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

$$\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A \tag{1}$$

where ${1}_A$ is the characteristic function of $A$ defined as

$${1}_A(x)=\begin{cases}1 & \text{if x\in A,} \\ 0 &\text{if x\notin A.} \end{cases} \tag{2}$$

and $\int\limits_{A}f$ is the Lebesgue integral of $f$ on $A$ defined as:

$$\int\limits_{A}f=\sup\left\{\int\limits_{A}s:0\le s\le f\text{ and }s\text{ is simple}\right\} \tag{3}$$

I can easily prove this property for simple functions so take this for granted:

$$\int\limits_{A}s=\int\limits_{\mathbb{R}}s{1}_A \tag{4}$$

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 $A\subset \mathbb{R}$, $$\int\limits_{A}f=\int\limits_{A}f1_A\le \int\limits_{\mathbb{R}}f1_A \tag{6}$$ We just have to show that $$\int\limits_{A}f\ge\int\limits_{\mathbb{R}}f{1}_A$$ The last inequality is proven in the answer given by Thomas.E

For a completely different approach you can look at my answer

• Hint: If $s\colon\mathbb R \to[0,\infty)$ is simple, so is $s|_A$, on the other side, if $s \colon A \to [0,\infty)$ is simple, extend $s$ by zero to a simple function $\sigma\colon \mathbb R \to [0,\infty)$... Jun 1, 2012 at 7:04
• I edited it to number the equations as I gathered that you had intended. Jun 1, 2012 at 7:49
• I am back. Dear martini, I agree with your observation but how does it continue from there? Jun 1, 2012 at 9:53
• If martini isn't Someone, that Continuous ly monitors your question, you should write @martini instead... and +1, interesting question. Jun 1, 2012 at 11:00
• Thx @draks, now if $f \colon \mathbb R \to [0,\infty]$ is given, suppose $s \le f1_A$ on $\mathbb R$, then $s|_A \le f|_A$, giving you one inequality, for the other suppose $s \le f|_A$, then $\sigma \le f1_A$, giving you the other one ... does this help? Jun 1, 2012 at 11:05

In the comments section ThomasE. proposed a completely different yet beautiful approach that I will now present here. First a lemma: $$\int\limits_{\mathbb{R}}f=\int\limits_{A}f+\int\limits_{A^c}f$$ 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 $$\int\limits_{A}f=\int\limits_{A}f+\int\limits_{A^c}0=\int\limits_{A}f1_A+\int\limits_{A^c}f1_A=\int\limits_{\mathbb{R}}f1_A$$ This all seems to be correct to my eyes, but is it?
• With this lemma it works. If you want to avoid working suprumems over sets in equalities, you may do something like the following too (you already got the idea). $"\Rightarrow"$: Let $0\leq s\leq f$ be arbitrary simple function, and define $s_{1}=s 1_{A}$ and $s_{2}=s 1_{A^{c}}$, whence $s=s_{1}+s_{2}$ and $s_{1}(x)\leq f(x)$ for $x\in A$ and $s_{2}(x)\leq f(x)$ for $x\in A^{c}$. Now using what you had already proven $\int_{R} s=\int_{R} s 1_{A}+s 1_{A^{c}}=\int_{R}s 1_{A}+\int_{R}s 1_{A^{c}}=\int_{A} s_{1}+\int_{A^{c}} s_{2}\leq \int_{A}f+\int_{A^{c}}f$. (Continues below) Jun 1, 2012 at 12:55
• ... by taking sup over all such $s$ we obtain $\int_{\mathbb{R}} f\leq \int_{A}f+\int_{A^{c}}f$. Can you show the other direction similarly? By choosing arbitrary simple functions $0\leq s_{1}\leq f|_{A}$ and $0\leq s_{2}\leq f|_{A^{c}}$, and showing $\int_{A}s_{1}+\int_{A}s_{2}\leq \int_{\mathbb{R}}f$, and taking supremum over all such $s_{1}$ and $s_{2}$ obtaining the other inequality. Jun 1, 2012 at 12:58
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.