\chi_A^{-1}[a,\infty) =\begin{cases}
\emptyset,&a>1,\\
A,& a\in (0,1],\\
\R,&a\leq 0.
\end{cases}
\] Therefore $\chi_A$ is measurable if and only if $\emptyset,A$ and $\R$ are all measurable if and only if $A$ is measurable.
Example 2. We check that $\lims f_n$ is measurable by showing that for every $a\in \R$, the set \[A:=\{x\in E:\lims f_n(x)<a\}
\] is measurable. Let $a\in \R$, then for $x\in A$, we try to find another description of $x$ in order to express $A$ in another form. Equivalently, since $x\in A$ iff $\lims f_n(x)<a$, we try to modify the statement that $\lims f_n(x)<a$.
Unsuccessful but Necessary Trial. Specifically, $\lims f_n(x)<a$ implies \[\exists N\ge 1,\forall n\ge N,f_n(x)<a.
\] Note that the last statement cannot be reversed. Since if we take $\lims$ on both sides, ``$<$'' becomes ``$\leq$''. But still we can proceed by modifying the bound $a$.
Correct Way. We have \begin{align*}
\lims f_n(x)<a&\implies \exists p\in \N,\lims f_n(x)<a-\frac{1}{p}\\&\implies \exists p\in \N,\exists N\in \N,\forall n\ge N,f_n(x)<a-\frac{1}{p}.
\end{align*} Fortunately the last statement can be reversed to $\lims f_n(x)<a$, so we have \begin{align*}
A&=\{x\in E:\lims f_n(x)<a\}\\
&=\bigg\{x\in E:\exists p\in \N,\exists N\in \N,\forall n\ge N,f_n(x)<a-\frac{1}{p}\bigg\}\\
&=\cupp_{p=1}^\infty \cupp_{N=1}^\infty \capp_{n=N}^\infty \bigg\{x\in E:f_n(x)<a-\frac{1}{p}\bigg\}\\
&=\cupp_{p=1}^\infty \cupp_{N=1}^\infty \capp_{n=N}^\infty f_n^{-1}\bigg(-\infty,a-\frac{1}{p}\bigg).
\end{align*} As every union and intersection is countable, by the hypothesis that $f_n$'s are measurable, we are done.
Now $\limi f_n=-\lims (-f_n)$ is measurable by the last paragraph.
Example 3. Let $a\in \R$, then \[
f(x)<a\iff \sup_{s\in S}f_s(x)<a\implies \forall s\in S,f_s(x)<a.
\] The last statement cannot be reversed to the first since $<$ becomes $\leq$ when taking supremum. Then what to do? Either we shrink the bound $a$ as in Example 2 (left as exercise) or we try $\leq a,>a$ or $\ge a$ instead.
Note that when $S$ is countable the statement is trivial since we have already such a result that \[
f_1,f_2,\dots \text{ measurable}\implies \sup_{n\ge 1}f_n\text{ measurable}.
\] We expect the proof is a bit different. Thus continuity must be brought into consideration.
Method 1. Let's consider $\leq a$. We have for any $a\in \R$, \[
f(x)\leq a\iff f_s(x)\leq a,\forall s\in S,
\] therefore \[
f^{-1}(-\infty,a]=\capp_{s\in S}\overbrace{f_s^{-1}\underbrace{(-\infty,a]}_{\text{closed}}}^{\text{closed by continuity}},
\] thus $f^{-1}(-\infty,a]$ is an intersection of closed set, it must be closed and hence measurable.
Method 2. Let's consider $>a$, from Supremum Limit Theorem we have \[
f(x)>a\iff \exists s\in S,f_s(x)>a,
\] therefore we have \[
\{x\in \R:f(x)>a\}= \cupp_{s\in S} \{x\in \R:f_s(x)>a\}=\cupp_{s\in S}f_s^{-1}(a,\infty).
\] Since $\cupp_{s\in S}f_s^{-1}(a,\infty)$ is a union of open sets, which is open and hence measurable.
Remark. We wouldn't expect to argue like ``$\cupp_{s\in S}\text{(measurable)}$ is measurable'' since the union $\cupp_{s\in S}$ is uncountable.
Example 4. (Uploaded on 26/11/2013) For every $n\in \N$ we may set $\epsilon=\frac{1}{n}$, then by hypothesis there is $g_n\in C(\R)$ and a measurable $A_n$ such that $f|_{A_n}=g|_{A_n}$ and $m(\R\setminus A_n)<\frac{1}{n}$. We expect $A:=\cupp_{n=1}^\infty A_n$ is so ``huge'' that $m(\R\setminus A)=0$. Indeed, \[
\forall n\in \N,\quad m(\R\setminus A)\leq m(\R\setminus A_n)<\frac{1}{n},
\] by taking $n\to\infty$, $m(\R\setminus A)=0$. Let's show that $f|_A$ is measurable as $\R\setminus A$ is negligible, more precisely, by Exercise 1, $f$ is measurable if and only if $f|_{A}$ is measurable.
