Proof of $\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A$ for the Lebesgue integral Let $f:\mathbb{R}\to [0,\infty]$ be a measurable function and $A\subset \mathbb{R}$. Then, show that
\begin{equation}
  \int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A \tag{1}
\end{equation}
where ${1}_A$ is the characteristic function of $A$ defined as 
\begin{equation}
{1}_A(x)=\begin{cases}1 & \text{if $x\in A$,}
\\
0 &\text{if $x\notin A$.}
\end{cases} \tag{2}
\end{equation}
and $\int\limits_{A}f$ is the Lebesgue integral of $f$ on $A$ defined as:
\begin{equation}
\int\limits_{A}f=\sup\left\{\int\limits_{A}s:0\le s\le f\text{ and }s\text{ is simple}\right\} \tag{3}
\end{equation}
I can easily prove this property for simple functions so take this for granted:
\begin{equation}
\int\limits_{A}s=\int\limits_{\mathbb{R}}s{1}_A \tag{4}
\end{equation}
where $s:\mathbb{R}\to [0,\infty]$ is a simple function.
Thus to prove (1) we need to show that:
\begin{gather} %omg wall of text code - mixedmath
\sup\left\{\int\limits_{A}s:0\le s\le f\text{ and }s\text{ is simple}\right\}=\sup\left\{\int\limits_{\mathbb{R}}s:0\le s\le f{1}_A\text{ and }s\text{ is simple}\right\}\notag\\
\sup\left\{\int\limits_{\mathbb{R}}s{1}_A:0\le s\le f\text{ and }s\text{ is simple}\right\}=\sup\left\{\int\limits_{\mathbb{R}}s:0\le s\le f{1}_A\text{ and }s\text{ is simple}\right\} \tag{5}
\end{gather}
My question is how do we prove (5)?
PROOF: 
It can be easily shown that
$\int\limits_{A}f=\int\limits_{A}f1_A$ and since $A\subset \mathbb{R}$,
\begin{equation}
\int\limits_{A}f=\int\limits_{A}f1_A\le \int\limits_{\mathbb{R}}f1_A
\tag{6}
\end{equation}
We just have to show that \begin{equation}
  \int\limits_{A}f\ge\int\limits_{\mathbb{R}}f{1}_A
\end{equation}
The last inequality is proven in the answer given by Thomas.E 
For a completely different approach you can look at my answer
 A: In the comments section ThomasE. proposed a completely different yet beautiful approach that I will now present here. 
First a lemma: \begin{equation}\int\limits_{\mathbb{R}}f=\int\limits_{A}f+\int\limits_{A^c}f\end{equation}
Proof: Let $s_1,s_2$ be any simple functions on $A$ and $A^c$ respectively and define $s(x)=\begin{cases}s_1(x) & \text{if $x\in A$,}
\\
s_2(x) &\text{if $x\in A^c$}
\end{cases}
$. Then, \begin{gather}\int\limits_{A}f+\int\limits_{A^c}f=\sup\left\{\int\limits_{A}s_1:0\le s_1\le f|_A\right\}+
\sup\left\{\int\limits_{A^c}s_2:0\le s_2\le f|_{A^c}\right\}\\
\int\limits_{A}f+\int\limits_{A^c}f=\sup\left\{\int\limits_{A}s_1+\int\limits_{A^c}s_2:0\le s_1\le f|_A\text{ and }0\le s_2\le f|_{A^c}\right\}
=\sup\left\{\int\limits_{\mathbb{R}}s:0\le s\le f\right\}\\
\int\limits_{A}f+\int\limits_{A^c}f\le \int\limits_{\mathbb{R}}f
\end{gather}
and 
\begin{gather} \int\limits_{\mathbb{R}}f=\sup\left\{\int\limits_{\mathbb{R}}s:0\le s\le f\text{ and }s\text{ is simple}\right\}=
\sup\left\{\int\limits_{A}s+\int\limits_{A^c}s:0\le s_A\le f\text{ and }0\le s|A^c\le f\right\}\\
\int\limits_{\mathbb{R}}f=\sup\left\{\int\limits_{A}s:0\le s|A\le f\right\}+\sup\left\{\int\limits_{A^c}s:0\le s\le f^c\right\}\le\int\limits_{A}f+\int\limits_{A^c}f
\end{gather}
Thus, the Lemma is proven. Now
\begin{equation}\int\limits_{A}f=\int\limits_{A}f+\int\limits_{A^c}0=\int\limits_{A}f1_A+\int\limits_{A^c}f1_A=\int\limits_{\mathbb{R}}f1_A\end{equation}
This all seems to be correct to my eyes, but is it?
A: Let $\varepsilon>0$. By definition of the Lebesgue integral you find a simple function $0\leq s_{\varepsilon}\leq f 1_{A}$ with $\int_{\mathbb{R}}f 1_{A}\leq \varepsilon+\int_{\mathbb{R}} s_{\varepsilon}$. Note that this implies $s_{\varepsilon}=s_{\varepsilon}1_{A}$ since $0\leq s_{\varepsilon}(x)\leq f(x) 1_{A}(x)=0$ for all $x\in A^{c}$. Using what you have proven to apply for simple functions $(*)$ and the choice of $s_{\varepsilon}$ $(**)$ it follows that 
\begin{align*}
\int_{\mathbb{R}} f1_{A}\leq \varepsilon+\int_{\mathbb{R}} s_{\varepsilon}=\varepsilon+\int_{\mathbb{R}}s_{\varepsilon}1_{A}\overset{(*)}{=} \varepsilon +\int_{\mathbb{A}} s_{\varepsilon} \overset{(**)}{\leq} \varepsilon+\int_{A}f 1_{A}=\varepsilon +\int_{A}f
\end{align*}
since $1_{A}(x)= 1$ for $x\in A$. Since the choice of $\varepsilon>0$ was arbitrary it follows that
\begin{align*}
\int_{\mathbb{R}} f1_{A}\leq \int_{A}f.
\end{align*}
And the other inequality you have already proven.
