Lebesgue integration of step function Let $J$ be a finite subinterval of the real line and $f:J\rightarrow\mathbb{R}$ a simple function taking on values $c_1,...,c_n$. The function $f$ is called a step function if $f^{-1}(c_i)$ is a finite union of intervals for each $i$. Given a simple function $s:J\rightarrow\mathbb{R}$ and a positive number $\varepsilon$, show that there exists a step function $f$ such that $$\int_J|s-f|\,d\mu_L<\varepsilon$$ Hint: Show that, if $A$ is a measurable subset of $J$, there exists a finite union of intervals $B$ such that $d(A,B)=\mu(S(A,B))<\varepsilon$. Now prove the above for $s=1_A$. Proceed.
I'm not sure how to do this problem, even with the hint. I don't know how to show the statement in the hint or how to get from there to the question itself. 
 A: Assuming that we are talking about the Lebesgue measure, let us use the letter $\lambda$, we remind its definition, for every $\lambda-$measurable set $A$:
$$\lambda(A):=\inf\left\{\sum_{i=1}^\infty l(I_i):I_i\text{ is an open interval, }i=1,2,\dots,\ A\subseteq\bigcup_{i=1}^\infty I_i\right\}$$
where $l(I)$ is the length of an iterval $I$ - so, if the interval has boundaries $a,b\in\mathbb{R}$ with $a<b$ then $l(I)=b-a$.
Note now that, if $I_i$, $i=1,2,\dots$ are all open intervals, and $I=\bigcup_i I_i$ is their union, we can see that the following sequence of sets:
$$\begin{align*}
B_1&=I_1\\
B_2&=I_1\setminus B_1\\
&\vdots\\
B_n&=I_n\setminus\bigcup_{i=1}^{n-1}B_i\\
&\vdots
\end{align*}$$
It is clear that $B_k\cap B_l=\varnothing$ for every $k,l\in\mathbb{N}$ and that:
$$\bigcup_{i=1}^\infty B_i=\bigcup_{i=1}^\infty I_i$$
Moreover, for every $k=1,2,\dots$, we can see, from the definition of $B_k$, that $B_k$ is a finite union of (not necessarily open) intervals - it is an open interval "minus" a finite number of open intervals. Also, note that:
$$\sum_{i=1}^\infty l(I_i)\geq\lambda\left(\bigcup_{i=1}^\infty I_i\right)=\lambda\left(\bigcup_{i=1}^\infty B_i\right)=\sum_{i=1}^\infty\lambda(B_i)=\sum_{i=1}^\infty l(B_i)\tag{$\star$}$$
Taking infima to $(\star)$, we can easily see that:
$$\lambda(A)=\inf\left\{\sum_{i=1}^\infty l(B_i):B_i\text{ is a finite union of  intrv., }B_i\cap B_j=\varnothing,\ \forall\ i,j\in\mathbb{N},\ A\subseteq\bigcup_{i=1}^\infty B_i\right\}$$
Can you now prove the hint, after this sub-hint? I'll come up with a complete solution, later on, if you cannot get it till the end! :)
Edit: Note at first that, if $\lambda(A)=0$, then is is easy to find such a set - e.g. a subinterval of $J$ with length lesser than $\epsilon$. So let us assume that $\lambda(A)>0$.
Let $\epsilon>0$. By definition of the Lebesgue measure - the "alternative" definition - we have that there exists a sequence $B_n$, where $B_n$ is a finite union of intervals, $B_k\cap B_l=\varnothing$, $\forall\ k,l=1,2,\dots$ such that:
$$\sum_{n=1}^\infty l(B_n)-\lambda(A)<\frac{\epsilon}{2}$$
Also, since $B_n$ are not intersecting, we have that:
$$\lambda\left(\bigcup_{n=1}^\infty B_n\setminus A\right)<\frac{\epsilon}{2}$$
Consider now the set sequence $$S_n=\bigcup_{i=1}^nB_i\setminus A$$ It is clear that $S_n$ is increasing and, hence:
$$\lim_{n\to\infty}\lambda(S_n)=\lambda\left(\bigcup_{i=1}^\infty B_i\setminus A\right)$$
So, for every $n\in\mathbb{N}$
$$\lambda(S_n)<\frac{\epsilon}{2}$$
Also, note that, if $$a_n=\lambda\left(A\setminus\bigcup_{i=1}^nB_i\right)$$
then $a_n$ is decreasing and, since it is bounded, it is convergent. Since $a_1<\infty$, we have that:
$$\lim_{n\to\infty}a_n=\lambda\left(A\setminus\bigcup_{i=1}^\infty B_i\right)=\lambda(\varnothing)=0$$
So, there exists a $n_0\in\mathbb{N}$ such that, for every $n\geq n_0$:
$$0\leq a_{n}<\frac{\epsilon}{2}\Rightarrow0\leq\lambda\left(A\setminus\bigcup_{i=1}^n B_i\right)<\frac{\epsilon}{2}$$
Now, let $$B=\bigcup_{i=1}^{n_0}B_i$$ so:
$$\lambda(S(A,B))=\lambda(A\setminus B)+\lambda(B\setminus A)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$
Since $B_i$ are all finite unions of intervals, it is obvious that $B$ is also a finite union of intervals.
