# 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

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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 $\sigma\colon \mathbb R \to [0,\infty)$... – martini Jun 1 '12 at 7:04
I edited it to number the equations as I gathered that you had intended. – mixedmath Jun 1 '12 at 7:49
I am back. Dear martini, I agree with your observation but how does it continue from there? – SomeoneContinuous Jun 1 '12 at 9:53
If martini isn't Someone, that Continuous ly monitors your question, you should write @martini instead... and +1, interesting question. – draks ... Jun 1 '12 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? – martini Jun 1 '12 at 11:05

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.
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) – T. Eskin Jun 1 '12 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. – T. Eskin Jun 1 '12 at 12:58