Method 1. We have \[
\{x\in A:f|_A(x)>a\}=\cupp_{n=1}^\infty \{x\in A_n:g_n|_{A_n}(x)>a\}=\cupp_{n=1}^\infty \overbrace{\underbrace{\big(g_n^{-1}(a,\infty)\big)}_{\text{open}}\cap A_n}^{\text{measurable}},
\] so $\{x\in A:f|_A(x)>a\}$ is a countable union of measurable sets, which must be measurable.
Method 2. Define $A_1'=A_1$ and $A_n'=A_n\setminus \cupp_{i=1}^{n-1}A_i$ for $n\ge 2$, then $\{A_n'\}$ is disjoint and $\cupp_{n=1}^\infty A_n'=\cupp_{n=1}^\infty A_n=A$, hence \[
f|_A=\sum_{n=1}^\infty f|_{A_n'}\chi_{A_n'} = \sum_{n=1}^\infty g_n\chi_{A_n'},
\] so $f|_A$ is a pointwise limit of measurable functions, and thus measurable.
Example 5. We note that the Cauchy criterion for the convergence of $\sum_{i=1}^\infty f(a+i)$ can be written as \[\forall k\in \N,\exists N\in \N, \forall n\ge N,\forall m\ge n,\sum_{i=n}^m f(a+i)<\frac{1}{k}.\] Therefore \[
\bigg\{a\in [0,\infty):\sum_{i=1}^\infty f(a+i)\in \R\bigg\}=\capp_{k=1}^\infty \cupp_{N=1}^\infty \capp_{n=N}^\infty \capp_{m=n}^\infty\bigg\{ a\in [0,\infty) : \sum_{i=n}^m f(a+i)<\frac{1}{k}\bigg\}.
\] Clearly it remains to check the measurability of $f_i(x):=f(x+i)$. For every interval $(a,b)$ we have \begin{align*}
x\in f_i^{-1}(a,b)&\iff f_i(x)= f(x+i)\in (a,b)\\
&\iff x+i\in f^{-1}(a,b)\iff x\in -i+f^{-1}(a,b),
\end{align*} hence \[f_i^{-1}(a,b)= -i + f^{-1}(a.b).
\] Since a translation of a measurable set is still measurable, $f_i$ is measurable. Now \[S=\capp_{k=1}^\infty\cupp_{N=1}^\infty\capp_{n=N}^\infty \capp_{m=n}^\infty\underbrace{(f_n+f_{n+1}+\cdots +f_m)^{-1}\big[0,\tfrac{1}{k}\big )}_\text{measurable}\] is measurable.
Example 6. Method 1. By Exercise 5, $f(x)=9$ a.e., and constant function $9$ is measurable, so $f$ is measurable.
Method 2. Let \[A_k=\{x\in (0,1]:x=0.a_1a_2\dots_\times,a_i\ge k,\exists\, i\}.\] Then observe that when $f(x)=\ell$, we have \[x\in A_1,x\in A_2,\dots,x\in A_\ell,x\not\in A_{\ell+1},\dots,x\not\in A_{9}.\] Then we find that $\sum_{i=1}^9\chi_{A_i}(x)=\ell$, therefore we have \[
f=\sum_{i=1}^9\chi_{A_i}.
\] It remains to check that each $A_i$ is measurable, we leave it as a practice in Exercise 6.
Method 3. In method 2 the construction is less intuitive (originally I want to have a ``direct counting''). Let's redo it as follows.
If $f(x)=\ell$, then $a_i\leq \ell$ for all $i$, and thus $a_i\leq j$ for all $i$, where $j=\ell,\dots,9$. If we define \[
A_j=\{0.a_1a_2\dots_\times :a_i\leq j ,\forall i\},
\] then by definition we have \[
x\in A_\ell,\dots, x\in A_{9}\quad \text{and}\quad x\not\in A_i,\text{ for all $i<\ell$},
\] therefore \[
\sum_{j=1}^9 \chi_{A_j}(x)=10-\ell\iff f(x)=10-\sum_{j=1}^9 \chi_{A_j}(x).
\] Now it is easy to check that each $A_j$ is measurable by considering its first $n$ digits and take intersection.
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