here is $f$ be bounded.
I don't have the book right by my hand. But if my memory serves me well, $t_n(x)$ is essentially the $n$-th digit (counting from left) of $f$ in binary representation. The function $2^nt_n$ is the characteristic function of some measurable set $T_n$. If $x\in K_n\subseteq T_n$, then \begin{align} 2^nt_n(x)=h_n(x)=1. \end{align} If $x\in X-V_n\subseteq X-T_n$, then \begin{align} 2^nt_n(x)=h_n(x)=0. \end{align}
You are in the right direction. Just let $\epsilon_0=\epsilon/2$. Then $g$ differ from $\tilde{f}$ on a set $E$ of measure less than $\epsilon/2$, and $\tilde{f}$ differ from $f$ on $A-K$ and $\mu(A-K)<\epsilon/2$. Now $g$ agrees with $f$ outside $(A-K)\cup E$, and $\mu((A-K)\cup E)\leq\mu(A-K)+\mu(E)<\epsilon$.
Are you asking about $\mu(\{g(x)\neq\tilde f_n(x)\})$? Because you just concluded yourself that \begin{align} \mu(\{f(x)\neq\tilde f_n(x)\})<\epsilon/2. \end{align} Anyway the idea is that $f$ differ from the bounded function $\tilde{f}_n$ on a set $X-B_n$ of measure $<\epsilon/2$, and some $g\in C_c(X)$ differ from $\tilde{f}_n$ on a set of measure $<\epsilon/2$ by the previous design. Hence $g$ differ from $f$ on a set of measure $<\epsilon$. Also don't forget that since $\tilde{f}_n$ takes on complex values, we actually need to approximate its real and imaginary parts separately.
If $g(x)=f(x)$, then $g(x)\leq R$, so $\varphi\circ g(x)=g(x)=f(x)$. In other words, \begin{align*} \{x\in X\colon g(x)=f(x)\}\subseteq\{x\in X\colon\varphi\circ g(x)=f(x)\}\,. \end{align*} Taking their complements in $X$, we get \begin{align*} \{x\in X\colon g(x)\neq f(x)\}\supseteq\{x\in X\colon\varphi\circ g(x)\neq f(x)\}\,. \end{align*} Hence \begin{align*} \mu(\{f(x)\neq\varphi\circ g(x)\})\leq\mu(\{f(x)\neq g(x)\})\,. \end{align*}