Good morning, I was trying to analyze the demonstration of Lusin's Theorem of the book "Measure Theory and Probability Theory - Krishna A. pg.69" and I found myself having three difficulties, I would appreciate if you could help me.
Lusin's Theorem:
If $F:\mathbb{R\rightarrow{\mathbb{R}}}$ is a decreasing function and $\mu=\mu_{F}^{*}$ the corresponding Lebesgue Stieltjes measure on $(\mathbb{R},M_{\mu_{F}^{*}})$. $f:\mathbb{R}\rightarrow{\mathbb{R}} $ $(M_{\mu_{F}^{*}},B(\mathbb{R}))$ measurable and $\mu({x: \left |{f(x)}\right |=\infty})=0$. Then $\forall{\epsilon>0}$, there is a continuous function $g:\mathbb{R}\rightarrow{\mathbb{R}}$ such that $\mu({x: f(x)\neq{g(x)}})<\epsilon$.
The proof begins by setting $-\infty<a<b<\infty$ and proves that $\mu({x: a\leq{x}\leq{b},\left |{f(x)}\right |>K})<\epsilon/2$ $K\in{(0,\infty)}$, then define $f_{K}(x)=f(x)I_{[a,b]}(x)I_{\left |{f}\right |\leq{K}}$ that is bounded for being a product of bounded functions and measurable for the same reason.
Then it makes the following statement which will prove later: For all $\epsilon>0$ , there is a continuous function $g:[a,b]\rightarrow{\mathbb{R}}$ such that $\mu({x:f_K(x)\neq{g(x)},a\leq{x}\leq{b}})<\epsilon/2$
1) My first question is that it says this implies $\mu({x: a\leq{x}\leq{b},f(x)\neq{g(x)},})<\epsilon$ . What I suppose is that we must use the two inequalities we have that are less than $\epsilon/2$. And maybe separate the values when $\left |{f}\right |>K$ y $\left |{f}\right |\leq{K}$
Then he begins to demonstrate the statement he made and how $f_K$ is measurable, for each $\epsilon>0$ there is a simple function $s(x)=\displaystyle\sum_{i=1}^k{c_iI_{A_i}(x)}$ with $A_i\subset{[a,b]}$, $A_i\in{M_{\mu_{F}^{*}}} $ all disjoint, with $\mu(A_i)<\infty$ and $c_i\in{\mathbb{R}}$. Then by a theorem indicating for each $A_i$ and $\eta>0$ there exists a finite number of disjoint open intervals $I_{ij}=(a_{ij},b_{ij}), j=1,...,n_{i}$ such that $\mu(A_i \Delta \displaystyle\bigcup_{j=1}^{n_{i}}{I_{ij} })<\eta/2k$ and for an exercise there exists a continuous function $g_{ij}$ such that $\mu(g_{ij}^{-1}(1) \Delta I_{ij})<\eta/kn_{i}$, $j=1,2,...,n_{i}, i=1,2,...,k.$. where $ g_{ij}(x) = \left \{ \begin{matrix} 1 & \mbox{if } a+\eta\leq{x}\leq{b-\eta} \\ 0 & \mbox{if } x\not\in{(a,b)} \\ lineal & \mbox{in} [a,a+\eta]\cup{[b-\eta,b]}\end{matrix}\right. $. That is to say $g_{ij}^{-1}(1)=[a+\eta,b-\eta]$ Defines $g_{i}=\displaystyle\sum_{j=1}^{n_{i}}{g_{ij}}, 1\leq{i}\leq{k}$ 2) Why from here $\mu(A_i \Delta g_{i}^{-1}(1))<\eta/k$
Then define for each $n\geq{1}, h_{n}(.)=g_{\displaystyle\frac{1}{2^n},\displaystyle\frac{1}{2^n}}(.)$ y $A_{n}={x: a\leq{x}\leq{b}, \left |{f_{K}(x)-h_{n}(x)}\right |>1/2^n}$
3) Why then $\mu{A_{n}}<1/2^n$ ?
Thank you The link of the book is Measure Theory and Probability Theory - Krishna A.