Now, if $s$ is a simple function with values $c_1,\dots,c_n$, and $A_i=s^{-1}(\{c_i\})$ for every $i=1,2,\dots,n$, then:
$$s(x)=\sum_{i=1}^nc_i1_{A_i}(x)$$
Due to the triangular inequality and the linearity of Lebesgue's integral, we shall prove the requested only in the case when $s=1_A$, where $A$ is some measurable set - note that a set $A$ is measurable if and only if $1_A$ is a measurable function.
Let $\epsilon>0$. If $s=1_A$, and $s$ is measurable, then $A$ is also measurable. So, there exists a union of intervals $B$, such that:
$$\lambda(S(A,B))<\epsilon$$
Let $f=1_B$. Note now that, on every $x\in A\cap B$ we have $s=f=1\Rightarrow s-f=0$. On every $x\in J\setminus(A\cup B)$ we have that $s=f=0\Rightarrow s-f=0$. Now, on every $x\in S(A,B)$ we have that $|s-f|=1$ (explain why), so $$|s-f|=!_{S(A,B)}$$
Now it follows that:
$$\int_J|s-f|d\lambda=\int_J1_{S(A,B)}d\lambda=\int_{S(A,B)}d\lambda=\lambda(S(A,B))<\epsilon$$.
Try to complete this proof - all the details missing. Ask whatever you may need! :)
A: There is a alternative way of doing your problem without using given hint.
Given any integrable function $f$. Then $f$ can be decomposible with respect to non-negative fuctions $f^+$ and $f^-$ such that $f= f^+ - f^-$. Now we know that, if $f$ is a real and measurable $\mathcal{F}$, then there exists a sequence $\{f_n\}$ of simple function,each measurable $\mathcal{F}$, such that 
$$
0\leq f_n \uparrow f \ \mbox{if} \  f \geq 0
$$
$$
0\geq f_n \downarrow f \ \mbox{if} \  f \leq 0
$$
So, Now we consider $\{f_{n}^+\}$ is a sequence of simple fuction such that $0\leq f_{n}^+ \uparrow f^+$
and $\{f_{n}^-\}$ is a sequence of simple fuction such that $0\leq f_{n}^- \uparrow f^-$.   Now using DCT,
 We have $f_{n}^+ \leq f^+$ and $f_{n}^- \leq f^-$
\begin{equation}
    \begin{split}
      |f_{n}^+ - f^+| \leq |f_{n}^+| + |f^+| \leq 2|f^+|\\
      |f_{n}^- - f^-| \leq |f_{n}^-| + |f^-| \leq 2|f^-|\\
    \end{split}
\end{equation}
and
\begin{equation}
    \begin{split}
        \lim_n |f_{n}^+ - f^+| = 0\\
        \lim_n |f_{n}^- - f^-| = 0\\
    \end{split}
\end{equation}
$f$ is integrable therefore $f^+$ and $f^-$ also integrable. Then DCT apply and  we get,
\begin{equation}
    \begin{split}
     \lim_n \int|f_{n}^+ - f^+| = 0\\
    \lim_n \int|f_{n}^- - f^-| = 0\\  
    \end{split}
\end{equation}
Now,
\begin{equation}
    \begin{split}
      \int |f - f_{\epsilon}| d\mu\\
      = \int|f^+ - f^- - (f_{n}^+ -  f_{n}^-)|d\mu\\
      =\int |(f^+ -f_{n}^+) + (f_{n}^- - f^-)|d\mu \\
      \leq \int |f^+ -f_{n}^+|d\mu + \int |f_{n}^- - f^-|d\mu
    \end{split}
\end{equation}
Therefore, we say that for a fix $\epsilon>0$ there exists a $k = \max\{k_1,k_2\} $(where $k_1\ \mbox{and} \ k_2 $comes from the definition) such that 
$$
\int |f d\mu - \int f_{\epsilon}|d\mu < \epsilon, \ n \geq k
$$
So, this result is true for general measure $\mu$.
You can take $\mu$ as Lebesgue measure $\mu_L$. Here simple function become step function. Hence your desired result.
